1.1 The present paper is the outcome of an attempt to apply to the principal problems of the theory of partitions the methods, depending upon the theory analytic functions, which have proved so fruitful in the theory of the distribution of primes and allied branches of the analytic theory of numbers.

The most interesting functions of the theory of partitions appear as the coefficients in the power-series which represents certain elliptic modular functions. Thus $p(n)$, the number of unrestricted partitions of $n$, is the coefficient of $x^n$ in the expansion of the function

If we write where the imaginary part of $\tau$ is positive, we see that $f(x)$ is substantially the reciprocal of the modular function called by Tannery and Molk³ $h(\tau)$; that, in fact,

The theory of partitions has, from the time of Euler onwards, been developed from an almost exclusively algebraical point of view. It consists of an assemblage of formal identities – many of them, it need hardly be said, of an exceedingly ingenious and beautiful character. Of asymptotic formluæ, one may fairly say, there are none⁴. So true is this, in fact, that we have been unable to discover in the literature of the subject any allusion whatever to the question of the order of magnitude of $p(n)$.

1.2 The function $p(n)$ may, of course, be expressed in the form of an integral

by means of Cauchy's theorem, the path $\Gamma$ enclosing the origin and lying entirely inside the unit circle. The idea which dominates this paper is that of obtaining asymptotic formulæ for $p(n)$ by a detailed study of the integral (1.21). This idea is an extremely obvious one; it is the idea which has dominated nine-tenths of modern research in analytic theory of numbers: and it may seem very strange that it should never have been applied to this particular problem before. Of this there are no doubt two explanations. The first is that the theory of partitions has received its most important developments, since its foundation by Euler, at the hands of a series of mathematicians whose interests have lain primarily in algebra. The second and more fundamental reason is to be found in the extreme complexity of the behavior of the generating function $f(x)$ near a point of the unit circle.

It is instructive to contrast this problem with the corresponding problems which arise for the arithmetical functions $\pi (n), \vartheta (n),\Psi (n), \mu (n), d(n), \ldots $ which have their genesis in Riemann's Zeta-function and the functions allied to it. In the latter problems we are dealing with functions defined by Dirichlet's series. The study of such functions presents difficulties far more fundamental than any which confront us in the theory of the modular functions. These difficulties, however, relate to the distribution of the zeros of the functions and their general behavior at infinity: no difficulties whatever are occasioned by the crude singularities of the functions in the finite part of the plane. The single finite singularity of $\zeta (s)$, for example, the pole at $s=1$, is a singularity of the simplest possible character. It is this pole which gives rise to the dominant terms in the asymptotic formulæ for the arithmetical functions associated with $\zeta (s)$. To prove such a formula rigorously is often exceedingly difficult; to determine precisely the order of the error which it involves is in many cases a problem which still defies the utmost resources of analysis. But to write down the dominant terms involves, as a rule, no difficulty more formidable than that of deforming a path of integration over a pole of the subject of integration and calculating the corresponding residue.

In the theory of partitions, on the other hand, we are dealing with functions which do not exist at all outside the unit circle. Every point of the circle is an essential singularity of the function, and no part of the contour of integration can be deformed in such a manner as to make its contribution obviously negligible. Every element of the contour requires special study; and there is no obvious method of writing down a ``dominant term''.

The difficulties of the problem appear then, at first sight, to be very serious. We possess, however, in the formulæ of the theory of linear transformation of the elliptic functions, an extremely powerful analytical weapon by means of which we can study the behavior of $f(x)$ near any assigned point of the unit circle⁵. It is to an appropriate use of these formulæ that the accuracy of our final results, an accuracy which will, we think, be found to be quite startling, is due.

1.3 It is very important, in dealing with such a problem as this, to distinguish clearly the various stages to which we can progress by arguments of progressively ``deeper'' and less elementary character. The earlier results are naturally (so far as the particular problem is concerned) superseded by the later. But the more elementary methods are likely to be applicable to other problems in which the more subtle analysis is impracticable.

We have attacked this particular problem by a considerable number of different methods, and cannot profess to have reached any very precise conclusions as to the possibilities of each. A detailed comparison of the results to which they lead would moreover expand this paper to a quite unreasonable length. But we have thought it worth while to include a short account of two of them. The first is quite elementary; it depends only on Euler's identity

-- an identity capable of wide generalisation – and on elementary algebraical reasoning. By these means we shew, in section 2, that where $A$ and $B$ are positive constants, for all sufficiently large values of $n$.

It follows that

and the next question which arises is the question whether a constant $C$ exists such that We prove that this is so in section 3. Our proof is still, in a sense, ``elementary.'' It does not appeal to the theory of analytic functions, depending only on a general arithmetic theorem concerning infinite series; but this theorem is of the difficult and delicate type which Messrs Hardy and Littlewood have called ``Tauberian.'' The actual theorem required was proved by us in a paper recently printed in these Proceedings⁶. It shews that in other words that where $\epsilon$ is small when $n$ is large. This method is one of very wide application. It may be used, for example, to prove that, if $p^{(s)}(n)$ denotes the number of partitions of $n$ into perfect $s$-th powers, then $$\mbox{log}~p^{(s)}(n)\sim (s+1)\left\{\frac{1}{s}\Gamma\left (1+\frac{1}{s}\right )\zeta \left (1+\frac{1}{s}\right)\right\}^{s/(s+1)}n^{1/(s+1)}.$$

It is certainly possible to obtain, by means of arguments of this general character, information about $p(n)$ more precise than that furnished by the formula (1.36). And it is equally possible to prove (1.36) by reasoning of a more elementary, though more special, character: we have a proof, for example, based on the identity $$np(n)=\sum_{\nu =1}^{n}\sigma (\nu )p(n-\nu ),$$ where $\sigma (\nu )$ is the sum of divisors of $\nu$, and a process of induction. But we are at present unable to obtain, by any method which does not depend upon Cauchy's theorem, a result as precise as that which we state in the next paragraph, a result, that is to say, which is ``vraiment asymptotique.''

1.4~~ Our next step was to replace (1.36) by the much more precise formula

The proof of this formula appears to necessitate the use of much more powerful machinery, Cauchy's integral (1.21) and the functional relation where This formula is a merely a statement in different notation of the relation between $h(\tau)$ and $h(T)$ where $$T=\frac{c+d\tau}{a+b\tau},\quad a=d=0, b=1, c=-1;$$ viz. $$h(\tau )=\sqrt{\left (\frac{i}{\tau}\right )}h(T).\href{#p36_en7}{^7}$$

It is interesting to observe the correspondence between (1.41) and the results of numerical computation. Numerical data furnished to us by Major MacMahon gave the following results: we denote the right-had side of (1.41) by $\varpi (n)$.

$$ \begin{array}{|c|c|c|c|} \hline n & p(n) & \varpi (n) & \varpi /p\\ \hline 10& 42 & 48.104 & 1.145 \\ \hline 20 & 627 & 692.385 & 1.104 \\ \hline 50 & 204226 & 217590.499 & 1.065 \\ \hline 80 & 15796476 & 16606781.567 & 1.051 \\ \hline \end{array} $$

It will be observed that the progress of $\varpi /p$ towards its limit unity is not very rapid, and that $\varpi - p$ is always positive and appears to tend rapidly to infinity.

1.5 In order to obtain more satisfactory results it is necessary to construct some auxiliary function $F(x)$ which is regular at all points of the unit circle save $x=1$, and has there a singularity of a type as near as possible to that of the singularity of $f(x)$. We may then hope to obtain a much more precise approximation by applying Cauchy's theorem to $f-F$ instead of to $f$. For although every point of the circle is a singular point of $f$, $x=1$ is, to put it roughly, much the heaviest singularity. When $x\longrightarrow 1$ by real values, $f(x)$ tends to infinity like an exponential $$\exp\left\{\frac{\pi^2}{6(1-x)}\right\};$$ when $$x=re^{2p\pi i /q},$$ $p$ and $q$ being co-prime integers, and $r\longrightarrow 1$, $|f(x)|$ tends to infinity like an exponential $$\exp\left\{\frac{\pi^2}{6q^2(1-r)}\right\};$$ while, if $$x=re^{2\theta\pi i},$$ where $\theta$ is irrational, $|f(x)|$ can become infinite at most like an exponential of the type $$\exp\left\{o\left (\frac{1}{1-r}\right )\right\}.\href{#p36_en8}{^8}$$

The function required is

where This function may be transformed into an integral by means of a general formula given by Lindelöf⁹; and it is then easy to prove that the ``principal branch'' of $F(x)$ is regular all over the plane except at $x=1$¹⁰; and that $$F(x)-\chi (x),$$ where is regular for $x=1$. If we compare (1.42) and (1.54), and observe that $f(x')$ tends to unity with extreme rapidity when $x$ tends to 1 along any regular path which does not touch the circle of convergence, we can see at once the very close similarity between the behaviour of $f$ and $F$ inside the unit circle and in the neighbourhood of $x=1$.

It should be observed that the term $-1$ in (1.52) and (1.54) is-so far as our present assertions are concerned-otiose: all that we have said remains true if it is omitted; the resemblance between the singularities of $f$ and $F$ becomes indeed even closer. The term is inserted merely in order to facilitate some of our preliminary analysis, and will prove to be without influence on the final result.

Applying Cauchy's theorem to $f-F$, we obtain

where $D$ is any number greater than $$\frac{1}{2}C=\mbox{$\frac{1}{2}$}\pi \sqrt{(\mbox{$\frac{2}{3}$})}.$$

1.6 The formula (1.55) is an asymptotic formula of a type far more precise than that of (1.41). The error term is, however, of an exponential type, and may be expected ultimately to increase with very great rapidity. It was therefore with considerable surprise that we found what exceedingly good results the formula gives for fairly large values of $n$. For $n=61,62,63$ it gives $$1121538.972,\quad 1300121.359,\quad 1505535.606,$$ while the correct values are $$1121505,\quad 1300156,\quad 1505499.$$ The errors $$33.972,\quad -34.641,\quad 36.606$$ are relatively very small, and alternate in sign.

The next step is naturally to direct our attention to the singular point of $f(x)$ next in importance after that at $x=1$, viz., that at $x=-1$; and to subtract from $f(x)$ a second auxiliary function, related to this point as $F(x)$ is to $x=1$. No new difficulty of principle is involved, and we find that

where $D$ is now any number greater than $\frac{1}{3}C$. It now becomes obvious why our earlier approximation gave errors alternately of excess and of defect.

It is obvious that this process may be repeated indefinitely. The singularities next in importance are those at $x=e^{\frac{2}{3}\pi i}$ and $x=e^{\frac{4}{3}\pi i}$; the next those at $x=i$ and $x=-i$; and so on. The next two terms in the approximate formula are found to be $$\frac{\sqrt{3}}{\pi\sqrt{2}}\cos\left (\mbox{$\frac{2}{3}$}n\pi -\mbox{$\frac{1}{18}$}\pi\right )\frac{d}{dn}\left (\frac{e^{\frac{1}{3}C\lambda _n}}{\lambda_n}\right )$$ and $$\frac{\sqrt{2}}{\pi}\cos\left (\half n\pi -\mbox{$\frac{1}{8}$}\pi\right ) \frac{d}{dn}\left (\frac{e^{\frac{1}{4} C\lambda _n}}{\lambda_n}\right ).$$ As we proceed further, the complexity of the calculations increases. The auxiliary function associated with the point $x=e^{2p\pi i/q}$ involves a certain $24q$-th root of unity, connected with the linear transformation which must be used in order to elucidate the behaviour of $f(x)$ near the point; and the explicit expression of this root in terms of $p$ and $q$, though known, is somewhat complex. But it is plain that, by taking a sufficient number of terms, we can find a formula in which the error is $$O(e^{C\lambda_n / \nu}),$$ where $\nu$ is a fixed but arbitrarily large integer.

1.7 A final question remains. We have still the resource of making $\nu$ a function of $n$, that is to say of making the number of terms in our approximate formula itself a function of $n$. In this way we may reasonably hope, at any rate, to find a formula in which the error is of order less than that of any exponential of the type $e^{an}$; of the order of a power of $n$, for example, or even bounded.

When, however, we proceeded to test this hypothesis by means of the numerical data most kindly provided for us by Major MacMahon, we found a correspondence between the real and the approximate values of such astonishing accuracy as to lead us to hope for even more. Taking $n=100$, we found that the first six terms of our formula gave $$\begin{array}{r} 190568944.783\phantom{,}\\ +348.872\phantom{,}\\ -2.598\phantom{,}\\ +.685\phantom{,}\\ +.318\phantom{,}\\ -.064\phantom{,}\\ \hline 190569291.996, \end{array}$$ while $p(100)=190569292$; so that the error after six terms is only .004. We then proceeded to calculate $p(200)$ and found $$\begin{array}{r} 3, 972, 998, 993, 185.896\phantom{,}\\ +36, 282.978\phantom{,}\\ -87.555\phantom{,}\\ +5.147\phantom{,} \\ +1.424\phantom{,}\\ +0.071\phantom{,}\\ +0.000\phantom{,}\href{#p36_en11}{^{11}}\\ +0.043\phantom{,}\\ \hline 3, 972, 999, 029, 388.004, \end{array}$$ and Major MacMahon's subsequent calculations shewed that $p(200)$ is, in fact, $$3, 972, 999, 029, 388.$$ These results suggest very forcibly that it is possible to obtain a formula for $p(n)$, which not only exhibits its order of magnitude and structure, but may be used to calculate its exact value for any value of $n$. That this is in fact so is shewn by the following theorem.

It should be observed that all the numbers $A_{q}$ are real. A table of $A_q$ from $q=1$ to $q=18$ is given at the end of the paper (Table II).

The proof of this main theorem is given in section 5; section 4 being devoted to a number of preliminary lemmas. The proof is naturally somewhat intricate; and we trust that we have arranged it in such a form as to be readily intelligible. In section 6 we draw attention to one or two questions which our theorem, in spite of its apparent completeness, still leaves open. In section 7 we indicate some other problems in combinatory analysis and the analytic theory of numbers to which our method may be applied; and we conclude by giving some functional and numerical tables: for the latter we are indebted to Major MacMahon and Mr. H. B. C. Darling. To Major MacMahon in particular we owe many thanks for the amount of trouble he has taken over very tedious calculations. It is certain that, without the encouragement given by the results of these calculations, we should never have attempted to prove theoretical results at all comparable in precision with those which we have enunciated.

2.1 In this section we give the elementary proof of the inequalities (1.32). We prove, in fact, rather more, viz., that positive constants $H$ and $K$ exist such that

for $n\geq 1$¹³. We shall use in our proof only Euler's formula (1.31) and a debased form of Stirling's theorem, easily demonstrable by quite elementary methods: the proposition that $$n! e^n/n^{n+\half}$$ lies between two positive constants for all positive integral values of $n$.

2.2 The proof of the first of the two inequalities is slightly the simpler. It is obvious that if $$\sum p_r(n)x^n=\frac{1}{(1-x)(1-x^2)\cdots (1-x^r)}$$ so that $p_r(n)$ is the number of partitions of $n$ into parts not exceeding $r$, then

We shall use this equation to prove, by induction, that It is obvious that (2.22) is true for $r=1$. Assuming it to be true for $r=s$, and using (2.21), we obtain \begin{eqnarray*} p_{s+1}(n)&\geq& \frac{s}{(s!)^2}\{n^{s-1}+(n-s-1)^{s-1}+(n-2s-2)^{s-1}+\cdots \}\\ &\geq&\frac{s}{(s!)^2}\left\{\frac{n^s-(n-s-1)^s}{s(s+1)} +\frac{(n-s-1)^s-(n-2s-2)^s}{s(s+1)}+\cdots \right\}\\ &=& \frac{n^s}{(s+1)(s!)^2}=\frac{(s+1)n^s}{\{(s+1)!\}^2}. \end{eqnarray*} This proves (2.22). Now $p(n)$ is obviously not less than $p_r(n)$, whatever the value of $r$. Take $r=[\sqrt{n}]$: then $$p(n)\geq p_{[\sqrt{n}]}(n)\geq \frac{[\sqrt{n}]}{n}\frac{n^{[\sqrt{n}]}}{\{[\sqrt{n}]!\}^2}>\frac{H}{n}e^{2\sqrt{n}},$$ by a simple application of the degenerate form of Stirling's theorem mentioned above.

2.3 The proof of the second inequality depends upon Euler's identity. If we write $$\sum q_r(n)x^n=\frac{1}{(1-x)^2(1-x^2)^2\cdots (1-x^r)^2},$$ we have

and We shall first prove by induction that This is obviously true for $r=1$. Assuming it to be true for $r=s$, and using (2.31), we obtain \begin{eqnarray*} q_{s+1}(n)\leq \frac{1}{(2s-1)!(s!)^2}\left\{(n+s^2)^{2s-1}+2(n+s^2-s-1)^{2s-1} +3(n+s^2-2s-2)^{2s-1}+\cdots\right\}. \end{eqnarray*} Now $$m(m-1)a^{m-2}b^2\leq (a+b)^m-2a^m+(a-b)^m,$$ if $m$ is a positive integer, and $a,b$, and $a-b$ are positive, while if $a-b\leq 0$, and $m$ is odd, the term $(a-b)^m$ may be omitted. In this inequality write $$m=2s+1,\quad a=n+s^2-ks-k~~~(k=0,1,2,\ldots ),\quad b=s+1,$$ and sum with respect to $k$. We find that $$(2s+1)2s(s+1)^2\left\{(n+s^2)^{2s-1}+2(n+s^2-s-1)^{2s-1} +\cdots\right\}\leq (n+s^2+s+1)^{2s+1};$$ and so $$q_{s+1}(n)\leq \frac{(n+s^2+s+1)^{2s+1}}{(2s+1)2s(s+1)^2(2s-1)!(s!)^2} \leq \frac{\{n+(s+1)^2\}^{2s+1}}{(2s+1)!\{(s+1)!\}^2}.$$ Hence (2.33) is true generally.

It follows from (2.32) that $$p(n)=q_1(n-1)+q_2(n-4)+\cdots \leq \sum_1^{\infty}\frac{n^{2r-1}}{(2r-1)!(r!)^2}.$$ But, using the degenerate form of Stirling's theorem once more, we find without difficulty that $$\frac{1}{(2r-1)!(r!)^2}\lt \frac{2^{6r}K}{4r!},$$ where $K$ is a constant. Hence $$p(n)\lt 8K\sum_1^{\infty}\frac{(8n)^{2r-1}}{4r!} \lt 8K\sum_1^{\infty}\frac{(8n)^{\frac{1}{2} r-1}}{r!} \lt\frac{K}{n}e^{2\sqrt{2n}}.$$ This is the second of the inequalities (2.11).

3.1 The value of the constant $$C=\lim \frac{\mbox{log}~p(n)}{\sqrt{n}},$$ is most naturally determined by the use of the following theorem.

If $g(x)=\sum a_nx^n$ is a power-series with positive coefficients, and $$\mbox{log}~g(x)\sim \frac{A}{1-x}$$ when $x\to 1$, then $$\mbox{log}~s_n=\mbox{log}~(a_0+a_1+\cdots +a_n)\sim 2\sqrt{An}$$ when $n\to \infty$.

This theorem is a special case¹⁴ of Theorem C in our paper already referred to.

Now suppose that $$g(x)=(1-x)f(x)=\sum \{p(n)-p(n-1)\}x^n =\frac{1}{(1-x^2)(1-x^3)(1-x^4)\cdots}.$$ Then $$a_n=p(n)-p(n-1)$$ is plainly positive. And

when $x\to 1$¹⁵. Hence where $C=2\pi /\sqrt{6}=\pi\sqrt{\mbox{$\frac{2}{3}$}}$, as in (1.53).

3.2 There is no doubt that it is possible, by ``Tauberian'' arguments, to prove a good deal more about $p(n)$ than is asserted by (3.12). The functional equation satisfied by $f(x)$ shews, for example, that $$g(x)\sim \frac{(1-x)^{3/2}}{\sqrt{2\pi}}\exp\left\{\frac{\pi^2}{6(1-x)}\right\},$$ a relation far more precise than (3.11). From this relation, and the fact that the coefficients in $g(x)$ are positive, it is certainly possible to deduce more than (3.12). But it hardly seems likely that arguments of this character will lead us to a proof of (1.41). It would be exceedingly interesting to know exactly how far they will carry us, since the method is comparatively elementary, and has a much wider range of application than the more powerful methods employed later in this paper. We must, however, reserve the discussion of this question for some future occasion.

4.1 We proceed now to the proof of our main theorem. The proof is somewhat intricate, and depends on a number of subsidiary theorems which we shall state as lemmas.

4.21 The Farey's series of order $m$ is the aggregate of irreducible rational fractions $$p/q\quad (0\leq p\leq q\leq m),$$ arranged in ascending order of magnitude. Thus $$\frac{0}{1}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{1}{1}$$ is the Farey's series of order 7.

Lemma 4.21. If $p/q, p'/q'$ are two successive terms of a Farey's series, then

This is, of course, a well-known theorem, first observed by Farey and first proved by Cauchy¹⁶. Th following exceedingly simple proof is due to Hurwitz¹⁷.

The result is plainly true when $m=1$. Let us suppose it true for $m=k$; and let $p/q, p'/q'$ be two consecutive terms in the series of order $k$.

Suppose now that $p''/q''$ is a term of the series of order $k+1$ which falls between $p/q, p'/q'$. Let $$p''q-pq''=\lambda >0,\quad p'q''-p''q'=\mu >0.$$ Solving these equations for $p'', q''$ and observing that $p'q-pq'=1$, we obtain $$p''=\mu p+\lambda p', \quad q''=\mu q+\lambda q'.$$

Consider now the aggregate of fractions $$(\mu p+\lambda p')/(\mu q+\lambda q'),$$ where $\lambda$ and $\mu$ are positive integers without common factor. All of these fractions lie between $p/q$ and $p'/q'$; and all are in their lowest terms, since a factor common to numerator and denominator would divide $$\lambda =q(\mu p+\lambda p')-p(\mu q+\lambda q'),$$ and $$ \mu =p'(\mu q+\lambda q')-q'(\mu p+\lambda p'). $$

Each of them first makes its appearance in the Farey's series of order $\mu q+\lambda q'$, and the first of them to make its appearance must be that for which $\lambda=1, \mu =1$. Hence $$p''=p+p', \quad q''=q+ q',$$ $$p''q-pq''=p'q''-p''q'=1.$$ The lemma is consequently proved by induction.

(i) Since $$\frac{1}{q(q+q')}+\frac{1}{q'(q'+q)}=\frac{1}{qq'} =\frac{p'q-pq'}{qq'}=\frac{p'}{q'}-\frac{p}{q},$$ the intervals just fill up the continuum.

(ii) Since neither $q$ nor $q'$ exceeds $m$, and one at least must be less than $m$, we have $$\frac{1}{q(q+q')}\gt\frac{1}{2mq}.$$ Also $q+q'>m$, since otherwise $(p+p')/(q+q')$ would be a term in the series between $p/q$ and $p'/q'$. Hence $$\frac{1}{q(q+q')}\lt\frac{1}{mq}.$$

4.22 The following mode of dissection of a circle, based upon Lemma 4.22, is of fundamental importance for our analysis.

Suppose that the circle is defined by $$x=Re^{2\pi i\theta}\quad (0\leq \theta\leq 1).$$ Construct the Farey's series of order $m$, and the corresponding intervals $j_{p,q}$. When these intervals are considered as intervals of variation of $\theta$, and the two extreme intervals, which correspond to abutting arcs on the circle, are regarded as constituting a single interval $\xi_{1,1}$, the circle is divided into a number of arcs $$\xi_{p,q},$$ where $q$ ranges from 1 to $m$ and $p$ through the numbers not exceeding and prime to $q$²⁰. We call this dissection of the circle the dissection $\Xi_{m}$. $\label{36f2}$

4.3 Lemma 4.31. Suppose that $q$ is a positive integer; that $p$ is a positive integer not exceeding and prime to $q$; that $p'$ is a positive integer such that $1+pp'$ is divisible by $q$; that $\omega_{p,q}$ is defined by the formulæ (1.721) or (1.722); that $$x=\exp \left (-\frac{2\pi z}{q}+\frac{2p\pi i}{q}\right ),\quad x'=\exp \left (-\frac{2\pi}{qz}+\frac{2p'\pi i}{q}\right ),$$ where the real part of $z$ is positive; and that $$f(x)=\frac{1}{(1-x)(1-x^2)(1-x^3)\cdots}.$$ Then $$f(x)=\omega_{p,q}\sqrt{z}~\exp \left (\frac{\pi}{12qz}- \frac{\pi z}{12q}\right ) f(x').$$

This lemma is merely a restatement in a different notation of well-known formulæ in the transformation theory.

Suppose, for example, that $p$ is odd. If we take $$a=p, b=-q,c=\frac{1+pp'}{q}, d=-p',$$ so that $ad-bc=1$; and write $$x=q^2=e^{2\pi i\tau},\quad x'=Q^2=e^{2\pi iT},$$ so that $$\tau = \frac{p}{q}+\frac{iz}{q}, \quad T=\frac{p'}{q}+\frac{i}{qz};$$ then we can easily verify that $$T=\frac{c+d\tau}{a+b\tau}.$$ Also, in the notation of Tanner and Molk, we have $$f(x)=\frac{q^{\frac{1}{12}}}{h(\tau )},\quad f(x')=\frac{Q^{\frac{1}{12}}}{h(T)};$$ and the formula for the linear transformation of $h(\tau)$ is $$h(T)=\left (\frac{b}{a}\right )\exp \left [\left\{\mbox{$\frac{1}{4}$}(a-1)-\mbox{$\frac{1}{12}$}[a(b-c)+bd(a^2-1)]\right\}\pi i\right ]\sqrt{(a+b\tau )}~h(\tau),$$ where $\sqrt{(a+b\tau )}$ has its real part positive²¹. A little elementary algebra will shew the equivalence of this result and ours.

The other formula for $\omega_{p,q}$ may be verified similarly, but in this case we must take $$a=-p, b=q,c=\frac{1+pp'}{q}, d=p'.$$ We have included in the Appendix (Table I) a short table of some values of $\omega_{p,q}$, or rather of $(\log \omega_{p,q})/\pi i$.

This is an immediate corollary from Lemma 4.31, since $$z=\frac{q}{2\pi}\log \left (\frac{1}{x_{p,q}}\right ),\quad e^{-\pi z/12q}=x^{\frac{1}{24}}_{p,q},$$ $$\frac{\pi}{12qz}=\frac{\pi^2}{6q^2\log (1/x_{p,q})}, \quad x'=\exp \left (-\frac{2\pi}{qz}+\frac{2p'\pi i}{q}\right )=x'_{p,q}.$$

If we observe that $$f(x'_{p,q})=1+p(1)x'_{p,q}+\cdots ,$$ we see that, if $x$ tends to $e^{2p\pi i/q}$ along a radius vector, or indeed any regular path which does not touch the circle of convergence, the difference $$f(x)-\omega_{p,q}\sqrt{\left\{\frac{q}{2\pi}\log \left (\frac{1}{x_{p,q}}\right )\right\}}x^{\frac{1}{24}}_{p,q} \exp\left\{\frac{\pi^2}{6q^2\log (1/x_{p,q})}\right\}$$ tends to zero with great rapidity. It is on this fact that our analysis is based.

4.41. The auxiliary function $F_a(x)$ is defined by the equation $$F_a(x)=\sum_1^{\infty}\Psi_a(n)x^n,$$ where $$\Psi_a(n)=\frac{d}{dn}\frac{\cosh a\lambda_n-1}{\lambda_n},$$ $$\lambda_n=\sqrt{(n-\mbox{$\frac{1}{24}$})}, \quad a>0.$$

Lemma 4.41. Suppose that a cut is made along the segment $(1,\infty)$ in the plane f $x$. Then $F_a(x)$ is regular at all points inside the region thus defined.

This lemma is an immediate corollary of a general theorem proved by Lindelöf on pp. 109 et seq. of his Calcul des residus²².

The function $$\Psi_a(z)=\frac{d}{dz}\frac{\cosh a\sqrt{(z-\mbox{$\frac{1}{24}$})}-1}{\sqrt{(z-\mbox{$\frac{1}{24}$})}}$$ satisfies the conditions imposed upon it by Lindelöf, if the number which he calls $\alpha$ is greater than $\frac{1}{24}$; and

if $$x=re^{i\theta}, \quad 0\lt\theta \lt2\pi ,\quad x^z =\exp \{z(\log r+i\theta)\}.$$

4.42.

We observe first that, when$\theta$ has fixed value between 0 and $2\pi$, the integral on the right-hand side of (4.411) is uniformly convergent for $\mbox{$\frac{1}{24}$}\leq \alpha \leq \alpha_0$. Hence we may take $\alpha=\mbox{$\frac{1}{24}$}$ in (4.411). We thus obtain $$F_a(x)=ix^{\frac{1}{24}}\int_0^{\infty}\frac{x^{it}}{1- e^{\frac{1}{12}\pi i -2\pi t}}\Psi_a(\mbox{$\frac{1}{24}$}+it) \ dt+ix^{\frac{1}{24}} \int_0^{\infty}\frac{x^{-it}}{1-e^{\frac{1}{12}\pi i +2\pi t}}\Psi_a(\mbox{$\frac{1}{24}$}-it) \ dt,$$ where the $\sqrt{it}$ and $\sqrt{-it}$ which occur in $\Psi_a(\mbox{$\frac{1}{24}$}+it)$ and $\Psi_a(\mbox{$\frac{1}{24}$}-it)$ are to be interpreted as $e^{(1/4)\pi i}\sqrt{t}$ and $e^{-(1/4)\pi i}\sqrt{t}$ respectively. We write this in the form

say, where $$X_a(x)=ix^{\frac{1}{24}}\int_0^{\infty} {x^{it}}\Psi_a(\mbox{$\frac{1}{24}$}+it) \ dt. $$ Now, since $$|x^{it}|=e^{-\theta t},\quad |x^{-it}|=e^{\theta t},$$ the functions $\Theta$ are regular throughout the angle of Lemma 4.42. And $$X_a(x)=\frac{x^{\frac{1}{24}}}{\sqrt{i}}\int_0^{\infty}e^{-\lambda t}\frac{d}{dt}\left (\frac{\cosh \mu\sqrt{t}-1}{\sqrt{t}}\right ) \ dt,$$ where $$\lambda =i\log\frac{1}{x},\quad \mu =a\sqrt{i}.$$ The form of this integral may be calculated by supposing $\lambda$ and $\mu$ positive, when we obtain \begin{eqnarray*} \int_0^{\infty}e^{-\lambda w^2}\frac{d}{dw}\left (\frac{\cosh \mu w-1}{w}\right ) \ dw &=& 2\lambda \int_0^{\infty}e^{-\lambda w^2}\left ({\cosh \mu w-1}\right ) \ dw\\ &=& \sqrt{(\lambda\pi )}(e^{\mu^2/4\lambda}-1). \end{eqnarray*} Hence and the proof of the lemma is completed.

Lemmas 4.41 and 4.42 shew that $x=1$ is the sole finite singularity of the principal branch of $F_a(x)$.

4.43

We use $K$ generally to denote a positive number independent of $x$ and of $a$. We may employ the formula (4.411). It is plain that $$\left|\frac{x^z}{1-e^{2\pi i z}}\right|\lt Ke^{-\theta_1|\eta |},$$ $$|\Psi_a(z)|=\left |\frac{d}{dz}\left\{\frac{\cosh a \sqrt{(z-\mbox{$\frac{1}{24}$})}-1}{\sqrt{(z-\mbox{$\frac{1}{24}$})}}\right\}\right | \lt Ke^{K\sqrt{|\eta |}},$$ where $\eta$ is the imaginary part of $z$. Hence $$|F_a(x)|\lt K\int_{-\infty}^{\infty}e^{K\sqrt{|\eta |}-\theta_1|\eta|} \ d\eta\lt K.$$

4.44

If we refer back to (4.421) and (4.422), we see that it is sufficient to prove that $$|\Theta_1(x)|\lt Ka^2,\quad |\Theta_2(x)|\lt Ka^2;$$ and we may plainly confine ourselves to the first of these inequalities. We have $$\Theta_1(x)=\frac{x^{\frac{1}{24}}}{\sqrt{i}}\int_0^{\infty} \frac{x^{it}}{e^{-\frac{1}{12}\pi i+2\pi t}-1} \frac{d}{dt}\left\{\frac{\cosh a\sqrt{(it)}-1}{\sqrt{(t)}}\right\} \ dt.$$ Rejecting the extraneous factor, which is plainly without importance, and integrating by parts, we obtain $$\Theta (x)=\int_0^{\infty}\Phi (t)\frac{\cosh a\sqrt{(it)}-1}{\sqrt{(t)}}\ dt,$$ where $$\Phi (t)=-\frac{ix^{it}\log x}{e^{-\frac{1}{12}\pi i+2\pi t}-1}+ \frac{2\pi x^{it}e^{-\frac{1}{12}\pi i+2\pi t} } {(e^{-\frac{1}{12}\pi i+2\pi t}-1)^2}.$$ Now $|\theta |\lt\half \pi$ and $|x^{it}|\lt Ke^{\half \pi t}$. It follows that $$|\Phi (t)|\lt Ke^{-\pi t};$$ and \begin{eqnarray*} |\Theta (x)|& \lt & K\int_0^{\infty}\frac{e^{-\pi t}}{\sqrt{t}}|\sinh ^2\half a\sqrt{(it)}| \ dt \\ & \lt & K\int_0^{\infty}\frac{e^{-\pi t}}{\sqrt{t}}\{\cosh a\sqrt{(\half t)}-\cos a\sqrt{(\half t)}\} \ dt\\ & \lt &K\int_0^{\infty}{e^{-\pi w^2}}\left (\cosh\frac{aw}{\sqrt{2}}-\cos\frac{aw}{\sqrt{2}}\right ) \ dw \\ &=&K(e^{a^2/8\pi}-e^{-a^2/8\pi})\lt Ka^2. \end{eqnarray*}

5.1 We write

where $C=\pi\sqrt{\mbox{$\frac{2}{3}$}}, \ x_{p,q}=xe^{-2p\pi i/q}$; and where the summation applies to all values of $p$ not exceeding $q$ and prime to $q$, and to all values of $q$ such that $\alpha$ being positive and independent of $n$. If then we have where $\Gamma$ is a circle whose centre is the origin and whose radius $R$ is less than unity. We take where $\beta$ also is positive and independent of $n$.

Our object is to shew that the integral on the right-hand side of (5.15) is of the form $O(n^{-\frac{1}{4}})$; the constant implied in the $O$ will of course be a function of $\alpha$ and $\beta$. It is to be understood throughout that $O$'s are used in this sense; $O(1)$, for instance, stands for a function of $x,n,p,q,\alpha$, and $\beta$ (or some only of these variables) which is less in absolute value than a number $K=K(\alpha, \beta)$ independent of $x,n,p$, and $q$.

We divide up the circle $\Gamma$ by means of dissection $\Xi_{\nu}$ of 4.22, into arcs $\xi_{p,q}$ each associated with a point $Re^{2p\pi i/q}$; and we denote by $\eta_{p,q}$ the arc of $\Gamma$ complementary to $\xi_{p,q}$. This being so, we have

say. We shall prove that each of these sums is of the form $O(n^{-\frac{1}{4}})$; and we shall begin with the second sum, which only involves the auxiliary functions $F$.

5.21. We have, by Cauchy's theorem,

where $\zeta_{p,q}$ consists of the contour $LMNM'L'$ shewn in the figure. Here $L$ and $L'$ are the ends of $\xi_{p,q}$, $LM$ and $M'L'$ are radii vectors, and $MNM'$ is part of a circle $\Gamma_1$ whose radius $R_1$ is greater than 1. $P$ is the point $e^{2p\pi i/q}$; and we suppose that $R_1$ is small enough to ensure that all points of $LM$ and $M'L'$ are at a distance from $P$ less than $\mbox{$\frac{1}{2}$}$. The other circle $c$ shewn in the figure has $P$ as its centre and radius $\frac{1}{2}$. We denote $LM$ by $\varpi_{p,q}$, $M'L'$ by $\varpi_{p,q}'$ and $MNM'$ by $\gamma_{p,q}$: and we write

5.22. Suppose first that $x$ lies on $\gamma_{p,q}$ and outside $c$. Then, in virtue of (5.11) and Lemma 4.43, we have

If on the other hand $x$ lies on $\gamma_{p,q}$, but inside $c$, we have, by (5.11) and Lemma 4.44, where But, if we refer to the definition of $\chi_{a}(x)$ in Lemma 4.42, and observe that $$\left |\exp\frac{a^2}{4\log (1/x)}\right |=\exp\frac{a^2\log (1/r)}{4[\{\log (1/r)\}^2+\theta^2]}\lt 1$$ if $x=re^{i\theta}$ and $r>1$, we see that on the part of $\gamma_{p,q}$ in question. Hence (5.221) holds for all $\gamma_{p,q}$. It follows that $$j_{p,q}^1=O(R_1^{-n}\sqrt{q}),$$

This sum tends to zero more rapidly than any power of $n$, and is therefore completely trivial.

5.231. We must now consider the sums which arise from the integrals along $\varpi_{p,q}$ and $\varpi_{p,q}'$; and it is evident that we need consider in detail only the first of these two lines. We write

say.

In the first place we have, from (5.222), $$j'_{p,q}=O\left (q^{-\frac{3}{2}}\int_{R}^{R_1}\frac{dr}{r^{n+1}}\right)=O(q^{-\frac{3}{2}}n^{-1}),$$ since

Thus

5.232. In the second place we have $$j''_{p,q}=\omega_{p,q}\frac{\sqrt{q}}{\pi\sqrt{2}} \int_{\varpi_{p,q}}\frac{\chi_{C/q}(x_{p,q})}{x^{n+1}} \ dx.$$ It is plain that, if we substitute $y$ for $xe^{-2p\pi i/q}$, then write $x$ again for $y$, and finally substitute for $\chi_{C/q}$ its explicit expression as an elementary function, given in Lemma 4.42, we obtain

say, where and the path of integration is now a line related to $x=1$ as $\varpi_{p,q}$ is to $x=e^{2p\pi i/q}$: the line defined by $x=re^{i\theta}$, where $R\leq r\leq R_1$, and $\theta$ is fixed and (by Lemma 4.22) lies between $1/2q\nu$ and $1/q\nu$.

Integrating $J$ by parts, we find

say.

5.233. In estimating $J_1,J_2$, and $J_3$, we must bear the following facts in mind.

(1) Since $|x|\geq R$, it follows from (5.2312) that $|x|^{-n}=O(1)$ throughout the range of integration.

(2) Since $1-R=\beta /n$ and $1/2q\nu \lt\theta\lt1/q\nu$, where $\nu =[\alpha\sqrt{n}]$, we have $$\log\left (\frac{1}{x}\right )=O\left (\frac{1}{q\sqrt{n}}\right ),$$ when $r=R$, and $$\frac{1}{\log (1/x)}=O(q\sqrt{n}),$$ throughout the range of integration.

(3) Since $$|E(x)|=\exp\frac{\pi^2\log (1/r)}{6q^2[\{\log (1/r)\}^2+\theta^2]},$$ $E(x)$ is less than 1 in absolute value when $r>1$. And, on the part of the path for which $r\lt1$, it is of the form $$\exp O\left (\frac{1}{q^2n\theta^2}\right )=\exp O(1)=O(1).$$ It is accordingly of the form $O(1)$ throughout the range of integration.

5.234. Thus we have, first

secondly and thirdly From (5.2341), (5.2342), (5.2343), and (5.2322), we obtain $$J=O\left(q^{-\half }n^{-\frac{5}{4}}\right)+O\left(q^{\half}n^{-\frac{7}{4}}\right);$$ and, from (5.2321), $\ds j''_{p,q}=O\left(n^{-\frac{5}{4}}\right)+O\left(qn^{-\frac{7}{4}}\right)$.

Summing, we obtain

5.235. From (5.2311), (5.2313), and (5.2344), we obtain

and in exactly the same way we can prove And from (5.212), (5.224), (5.2351), and (5.2352), we obtain, finally,

5.31. We turn now to the discussion of

say, where $$\rho_{p,q}(x)=\omega_{p,q}\sqrt{\left (\frac{q}{2\pi}\log\frac{1}{x_{p,q}}\right )} x^{\frac{1}{24}}_{p,q},$$ $$X_{p,q}(x)=\chi_{p,q}(x)+\rho_{p,q}(x)=\rho_{p,q}(x)E(x_{p,q}),$$ $E(x)$ being defined as in (5.23211).

5.32. The discussion of these two sums is, after the analysis which precedes, a simple matter. The arc $\xi_{p,q}$ is less than a constant multiple of $1/q\sqrt{n}$; and $x^{-n}=O(1)$ on $\xi_{p,q}$. Also $$\ds |F_{p,q}(x)-\chi_{p,q}(x)|=O\left(q^{-\frac{3}{2}}\right),$$ by (5.222); and

since $\ds |x_{p,q}|=R=1-(\beta / n),\quad |amx_{p,q}|\lt 1/q\nu$.

Hence $$J_{p,q}^2=O\left(q^{-\frac{5}{2}}n^{-\half}\right),$$

and $$J_{p,q}^3=O\left(q^{-1}n^{-\frac{3}{4}}\right),$$

5.33. From (4.321) and (5.2221), we have

where $$\Omega (z)=f(z)-1=\prod_1^{\infty}\left (\frac{1}{1-z^{\nu}}\right )-1=\sum_1^{\infty}p(\nu )z^{\nu},$$ if $|z|\lt 1$ and $$\ds x'_{p,q}=\exp\left\{-\frac{4\pi^2}{q^2\log (1/x_{p,q})}+\frac{2\pi ip'}{q}\right\}.$$ Now $$\ds |x'_{p,q}|=\exp\left[-\frac{4\pi^2\log (1/R)}{q^2\{[\log (1/R)]^2+\theta^2\}}\right ],$$ where $\theta$ is the amplitude of $x_{p,q}$. Also $$q^2\{[\log (1/R)]^2+\theta^2\}=O\left\{q^2\left (\frac{1}{n^2}+\frac{1}{q^2n}\right)\right\}=O\left(\frac{1}{n}\right),$$ while $\log (1/R)$ is greater than a constant multiple of $1/n$. There is therefore a positive number $\delta$, less than unity and independent of $n$ and of $q$, such that $$|x'_{p,q}|\lt \delta;$$ and we may write $\ds\Omega (x'_{p,q})=O(|x'_{p,q}|)$.

We have therefore $$E(x_{p,q})\Omega(x'_{p,q})=O(|x'_{p,q}|^{-\frac{1}{24}})O(|x'_{p,q}|) =O(|x'_{p,q}|^{\frac{23}{24}})=O(1);$$ and so, by (5.321), $$f(x)-\chi_{p,q}(x)=O(\sqrt{q})O\left (\sqrt{\left |\log\frac{1}{x_{p,q}}\right |}\right )O(1)=O(n^{-\frac{1}{4}}).$$ And hence, as the length of $\xi_{p,q}$ is of the form $O(1/q\sqrt{n})$, we obtain $$J_{p,q}^1=O(q^{-1}n^{-\frac{3}{4}}),$$

5.34. From (5.311), (5.322), (5.323), and (5.332), we obtain

5.4. From (5.15), (5.17), (5.2353), and (5.341), we obtain

But $$\sum_p c_{p,q,n}=\frac{\sqrt{q}}{\pi\sqrt{2}}A_q \frac{d}{dn}\frac{\cosh (C\lambda_n/q)-1}{\lambda_n}, $$ where $$A_q=\sum_p\omega_{p,q}e^{-2np\pi i/q}.$$ All that remains, in order to complete the proof of the theorem, is to shew that $$\cosh (C\lambda_n/q)-1$$ may be replaced by $\half e^{C\lambda_n/q}$;

and in order to prove this it is only necessary to shew that $$\sum_{q\lt O(\sqrt{n})}q^{\frac{3}{2}} \frac{d}{dn}\frac{\half e^{C\lambda_n/q}-\cosh (C\lambda_n/q)+1}{\lambda_n}=O(n^{-\frac{1}{4}}).$$ On differentiating we find that the sum is of the form $$\sum_{q\lt O(\sqrt{n})}q^{\frac{3}{2}}\left\{O\left (\frac{1}{qn}\right )+O\left (\frac{1}{n^{\frac{3}{2}}} \right )\right\} =O\left\{\frac{1}{n}\sum_{q\lt O(\sqrt{n})}q^{\frac{1}{2}}\right\} =O(n^{-\frac{1}{4}}).$$ Thus the theorem is proved.

6.1. The theorem which we have proved gives information about $p(n)$ which is in some ways extraordinarily exact. We are for this reason the more anxious to point out explicitly two respects in which the results of our analysis are incomplete.

6.21. We have proved that $$p(n)=\sum A_q\phi_q+O(n^{-\frac{1}{4}}),$$ where the summation extends over the values of $q$ specified in the theorem, for every fixed value of $\alpha$; that is to say that, when $\alpha$ is given, a number $K=K(\alpha )$ can be found such that $$\left|p(n)-\sum A_q\phi_q \right|\lt Kn^{-\frac{1}{4}}$$ for every value of $n$. It follows that

where $\{x\}$ denotes the integer nearest to $x$, for $n\geq n_0$, where $n_0=n_0(\alpha)$ is a certain function of $\alpha$.

The question remains whether we can, by an appropriate choice of $\alpha$, secure the truth of (6.211) for all values of $n$, and not merely for all sufficiently large values. Our opinion is that this is possible, and that it could be proved to be possible without any fundamental change in our analysis. Such a proof would however involve a very careful revision of our argument. It would be necessary to replace all formulæ involving $O$'s by inequalities, containing only numbers expressed explicitly as functions of the various parameters employed. This process would certainly add very considerably to the length and the complexity of our argument. It is, as it stands, sufficient to prove what is, from our point of view, of the greatest interest; and we have not thought it worth while to elaborate it further.

6.22. The second point of incompleteness of our results is of much greater interest and importance. We have not proved either that the series $$\sum_1^{\infty}A_q\phi_q$$ is convergent, or that, if it is convergent, it represents $p(n)$. Nor does it seem likely that our method is one intrinsically capable of proving these results, if they are true–a point on which we are not prepared to express any definite opinion.

It should be observed in this connection that we have not even discovered anything definite concerning the order of magnitude of $A_q$ for large values of $q$. We can prove nothing better than the absolutely trivial equation $A_q=O(q)$. On the other hand we can assert that $A_q$ is, for an infinity of values of $q$, effectively of an order as great as $q$, or indeed even that it does not tend to zero (though of course this is most unlikely).

6.3. Our formula directs us, if we wish to obtain the exact value of $p(n)$ for a large value of $n$, to take a number of terms of order $\sqrt{n}$. The numerical data suggest that a considerably smaller number of terms will be equally effective; and it is easy to see that this conjecture is correct.

Let us write $$\beta =4\pi\sqrt{\left (\mbox{$\frac{2}{3}$}\right )}=4C, \quad \mu =\left [\frac{\beta\sqrt{n}}{\log n}\right ],$$ and let us suppose that $\alpha \lt 2$. Then \begin{eqnarray*} \sum_{\mu +1}^{\nu}A_q\phi_q&=&\sum_{\mu +1}^{\nu}O(q^{\frac{3}{2}})O\left (\frac{1}{qn}\right )O(e^{C\sqrt{n}/q})=O\left (\frac{1}{n}\sum_{\mu +1}^{\nu}\sqrt{q}e^{C\sqrt{n}/q}\right )\\ &=&O\left (\frac{1}{n}\int_{\mu}^{\nu}\sqrt{x}e^{C\sqrt{n}/x} \ dx\right ), \end{eqnarray*} since $\sqrt{q}e^{C\sqrt{n}/q}$ decreases steadily throughout the range of summation²⁵.

Writing $\sqrt{n}/y$ for $x$, we obtain \begin{eqnarray*} O\left(n^{-\frac{1}{4}}\int_{1/\alpha}^{\sqrt{n}/\mu}y^{-\frac{5}{2}}e^{Cy} \ dy\right) &=& O\left \{n^{-\frac{1}{4}}\left (\frac{\sqrt{n}}{\mu}\right )^{-\frac{5}{2}}e^{C\sqrt{n}/\mu}\right \}=O\left\{n^{-\frac{1}{4}}(\log n)^{-\frac{5}{2}}e^{\frac{1}{4}\log n}\right\}\\ &=& O(\log n)^{-\frac{5}{2}}=o(1). \end{eqnarray*} It follows that it is enough, when $n$ is sufficiently large, to take $$\left [\frac{\beta\sqrt{n}}{\log n}\right ]$$ terms of the series. It is probably also necessary to take a number of terms of order $\sqrt{n}/(\log n)$; but it is not possible to prove this rigorously without a more exact knowledge of the properties of $A_q$ than we possess.

6.4. We add a word on certain simple approximate formulæ for $\log p(n)$ found empirically by Major MacMahon and by ourselves. Major MacMahon found that if

then $a_n$ is approximately equal to 2 within the limits of his table of values of $p(n)$ (Table IV). This suggested to us that we should endeavour to find more accurate formulæ of the same type. The most striking that we have found is the mode of variation of $a_n$ is shewn in Table III.

In this connection it is interesting to observe that the function $$13^{-\sqrt{n}}p(n)$$ (which ultimately tends to infinity with exponential rapidity) is equal to .973 for $n=30000000000$.

7.1. We shall conclude with a few remarks concerning actual or possible applications of our method to other problems in Combinatory Analysis or the Analytic Theory of Numbers.

The class of problems in which the method gives the most striking results may be defined as follows. Suppose that $q(n)$ is the coefficient of $x^n$ in the expansion of $F(x)$, where $F(x)$ is a function of the form

$f(x)$ being the function considered in this paper, The $a$'s, $b$'s, $\alpha$'s, and $\beta$'s being positive integers, and the number of factors in numerator and denominator being finite; and suppose that $|F(x)|$ tends exponentially to infinity when $x$ tends in an appropriate manner to some or all the points $e^{2p\pi i/q}$. Then our method may be applied in its full power to the asymptotic study of $q(n)$, and yields results very similar to those which we have found concerning $p(n)$.

Thus, if $$F(x)=\frac{f(x)}{f(x^2)}=(1+x)(1+x^2)(1+x^3)\cdots =\frac{1}{(1-x)(1-x^3)(1-x^5)\cdots},$$ so that $q(n)$ is the number of partitions of $n$ into odd parts, or into unequal parts²⁷, we find that \begin{eqnarray*} q(n)=\frac{1}{\sqrt{2}}\frac{d}{dn} J_0\left[i\pi\sqrt{\{\mbox{$\frac{1}{3}$}(n+\mbox{$\frac{1}{24}$})\}}\right] +\sqrt{2}\cos (\mbox{$\frac{2}{3}$}n\pi-\mbox{$\frac{1}{9}$}\pi) \frac{d}{dn} J_0\left[\mbox{$\frac{1}{3}$}i\pi\sqrt{\{\mbox{$\frac{1}{3}$}(n+\mbox{$\frac{1}{24}$})\}}\right] +\cdots . \end{eqnarray*} The error after $[\alpha\sqrt{n}]$ terms is of the form $O(1)$. We are not in a position to assert that the exact value of $q(n)$ can always be obtained from the formula (though this is probable); but the error is certainly bounded.

If $$F(x)=\frac{f(x^2)}{f(-x)}=\frac{f(x)f(x^4)}{\{f(x^2)\}^2} =(1+x)(1+x^3)+x^5)\cdots ,$$ so that $q(n)$ is a number of partitions of $n$ into parts which are both odd and unequal, then \begin{eqnarray*} q(n)=\frac{d}{dn} J_0\left[i\pi\sqrt{\{\mbox{$\frac{1}{6}$}(n-\mbox{$\frac{1}{24}$})\}}\right] +{2}\cos (\mbox{$\frac{2}{3}$}n\pi-\mbox{$\frac{2}{9}$}\pi ) \frac{d}{dn} J_0\left[\mbox{$\frac{1}{3}$}i\pi\sqrt{\{\mbox{$\frac{1}{6}$}(n-\mbox{$\frac{1}{24}$})\}}\right] +\cdots . \end{eqnarray*} The error is again bounded (and probably tends to zero).

If $$F(x)=\frac{\{f(x)\}^2}{f(x^2)}=\frac{1}{1-2x+2x^4-2x^9+\cdots},$$ $q(n)$ has no very simple arithmetical interpretation; but the series is none the less, as the direct reciprocal of simple $\vartheta$-function, of particular interest. In this case we find $$q(n)=\frac{1}{4\pi}\frac{d}{dn}\frac{e^{\pi\sqrt{n}}}{\sqrt{n}} +\frac{\sqrt{3}}{2\pi}\cos (\mbox{$\frac{2}{3}$}n\pi -\mbox{$\frac{1}{6}$}\pi )\frac{d}{dn}\frac{e^{\frac{1}{3}\pi\sqrt{n}}}{\sqrt{n}}+\cdots .$$ The error here is (as in the partition problem) of order $O(n^{-\frac{1}{4}})$, and the exact value can always be found from the formula.

7.2. The method also be applied to product of form (7.11) which have (to put the matter roughly) no exponential infinities. In such cases the approximation is of much less exact character. On the other hand the problems of this character are of even greater arithmetical interest.

The standard problem of this category is that of the representation of a number as a sum of $s$ squares, $s$ being any positive integer odd or even²⁸. We must reserve the application of our method to this problem for another occasion; but we can indicate the character of our main result as follows.

If $r_s(n)$ is the number of representations of $n$ as the sum of $s$ squares we have $$F(x)=\sum r_s(n)x^n=(1+2x+2x^4+\cdots )^s=\frac{\{f(x^2)\}^s}{\{f(-x)\}^{2s}}=\frac{\{f(x)\}^{2s}\{f(x^4)\}^{2s}}{\{f(x^2)\}^{5s}}.$$

We find that

where $c_q$ is a function of $q$ and of $n$ of the same general type as the function $A_q$ of this paper. The series is absolutely convergent for sufficiently large values of $s$, and the summation in (7.21) may be regarded indifferently as extended over all values of $q$ or only over a range $1\leq q\leq \alpha \sqrt{n}$. It should be observed that the series (7.22) is quite different in form from any of the infinite series which are already known to occur in connection with this problem.

7.3. There is also a wide range of problems to which our methods are partly applicable. Suppose, for example, that $$F(x)=\sum p^2(n)x^n=\frac{1}{(1-x)(1-x^4)(1-x^9)\cdots},$$ so that $p^2(n)$ is the number of partitions of $n$ into squares. Then $F(x)$ is not an elliptic modular function; it possesses no general transformation theory: and the full force of our method can not be applied. We can still, however, apply some of our preliminary methods. Thus the ``Tauberian'' argument shews that $$\log p^2(n)\sim2^{-\frac{4}{3}}3\pi^{\frac{1}{3}}\{\zeta (\mbox{$\frac{3}{2}$})\}^{\frac{2}{3}}n^{\frac{1}{3}}.$$ And although there is no general transformation theory, there is a formula which enables us to specify the nature of the singularity at $x=1$. This formula is \begin{eqnarray*} \frac{1}{f(e^{-\pi z})}&=&2 \sqrt{\left (\frac{\pi}{z}\right )}\exp \left\{\frac{2\pi}{\sqrt{z}}\zeta (-\half )\right \}\\ &&\hspace{0.6in}\times \prod_1^{\infty}\left\{1-2e^{-2\pi\sqrt{(n/z)}}\cos 2\pi \sqrt{(n/z)}+e^{-4\pi\sqrt{(n/z)}}\right\}. \end{eqnarray*} By the use of this formula, in conjunction with Cauchy's theorem, it is certainly possible to obtain much more precise information about $p^2(n)$ and in particular the formula $$p^2(n)\sim 3^{-\half }(4\pi n)^{-\frac{7}{6}}\{\zeta (\mbox{$\frac{3}{2}$})\}^{\frac{2}{3}}e^{2^{-(4/3)}3\pi^{(1/3)}\{\zeta (\mbox{$\frac{3}{2}$})\}^{\frac{2}{3}}n^{\frac{1}{3}}}.$$ The corresponding formula for $p^s(n)$, the number of partitions of $n$ into perfect $s$-th powers, is $$p^s(n)\sim (2\pi )^{-\half (s+1)}\sqrt{\left (\frac{s}{s+1}\right )}kn^{\frac{1}{s+1}-\frac{3}{2}}e^{(s+1)kn^{1/(s+1)}},$$ where $$k=\left \{\frac{1}{s}\Gamma\left (1+\frac{1}{s}\right ) \zeta\left (1+\frac{1}{s}\right )\right\}^{\frac{s}{s+1}}.$$

The series (7.21) may be written in the form $$\frac{\pi^{\half s}}{\Gamma (\half s)}n^{\half s-1}\sum_{p,q}\frac{\omega_{p,q}^s}{q^{\half s}}e^{-np\pi i/q},$$ where $\omega_{p,q}$ is always one of the five numbers 0, $e^{\frac{1}{4}\pi i}$, $e^{-\frac{1}{4}\pi i}$, $-e^{\frac{1}{4}\pi i}$, $-e^{-\frac{1}{4}\pi i}$. When $s$ is even it begins $$\frac{\pi^{\half s}}{\Gamma (\half s)}n^{\half s-1}\left\{1^{-\half s}+2\cos (\half n\pi-\mbox{$\frac{1}{4}$}s\pi )2^{-\half s} +2\cos (\mbox{$\frac{2}{3}$} n\pi-\mbox{$\frac{1}{2}$}s\pi )3^{-\half s}+\cdots\right\}.$$ It has been proved by Ramanujan that the series gives an exact representation of $r_s(n)$ when $s=4,6,8$; and by Hardy that this also true when $s=3,5,7$. See Ramanujan, ``On certain trigonometrical sums and their applications in the Theory of Numbers''; Hardy, ``On the expression of a number as the sum of any number of squares, and in particular of five or seven²⁹''.