Irregular numbers
Journal of the Indian Mathematical Society, V, 1913, 105 – 106

1. Let $a_2, a_3, a_5, a_7, \ldots$ denote numbers less than unity, where the subscripts 2,3,5,7, $\ldots$ are the series of prime numbers. Then

\begin{eqnarray} \frac{1}{1-a_2} \cdot \frac{1}{1-a_3} \cdot \frac{1}{1-a_5} \ldots &=& 1 + a_2 + a_3 + a_2 \cdot a_2 + a_5 \nonumber \\ & &+ a_2 \cdot a_3 + a_7 + a_2 \cdot a_2 \cdot a_2+ a_3 \cdot a_3+ \ldots, \end{eqnarray}
the terms being so arranged that the products obtained by multiplying the subscripts are the series of natural numbers 2, 3, 4, 5, 6, 7, 8, 9, ${\ldots}$.

The above result is easily got if we remember that the natural numbers are formed by multiplying primes and their powers.

2. Similarly, we have

\begin{eqnarray} \frac{1}{1+a_2} \cdot \frac{1}{1+a_3} \cdot \frac{1}{1+a_5} \ldots &=& 1 - a_2 - a_3 + a_2 \cdot a_2 - a_5\nonumber\\ && + a_2 \cdot a_3 - a_7 - a_2 \cdot a_2 \cdot a_2 + a_3 \cdot a_3 + \ldots, \end{eqnarray}
where the sign is negative whenever a term contains an odd number of prime subscripts.

3. Put $a_2 = 1/2^n, a_3 = 1/3^n, a_5 = 1/5^n, \ldots$ in (1), and we get

$$\left(1-\frac{1}{2^n} \right) \left(1 - \frac{1}{3^n} \right) \left(1-\frac{1}{5^n}\right) \left(1-\frac{1}{7^n}\right) \ldots =\frac{1}{S_n},$$
where $S_n$ denotes $1/1^n + 1/2^n + 1/3^n + 1/4^n + \ldots$.

Changing $n$ into $2n$ in (3) and dividing by the original, we obtain

$$\left(1+\frac{1}{2^n}\right) \left(1+\frac{1}{3^n}\right) \left(1+\frac{1}{5^n}\right) \left(1+\frac{1}{7^n}\right) \ldots = \frac{S_n}{S_{2n}}.$$

Examples:

\begin{eqnarray} \mbox{(i) } \left(1+\frac{1}{2^2}\right) \left(1+\frac{1}{3^2}\right) \left(1+ \frac{1}{5^2}\right) \ldots = \frac{15}{\pi^2}, \end{eqnarray}
\begin{eqnarray} \mbox{(ii) } \left(1+\frac{1}{2^4}\right) \left(1+\frac{1}{3^4}\right) \left(1+\frac{1}{5^4}\right) \ldots = \frac{105}{\pi^4}, \end{eqnarray}
since $$S_2 = \pi^2/6, S_4 = \pi^4/90, S_8=\pi^8/9450.$$

4. Subtract (2) from (1) and put $a_2 = 2^{-n} \ldots ;$ then $$\frac{1}{2^n} + \frac{1}{3^n} + \frac{1}{5^n} + \frac{1}{7^n} + \frac{1}{8^n} + \frac{1}{11^n} + \frac{1}{12^n} + \ldots = \frac{S_n^2 - S_{2n}}{2S_n},$$ where the numbers 2, 3, 5, 7, 8, $\ldots$ contain an odd number of prime divisors.

Examples:

\begin{eqnarray} \mbox{(i) } \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{8^2} + \cdots = \frac{\pi^2}{20}, \end{eqnarray}
\begin{eqnarray} \mbox{(ii) } \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \frac{1}{8^4}+ \cdots = \frac{\pi^4}{1260}. \end{eqnarray}

5. Again (2,3,5,7, $\ldots$ being the prime numbers)

\begin{eqnarray} (1+a_2) (1+a_3) (1+a_5) (1+a_7) \cdots &=& 1 + a_2 + a_3 + a_5\nonumber\\ &&+ a_2 \cdot a_3 + a_7 + a_2 \cdot a_5 + a_{11} + a_{13} + \cdots, \end{eqnarray}
where the product of the subscripts in any term is a natural number containing dissimilar prime divisors; and

$$(1-a_2) (1-a_3) (1-a_5) (1-a_7) \ldots = 1 - a_2 - a_3 - a_5 + a_2 \cdot a_3 - a_7,$$
where the signs are negative whenever the number of factors is odd.

6. Replacing as before $a_2, a_3, a_5$, $\ldots$ by the values given in $\S 3$ and using (4), we deduce that

$$1 + \frac{1}{2^n} + \frac{1}{3^n} + \frac{1}{5^n} + \frac{1}{6^n}+ \ldots = \frac{S_n}{S_{2n}},$$
where 2, 3, 5, 6, 7, $\ldots$ are the numbers containing dissimilar prime divisors.

7. Also taking half the difference between (3) and (4),

\begin{eqnarray} \frac{1}{2^n} + \frac{1}{3^n} &+& \frac{1}{5^n} + \frac{1}{7^n} + \frac{1}{11^n} + \frac{1}{13^n} + \frac{1}{17^n} + \frac{1}{19^n} + \frac{1}{23^n}+\frac{1}{29^n} \nonumber\\ &+& \frac{1}{30^n} + \frac{1}{31^n} + \ldots = \frac{S_n^2-S_{2n}}{2S_n S_{2n}}, \end{eqnarray}
where 2, 3, 5, $\ldots$ are numbers containing an odd number of dissimilar prime divisors.

Examples:

\begin{eqnarray} \mbox{(i) } \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{9}{2 \pi^2}, \end{eqnarray}
\begin{eqnarray} \mbox{(ii) }\frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots = \frac{15}{2 \pi^4}. \end{eqnarray}

8. Subtracting (11) from $S_n$, we have

$$\frac{1}{4^n} + \frac{1}{8^n} + \frac{1}{9^n} + \frac{1}{12^n} + \ldots = \frac{S_n(S_{2n}-1)}{S_{2n}},$$
where 4, 8, 9, $\ldots$ are composite numbers having at least two equal prime divisors.