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Define $\tau(k)$ as follows :

$$\Delta(x)=x\prod_{n=1}^\infty(1-x^n)^{24}=\sum_{k=1}^\infty\tau(k)x^k.$$

This is equivalent to

$$\prod_{n=1}^\infty(1-x^n)^{24}=\sum_{k=1}^\infty\tau(k)x^{k-1}.$$

Let $a(n) = \tau(n)$ mod $(n-1)$ for $n > 1$.

For example (OEIS A299204),

$a(2) = 0, a(3) = 0, a(4) = 1, a(5) = 2$.

Surprisingly, except $ 4 $ and $ 16 $, there is no integer $n \le 10^7$ such that $a(n) = 1$.

Are the only solutions of the equation $a(n) = 1$ $n=4$ and $n=16$?

P.S.

Let $b(k)$ = number of the solution of the equation $a(n) = k$ for $n$ in $[2,10^7]$.

$ \{b(0), b(1), \cdots \} = \{631, 2, 544, 11, 18, 8, 14, 5, 20, 15, 14, 4, 23, 4, 7, 9, 17, 5, 25, 6, 21, 7, 18, 11, 15, 7, 16, 13, 19, 11, 12, 6, 16, 6, 7, 9, 20, 5, 12, 9, 21, 6, 17, 1, 11, 13, 11, 8, 26, 11, 12, 6, 18, 4, 22, 5, 21, 8, 13, 2, 18, 2, 14, 9, 15, 7, 16, 6, 11, 13, 11, 3, 39, 6, 12, 11, 14, 14, 14, 2, 12, 8, 10, 4, 31, 9, 7, 10, 10, 8, 20, 7, 18, 12, 14, 3, 17, 3, 17, 16, \cdots \}.$

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    $\begingroup$ Are the solution of a(n)=1 only 4 and 16? $\endgroup$ – TOM Feb 8 '18 at 14:08
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    $\begingroup$ I think this is a fine question, although the motivation is missing, and the wording (grammar) could be improved. I am saying this so that the question is not closed (but perhaps improved by some). $\endgroup$ – GH from MO Feb 8 '18 at 15:08
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    $\begingroup$ There is definitely something to explain here. The values of $a(n)$ appear t o be fairly uniformly distributed, but if that were the case then the probablility of having no solutions in $[17,10^7]$ would be about $10^{-5.8}$. $\endgroup$ – Neil Strickland Feb 8 '18 at 16:21
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    $\begingroup$ I believe it is open whether $\tau(n) = 1$ for any $n$, but it is known that there are only finitely many $n$ for which $\tau(n) = 1$ by work of Murty, Murty and Shorey from 1987. $\endgroup$ – Jeremy Rouse Feb 8 '18 at 21:27
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    $\begingroup$ Don't forget that $\Delta\equiv q+q^9+q^{25}+\dots\pmod2$, so the probability is very much biased in favor of even values. Nonetheless it is a fact that $a(n)=3$ occurs quite often for instance. $\endgroup$ – Henri Cohen Feb 8 '18 at 22:55

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