Dear Professor Zeilberger,

One of them I could find as eq 1.10 in the following paper:

http://www.sciencedirect.com/science/article/pii/S0377042703006186?np=y

For the other two a(5n+2) and a(5n+4) my program delivers the following
identity:

(\sum_{n=0}^{\infty} a(5n+2)q^n)(\sum_{n=0}^{\infty} a(5n+4)q^n)
=100\prod_{m=1}^{\infty} \frac{(1-q^{5n})^8}{(1-q^n)^{12}}
+2500q\prod_{m=1}^{\infty} \frac{(1-q^{5n})^{14}}{(1-q^n)^{18}}
+15625q^2\prod_{m=1}^{\infty} \frac{(1-q^{5n})^{20}}{(1-q^n)^{24}}

Note that the right hand side has the form 25C(q) where C(q) is a power
series in q with integer coefficients.
Next we call the product of the two power series on the left hand side
A(q)B(q).

Then the above identity can be written as A(q)B(q)=25C(q).
This gives the following possibilities:
either (a) every coefficient in A(q) is divisible by 25
or (b) every coefficient in B(q) is divisible by 25
or (c) every coefficient in A(q) is divisible by 5 and every coefficient
in B(q) is divisible by 5.

The cases (a) and (b) can be excluded because a(2)=5 and a(4)=20, or in
other words the first term in A(q) and in B(q) is not divisible by 5.

We conclude that case (c) is true, that is there are q-series D(q), E(q)
with integer coefficients such that
A(q)=5D(q) and B(q)=5E(q).

I do not know if this is well known, I suspect that it is hidden in some
paper however I doubt that the identity delivered by my program can be
found in some paper.

I feel that an elementary proof can be given and this is certainly
interesting, however it falls out of my area of expertise.

Best wishes,

Silviu
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Am Fr, 22.04.2016, 00:13, schrieb Doron Zeilberger:
> Dear Silviu,
>
>    Let a(n) be the coefficient of q^n in the power series expansion of
>
> 1/((1-q)(1-q^2)*(1-q^3)...   )^2
>
> 1, 2, 5, 10, 20, 36, 65, 110, 185, 300, 481, 752, 1165, ...
>
> https://oeis.org/A000712
>
> Then my class today noticed, empiricaly that
>
> a(5*n+2), a(5*n+3), and a(5*n+4)
>
> are all divisible by 5.
>
> Here  are my questions
>
>  1. Is it well-known?
>
>  2. Is it easy to prove by elementary human reasonings?
>
>  3. Can your amazing program prove this?
>
>      Best wishes,
>
>            Doron
>


Quick Reply
-----------------

Dear Silviu,

  Thanks! Very impressive.

My students and I found quite a few other examples of this type.

I have another question. Of course the explicit identity that your program found
for the generating function is much better, but as far as just proving
the congruence, e.g. a(5*n+2) mod 5 =0,
I understand that your algorithm supplies an a priori integer N_0, such that
if it is true for n<=N_0 it is always true.

Is there a quick way to find this (not necessarily sharp) upper bound,
for any given case of  subsequeces  of the  a(n*p+j) mod p, for
specific 1/((q;q)_infinity)^r?

  Best wishes,

           Doron
------------


Dear Professor Zeilberger,

For exacly your problem there is indeed a quick way.

In the paper below, Theorem 1 gives you exactly that bound and note that
c_h(n) appearing in theorem is defined on page 1 and is exactly the
function you are after.

http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/013.pdf

Best wishes,

Silviu
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