<s>[INST] Number of integer partitions modulo 3 Motivated by Parity of number of partitions of $n!/6$ and $n!/2$, I asked my computer (and my FriCAS package for guessing) for an algebraic differential equation for the number of integer partitions modulo 3.  This is what it answered:

(1) -> s := [partition n for n in 1..]

   (1)  [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, ...]
                                                        Type: Stream(Integer)
(32) -> guessADE([s.i for i in 1..400], safety==290, maxDerivative==2)$GUESSF PF 3

   (32)
   [
     [
         n
       [x ]f(x):
               3    2      2          ,,       4 ,   3       3          2  ,   2
             (x f(x)  + 2 x f(x) + x)f  (x) + x f (x)  + (2 x f(x) + 2 x )f (x)

           +
                            ,            2
             (2 x f(x) + 2)f (x) + 2 f(x)

         =
           0
       ,
                          3      4
      f(x) = 1 + 2 x + 2 x  + O(x )]
     ]

Of course, this is only a guess, but it seems fairly well tested.  Only 110 terms were needed to guess the recurrence, all the other 290 were used to check it.
My question is: is this known, and if not so, is this interesting?
 [/INST] Well, $f''f^2 +xf' ^3+2ff'^2 =0$ modulo 3 for $f=\prod(1-x^m)=\sum_{n\in \mathbb{Z} } (-1)^nx^{n(3n+1)/2}$ may be quickly seen as follows.
Differentiating the power series for $f$ and expanding the brackets we see that we should prove that
$$
\sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} 
(-1)^{a+b+c} \left(a(a-2)-abc+2ab\right) 
$$
is divisible by 3. Multiplying the sum by 3 and using cycling shift of variables we reduce it to proving that the sum
$$
\sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} 
(-1)^{a+b+c} \left( a(a-2)+b(b-2)+c(c-2)-3abc+2ab+2bc+2ac\right) 
$$
is divisible by 9.
The summand is $(a+b+c) (a+b+c-2)-3abc$. If $a=b=c$ this is $9a^2 - 3(a^3+2a)$, divisible by 9. Other triples partition by permuting the variables onto 3-tuples and 6-tuples, so the sum of $3abc$ is of course divisible by 9. As for $a+b+c$, it is congruent to $2n$ modulo 3, thus unless $n=3t+2$ the expression $(a+b+c) (a+b+c-2)$ is divisible by 3, as we need. 
It remains to show that for $n=3m+2$ the sum of $(-1)^{a+b+c}$ over our triples is divisible by 9. In other words, the coefficient of $x^{3m+2} $ in $f^3$ must be divisible by 9. We have 
$$
f^3=\prod (1-x^k)^3 =\prod (1-x^{3k}+3(x^k-x^{2k})).
$$
Expanding the brackets and reducing modulo 9 we should prove that the expression
$$
[x^{3m+2} ] 3 f(x^3) \sum_k \frac{x^k-x^{2k} } {1 - x^{3k} }
$$
is divisible by 9.
But in the latter sum $\sum_k (x^k-x^{2k}+x^{4k}-x^{5k}+\dots) $ all coefficients of powers $x^{3s+2} $ do cancel, since if $3s+2=kr$, the guys $x^{kr} $ and $x^{rk} $ go with different signs. 
</s>