Motivated by [Parity of number of partitions of $n!/6$ and $n!/2$](https://mathoverflow.net/q/323308), 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?



Number of integer partitions modulo 3
The function $\,h(x) := \prod\_{n=1}^\infty (1 - x^n)\,$ is known as a Ramanujan theta function. It is essentially the Dedekind $\eta$ function. The connection is $\,f(q) := q h(q^{24}) = \eta(24\tau)\,$ where $\,q=\exp(2\pi i \tau).\,$ Differentiating we get
$\, dq = (2\pi i)\, q\, d\tau,\,$ and
$$ \frac{d}{d\tau} f(q) = (2\pi i)\, q\, f'(q).$$
Since all powers of $\,q\,$ in $\,f(q)\,$ are of the form
 $\,q^{24n+1}\,$ then the congruence 
 $\,f'(q) - f(q) \equiv 0 \pmod 3\,$ holds.


If we use the original $\,h(x)\,$ then there is similar but more complicated result using modulo $3$ exponents. Let$\, A = A(x) := h(x)\,$ and let $\,A\_0\,$ be all the terms of $\,A\,$ where the exponent of $\,x\,$ is $0$ modulo $3$ and similarly for $\,A\_1\,$ and $\,A\_2\,$ so that $\,A = A\_0+A\_1+A\_2.\,$ This is the *trisection* of the power series $\,A.\,$ Refer to my essay [A Multisection of q-Series](http://grail.eecs.csuohio.edu/~somos/multiq.html) for the nice identity
 $$ 0 = A\_2 A\_0^2 + A\_0 A\_1^2 + A\_1 A\_2^2. \tag{1}$$
Now let $\, B = B(x) := A'(x)\,$ and let $\,B\_0,B\_1,B\_2\,$ be the trisection of $\,B.\,$
It is easy to show that 
$$ q\,B\_0 \equiv A\_1,\quad q\,B\_1 \equiv -A\_2,
\quad B\_2 \equiv 0 \pmod 3.$$
Now let $\, C = C(q) := A''(x)\,$ and let
$\,C\_0,C\_1,C\_2\,$ be the trisection of $\,C.\,$
It is easy to show that
$$ q^2\,C\_0 \equiv -A\_2, \quad C\_1 \equiv C\_2 \equiv 0 \pmod 3.$$
When we make these substitutions in the expression
$$ h''(x) h(x)^2 + x\, h'(x)^3 + 2\,h(x)h'(x)^2 \tag{2}$$
and reduce modulo $3$ we get the nice identity $(1)$.


