6
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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?

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  • $\begingroup$ I should add: for the parity of the partition function, no such differential equation was found - at least not using the first few hundred terms. $\endgroup$ Feb 15, 2019 at 22:20
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    $\begingroup$ Would you please rewrite the same equation for $1/f=\sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k+1)/2}$? It looks on the first glance that the key property of 3 is that the derivative of $1/f$ has simplified (compared to other localisations) formula: $(2/f)'=\sum_{k=-\infty}^{\infty} (-1)^k k x^{k(3k+1)/2}$. $\endgroup$ Feb 15, 2019 at 22:39
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    $\begingroup$ @FedorPetrov, indeed, this is quite remarkable. The equation for $1/f$ is simply $f(x)^2 f^{''}(x) + xf'(x)^3 + 2f(x)f'(x)^2=0$. $\endgroup$ Feb 15, 2019 at 22:56
  • $\begingroup$ I am sorry, but I can't verify your ADE. If $\,f(x) = a_0 + a_1\,x + O(x^2)\,$ then your differential operator produces $\,(2a_0^2+2a_1)+O(x).\,$ Your power series has $\,a_0=1, a_1=2\,$ which gives $\,6+O(x)\,$ by my calculations instead of $0$. $\endgroup$
    – Somos
    Feb 16, 2019 at 2:51
  • 1
    $\begingroup$ @Somos, I did not verify by hand, but your computation looks OK, since $6 + O(x) = O(x)\mod 3$. $\endgroup$ Feb 16, 2019 at 8:05

2 Answers 2

2
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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 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 $(1)$.

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6
  • $\begingroup$ Very interesting! Do other relations in your essay also produce differential equations modulo other primes? $\endgroup$ Feb 16, 2019 at 17:35
  • $\begingroup$ @FedorPetrov Good question!! I had not considered differential equation power series modulo primes before which is why the original question confused me. However, based on my experience with other kinds of congruences between power series I would not be surprised. For example, $ \theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)$ and $\theta_2^4(q) \equiv 0 \pmod {16}$ $\endgroup$
    – Somos
    Feb 16, 2019 at 17:52
  • $\begingroup$ as I understood from OP, there are difficulties already with differential equation modulo 2. On the other hand, it may appear, say, that there exists a non-trivial relation modulo 4 which is trivial modulo 2. $\endgroup$ Feb 16, 2019 at 18:06
  • $\begingroup$ @FedorPetrov I am not surprised at the modulo 2 problem. It has to do with modular functions I think. You may be correct about modulo 4 though. I am not sure. $\endgroup$
    – Somos
    Feb 16, 2019 at 18:11
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    $\begingroup$ @FedorPetrov Okay, I got some modulo 8 results. Let $f:=f(x), f_1:=f'(x), f_2:=f''(x),f_3:=f'''(x).$ Then $6 f f_1^3 +3f^2 f_1 f_2 + 3 f^3 f_3 + 2x f_1^4 + x f^2 f_2^2 \equiv 0 \pmod 8$ based on a 4th degree 4-section identity. $\endgroup$
    – Somos
    Feb 16, 2019 at 19:12
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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.

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1
  • $\begingroup$ Very very nice! Although I must say that it doesn't answer the question :-) $\endgroup$ Feb 16, 2019 at 11:44

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