The following is an addition to A function from partitions to natural numbers - is it familiar?; the function $f(\lambda)$ as defined there, when summed over all partitions of n, gives an unexpected hit in http://oeis.org/A232434. Checked up to n=36.

In short:
Define $g(x, q)$ by $$\frac{\partial g(x,q)}{\partial x}=g(x,q)^2 * g(q x,q),$$ with series $$g(x,q)=\sum _{k=0}^{\infty } \frac{ c_k(q) x^k}{k!},$$ define $t_{n}=n(n+1)/2$; then the coefficient of $q^{t_{n-1}}$ in $c_{n+1}(q)$ is $\sum{_{\lambda \in n}}{f(\lambda)}.$

f(n=2)=6 and $c_{3}(q) =q^3+2 q^2+6 q+6$ ; $q^{t_{1}} = q^1$ with coeff. =6.
Likewise f(n=3)=14 and $c_{4}(q) =q^6+2 q^5+6 q^4+14 q^3+22 q^2+36 q+24$ ;
$q^{t_{2}} = q^3$ with coeff. =14;

Remark that the differential equation has no closed form solution, but its series can be generated upto $x^n$ for any given n.

Question: are there any arguments to trust\distrust this conjecture? Can it be proved ?

correction 28/05/2014 22:02 CET: the differntial equation has a product in the RHS, not a sum. Thanks to Pietro for spotting it.

  • 1
    $\begingroup$ If you are going to post here regularly, it would be nice if you looked into the ways to format mathematics on this website. $\endgroup$ Apr 13, 2014 at 1:27
  • $\begingroup$ @GerryMyerson: should look better now? $\endgroup$
    – Wouter M.
    Apr 13, 2014 at 15:54
  • $\begingroup$ @NateEldredge: extra dollar signs left as an excercise to the poster? $\endgroup$
    – Wouter M.
    Apr 13, 2014 at 22:02
  • $\begingroup$ That was something I missed when I was editing it. It was my fault and not the OP's fault. Sorry. $\endgroup$ Apr 13, 2014 at 23:10
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    $\begingroup$ A typo? Comparing with the OEIS link, I guess the differential equation for $g$ has a product, not a sum, in the RHS. $\endgroup$ May 28, 2014 at 14:10


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