%I #24 Jul 20 2019 21:03:28
%S 1,2,6,14,32,68,142,276,542,1022,1876,3394,6066,10628,18412,31344,
%T 52868,88370,146180,239310,388370,624688,997586,1582640,2493908,
%U 3902574,6069194,9378078,14411150,22034860,33520642,50747992,76471200,114689926,171242092,254587046,376981800,556129468,817412048,1197096472,1747047580
%N Limit of rows in triangle A232433 when read in reverse order.
%C Conjecture: a(n) equals sum of f(lambda) over all partitions of n, where f is defined recursively as f({})=1; f(lambda)=binomial(i+j,j) f(mu)f(nu); with i and j the row and column of the box in the Young-Ferrers diagram of lambda such that i+j is maximized, and mu is lambda with the first i rows removed, and nu is lambda with the first j columns removed. See Math Overflow link. - _Wouter Meeussen_, Apr 07 2014
%H Matt Fayers, <a href="http://mathoverflow.net/questions/132338">A function from partitions to natural numbers - is it familiar?</a>, MathOverflow 30 may 2013. [From Wouter Meeussen, Apr 07 2014]
%F E.g.f. of triangle A232433 satisfies: G(x,q) = exp(Integral G(x,q)*G(q*x,q) dx).
%e The triangle A232433 of coefficients of x^n*q^k, n >= 0, k = 0..n*(n-1)/2, begins:
%e [1];
%e [1];
%e [2, 1];
%e [6, 6, 2, 1];
%e [24, 36, 22, 14, 6, 2, 1];
%e [120, 240, 210, 160, 104, 56, 32, 14, 6, 2, 1];
%e [720, 1800, 2040, 1830, 1448, 992, 674, 408, 232, 128, 68, 32, 14, 6, 2, 1]; ...
%e where this sequence is the limit of the rows read in reverse order.
%t Clear[c]; c[0] = 1; Table[f = Sum[c[k] x^k/k!, {k, 0, n}];
%t c[n + 1] = n! SeriesCoefficient[f^2 (f /. x -> q x), {x, 0, n}] // Simplify; Coefficient[q*c[n + 1], q^(1 + n*(n - 1)/2)], {n, 0, 64}]
%t (* or via combinatorics: *)
%t Clear[f];f[{}]:=1;f[\[Lambda]_?PartitionQ]:=f[\[Lambda]]=Block[{temp,i,j,\[Mu],\[Nu]},temp=\[Lambda]+Range[Length[\[Lambda]]];{i}=First@Position[temp,Max[temp],1,1];j=\[Lambda][[i]];\[Mu]=Drop[\[Lambda],i];\[Nu]=DeleteCases[\[Lambda]-j,q_/;(q<=0)];Binomial[i+j,j]f[\[Mu]]f[\[Nu]]];
%t Table[Total[f/@IntegerPartitions[n]],{n,0,24}] (* _Wouter Meeussen_, Apr 07 2014 *)
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=exp(intformal(A*subst(A,x,x*y +x*O(x^n)),x)));n!*polcoeff(polcoeff(A,n,x),(n-1)*(n-2)/2,y)}
%o for(n=1,20,print1(a(n),", "))
%Y Cf. A232433.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Nov 23 2013