A232434 Limit of rows in triangle A232433 when read in reverse order.

1, 2, 6, 14, 32, 68, 142, 276, 542, 1022, 1876, 3394, 6066, 10628, 18412, 31344, 52868, 88370, 146180, 239310, 388370, 624688, 997586, 1582640, 2493908, 3902574, 6069194, 9378078, 14411150, 22034860, 33520642, 50747992, 76471200, 114689926, 171242092, 254587046, 376981800, 556129468, 817412048, 1197096472, 1747047580

Offset: 1

Author

Paul D. Hanna, Nov 23 2013

Keywords

• nonn

Comments

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

Examples

			The triangle A232433 of coefficients of x^n*q^k, n >= 0, k = 0..n*(n-1)/2, begins:
[1];
[1];
[2, 1];
[6, 6, 2, 1];
[24, 36, 22, 14, 6, 2, 1];
[120, 240, 210, 160, 104, 56, 32, 14, 6, 2, 1];
[720, 1800, 2040, 1830, 1448, 992, 674, 408, 232, 128, 68, 32, 14, 6, 2, 1]; ...
where this sequence is the limit of the rows read in reverse order.
		

Links

• Matt Fayers, A function from partitions to natural numbers - is it familiar?, MathOverflow 30 may 2013. [From Wouter Meeussen, Apr 07 2014]

Crossrefs

Cf. A232433.

Programs

Mathematica

Clear[c]; c[0] = 1; Table[f = Sum[c[k] x^k/k!, {k, 0, n}];
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}]
(* or via combinatorics: *)
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]]];
Table[Total[f/@IntegerPartitions[n]],{n,0,24}] (* Wouter Meeussen, Apr 07 2014 *)

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)}
for(n=1,20,print1(a(n),", "))

Formula

E.g.f. of triangle A232433 satisfies: G(x,q) = exp(Integral G(x,q)*G(q*x,q) dx).