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%I #97 Oct 06 2018 05:39:34
%S 0,0,0,0,1,1,6,7,18,24,49,66,116,158,255,346,525,707,1030,1374,1936,
%T 2560,3519,4608,6207,8056,10673,13735,17942,22906,29569,37469,47864,
%U 60235,76249,95335,119705,148770,185447,229182,283810,348903,429498,525411,643244
%N a(n) = Sum_{p in P} binomial(H(2,p),2), where P is the set of partitions of n, and H(2,p) = number of hooks of size 2 in p.
%C This sequence is part of the contribution to the quadratic b^2 term of a 2-truncation of the Han/Nekrasov-Okounkov hooklength formula (2-truncation here being the limiting of hook sizes counted by the formula to only those of size 1 or 2). Exploring this sequence may lead to more general formulas regarding the hooklength formula for larger hooks, or the entire contribution to the quadratic term of the formula.
%H Vaclav Kotesovec, <a href="/A301313/b301313.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1900 from Alois P. Heinz)
%H Guo-Niu Han, <a href="https://arxiv.org/abs/0805.1398">The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications</a>, arXiv:0805.1398 [math.CO], 2008.
%H Guo-Niu Han, <a href="https://doi.org/10.5802/aif.2515">The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications</a>, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
%F G.f.: (q^4+3*q^6)/((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j). - _Emily Anible_, May 18 2018
%F a(n) ~ sqrt(3) * exp(Pi*sqrt((2*n)/3)) / (4*Pi^2). - _Vaclav Kotesovec_, Oct 06 2018
%e For n=6, we sum over the partitions of 6. For each partition, we calculate binomial(number of hooks of size 2 in partition, 2):
%e 6............binomial(1,2) = 0
%e 5,1..........binomial(1,2) = 0
%e 4,2..........binomial(2,2) = 1
%e 4,1,1........binomial(2,2) = 1
%e 3,3..........binomial(2,2) = 1
%e 3,2,1........binomial(0,2) = 0
%e 3,1,1,1......binomial(2,2) = 1
%e 2,2,2........binomial(2,2) = 1
%e 2,2,1,1......binomial(2,2) = 1
%e 2,1,1,1,1....binomial(1,2) = 0
%e 1,1,1,1,1,1..binomial(1,2) = 0
%e ------------------------------
%e Total........................6
%p b:= proc(n, i, p, l) option remember; `if`(n=0, p*(p-1)/2,
%p `if`(i>n, 0, b(n, i+1, p, 1)+add(b(n-i*j, i+1, p+
%p `if`(j>1, 1, 0)+l, 0), j=1..n/i)))
%p end:
%p a:= n-> b(n, 1, 0$2):
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Apr 05 2018
%t b[n_, i_, p_, l_] := b[n, i, p, l] = If[n == 0, p*(p-1)/2, If[i > n, 0, b[n, i+1, p, 1] + Sum[b[n-i*j, i+1, p+If[j>1, 1, 0]+l, 0], {j, 1, n/i}]] ];
%t a[n_] := b[n, 1, 0, 0];
%t Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, Apr 28 2018, after _Alois P. Heinz_ *)
%t Table[Sum[(2*k - 5 - (-1)^(k/2))*(1 + (-1)^k)/4 * PartitionsP[n-k], {k, 1, n}], {n, 0, 60}] (* _Vaclav Kotesovec_, Oct 06 2018 *)
%Y Cf. A024786, A138785.
%K nonn
%O 0,7
%A _Emily Anible_, Apr 03 2018
%E a(10)-a(44) from _Alois P. Heinz_, Apr 03 2018
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