A302348 - OEIS

%I #25 May 20 2018 18:13:48

%S 0,0,1,1,4,5,14,18,37,50,90,122,199,270,415,559,820,1096,1556,2060,

%T 2847,3736,5057,6576,8747,11279,14788,18916,24493,31097,39838,50225,

%U 63737,79833,100471,125076,156237,193394,239956,295443,364334,446349,547360,667440

%N a(n) = Sum_{p in P} (H(2,p)^2)/2, where P is the set of partitions of n, and H(2,p) is the number of hooks of length 2 in p.

%C This sequence is part of the contribution to the b^2 term of the Han/Nekrasov-Okounkov hooklength formula truncated at hooks of size two.

%C It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.

%H Alois P. Heinz, <a href="/A302348/b302348.txt">Table of n, a(n) for n = 0..1000</a>

%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.

%H W. J. Keith, <a href="https://doi.org/10.1007/s11139-015-9704-x">Restricted k-color partitions</a>, Ramanujan Journal (2016) 40: 71.

%F G.f: (q^2*(1+q^2+2*q^4))/((1-q^2)*(1-q^4)*Product_{i>0} (1-q^i)).

%e For a(6), we sum over partitions of six. For each partition, we count 1 for each hook of length 2, then square the total in each partition. We divide the final result in half to get a(6).

%e 6............1^2 = 1

%e 5,1..........1^2 = 1

%e 4,2..........2^2 = 4

%e 4,1,1........2^2 = 4

%e 3,3..........2^2 = 4

%e 3,2,1........0^2 = 0

%e 3,1,1,1......2^2 = 4

%e 2,2,2........2^2 = 4

%e 2,2,1,1......2^2 = 4

%e 2,1,1,1,1....1^2 = 1

%e 1,1,1,1,1,1..1^2 = 1

%e --------------------

%e Total.............28/2=14

%p b:= proc(n, i, p, l) option remember; `if`(n=0, p^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)/2:

%p seq(a(n), n=0..50);  # _Alois P. Heinz_, Apr 06 2018

%t b[n_, i_, p_, l_] := b[n, i, p, l] = If[n == 0, p^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]/2;

%t Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, May 18 2018, after _Alois P. Heinz_ *)

%Y Cf. A024786, A302347.

%K nonn

%O 0,5

%A _Emily Anible_, Apr 05 2018