A302348 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.
0, 0, 1, 1, 4, 5, 14, 18, 37, 50, 90, 122, 199, 270, 415, 559, 820, 1096, 1556, 2060, 2847, 3736, 5057, 6576, 8747, 11279, 14788, 18916, 24493, 31097, 39838, 50225, 63737, 79833, 100471, 125076, 156237, 193394, 239956, 295443, 364334, 446349, 547360, 667440
Offset: 0
Keywords
Examples
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). 6............1^2 = 1 5,1..........1^2 = 1 4,2..........2^2 = 4 4,1,1........2^2 = 4 3,3..........2^2 = 4 3,2,1........0^2 = 0 3,1,1,1......2^2 = 4 2,2,2........2^2 = 4 2,2,1,1......2^2 = 4 2,1,1,1,1....1^2 = 1 1,1,1,1,1,1..1^2 = 1 -------------------- Total.............28/2=14
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.
- Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
- W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71.
Programs
-
Maple
b:= proc(n, i, p, l) option remember; `if`(n=0, p^2, `if`(i>n, 0, b(n, i+1, p, 1)+add(b(n-i*j, i+1, p+ `if`(j>1, 1, 0)+l, 0), j=1..n/i))) end: a:= n-> b(n, 1, 0$2)/2: seq(a(n), n=0..50); # Alois P. Heinz, Apr 06 2018
-
Mathematica
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}]]]; a[n_] := b[n, 1, 0, 0]/2; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, May 18 2018, after Alois P. Heinz *)
Formula
G.f: (q^2*(1+q^2+2*q^4))/((1-q^2)*(1-q^4)*Product_{i>0} (1-q^i)).
Comments