A357637 Triangle read by rows where T(n,k) is the number of integer partitions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.
1, 0, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 1, 1, 3, 0, 0, 0, 2, 2, 3, 0, 0, 0, 0, 5, 2, 4, 0, 0, 0, 0, 2, 6, 3, 4, 0, 0, 0, 0, 2, 3, 9, 3, 5, 0, 0, 0, 0, 0, 4, 7, 10, 4, 5, 0, 0, 0, 0, 0, 0, 11, 8, 13, 4, 6, 0, 0, 0, 0, 0, 0, 4, 15, 12, 14, 5, 6, 0, 0, 0, 0, 0, 0, 3, 7, 25, 13, 17, 5, 7
Offset: 0
Examples
Triangle begins: 1 0 1 0 0 2 0 0 1 2 0 0 1 1 3 0 0 0 2 2 3 0 0 0 0 5 2 4 0 0 0 0 2 6 3 4 0 0 0 0 2 3 9 3 5 0 0 0 0 0 4 7 10 4 5 0 0 0 0 0 0 11 8 13 4 6 0 0 0 0 0 0 4 15 12 14 5 6 0 0 0 0 0 0 3 7 25 13 17 5 7 Row n = 9 counts the following partitions: (3222) (333) (432) (441) (9) (22221) (3321) (522) (531) (54) (21111111) (4221) (4311) (621) (63) (111111111) (32211) (5211) (711) (72) (222111) (6111) (81) (2211111) (33111) (3111111) (42111) (51111) (321111) (411111)
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Crossrefs
Programs
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Maple
b:= proc(n, i, s, t) option remember; `if`(n=0, x^s, `if`(i<1, 0, b(n, i-1, s, t)+b(n-i, min(n-i, i), s+`if`(t<2, i, -i), irem(t+1, 4)))) end: T:= n-> (p-> seq(coeff(p, x, i), i=-n..n, 2))(b(n$2, 0$2)): seq(T(n), n=0..15); # Alois P. Heinz, Oct 12 2022
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Mathematica
halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}]; Table[Length[Select[IntegerPartitions[n],halfats[#]==k&]],{n,0,12},{k,-n,n,2}]
Formula
Conjecture: The column sums are A029862.
Comments