A152146 Triangle read by rows: T(n,k) (n >= 0, 0 <= k <= n) = number of partitions of 2n into 2k odd parts.
1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 2, 2, 1, 1, 0, 3, 3, 2, 1, 1, 0, 3, 5, 3, 2, 1, 1, 0, 4, 6, 5, 3, 2, 1, 1, 0, 4, 9, 7, 5, 3, 2, 1, 1, 0, 5, 11, 11, 7, 5, 3, 2, 1, 1, 0, 5, 15, 14, 11, 7, 5, 3, 2, 1, 1, 0, 6, 18, 20, 15, 11, 7, 5, 3, 2, 1, 1, 0, 6, 23, 26, 22, 15, 11, 7, 5, 3, 2, 1, 1
Offset: 0
Examples
Triangle begins: 1 0 1 0 1 1 0 2 1 1 0 2 2 1 1 0 3 3 2 1 1 0 3 5 3 2 1 1 0 4 6 5 3 2 1 1 0 4 9 7 5 3 2 1 1 0 5 11 11 7 5 3 2 1 1 0 5 15 14 11 7 5 3 2 1 1 0 6 18 20 15 11 7 5 3 2 1 1 0 6 23 26 22 15 11 7 5 3 2 1 1 0 7 27 35 29 22 15 11 7 5 3 2 1 1 0 7 34 44 40 30 22 15 11 7 5 3 2 1 1 0 8 39 58 52 42 30 22 15 11 7 5 3 2 1 1 0 8 47 71 70 55 42 30 22 15 11 7 5 3 2 1 1 0 9 54 90 89 75 56 42 30 22 15 11 7 5 3 2 1 1 0 9 64 110 116 97 77 56 42 30 22 15 11 7 5 3 2 1 1 0 10 72 136 146 128 100 77 56 42 30 22 15 11 7 5 3 2 1 1 From _Gus Wiseman_, Jun 20 2021: (Start) For example, row n = 6 counts the following partitions (B = 11): (75) (3333) (333111) (33111111) (3111111111) (111111111111) (93) (5331) (531111) (51111111) (B1) (5511) (711111) (7311) (9111) The corresponding strict partitions are: (7,5) (8,4) (9,3) (10,2) (11,1) (12) (6,5,1) (5,4,3) (7,3,2) (9,2,1) (5,4,2,1) (6,4,2) (8,3,1) (7,4,1) (6,3,2,1) The corresponding normal partitions are: 43221 33321 3321111 321111111 21111111111 111111111111 322221 332211 32211111 2211111111 2222211 432111 222111111 3222111 22221111 (End)
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Crossrefs
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, b(n, i-2)+`if`(i>n, 0, expand(sqrt(x)*b(n-i, i))))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(2*n, 2*n-1)): seq(T(n), n=0..12); # Alois P. Heinz, Jun 21 2021
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Mathematica
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}]; Table[Length[Select[IntegerPartitions[n],UnsameQ@@#&&ats[#]==k&]],{n,0,30,2},{k,0,n,2}] (* Gus Wiseman, Jun 20 2021 *)
Formula
T(n,k) = A152140(2n,2k).
Comments