A152157 Triangle read by rows: T(n,k) (n>=0, 0<=k<=n) = number of partitions of 2n+1 into 2k+1 odd parts.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 2, 1, 1, 1, 5, 5, 3, 2, 1, 1, 1, 7, 7, 5, 3, 2, 1, 1, 1, 8, 10, 7, 5, 3, 2, 1, 1, 1, 10, 13, 11, 7, 5, 3, 2, 1, 1, 1, 12, 18, 15, 11, 7, 5, 3, 2, 1, 1, 1, 14, 23, 21, 15, 11, 7, 5, 3, 2, 1, 1, 1, 16, 30, 28, 22, 15, 11, 7, 5, 3, 2, 1, 1

Offset: 0

Author

R. J. Mathar, Sep 25 2009

Keywords

• nonn
• tabl

Comments

This number triangle satisfies T(n,k) = A008284(n+k+1,2*k+1), n,k >= 0. This means that T(n,k) is also the number of partitions of N:=n+k+1 into M:=2*k+1 parts. For the proof add 1 to every odd part of each partition of n':=2*n+1 into m:=2*k+1 parts which are all odd, and divide each part by a factor of 2, thus obtaining a partition of n+k+1 into m=2*k+1 parts. All partitions of N,for N>=1, into an odd number of parts M (M from {1,...,N}) are reached: just take k=(M-1)/2 and n=N-1-k. Each partition of N into an odd number of parts can only arise once from the given recipe (for given N and M the k and n values are unique). See also a comment by Franklin T. Adams-Watters on A152140. - Wolfdieter Lang, Jul 09 2012

Examples

			Triangle begins:
1
1  1
1  1   1
1  2   1   1
1  3   2   1   1
1  4   3   2   1   1
1  5   5   3   2   1   1
1  7   7   5   3   2   1  1
1  8  10   7   5   3   2  1  1
1 10  13  11   7   5   3  2  1  1
1 12  18  15  11   7   5  3  2  1  1
1 14  23  21  15  11   7  5  3  2  1  1
1 16  30  28  22  15  11  7  5  3  2  1  1
1 19  37  38  30  22  15 11  7  5  3  2  1  1
1 21  47  49  41  30  22 15 11  7  5  3  2  1 1
1 24  57  65  54  42  30 22 15 11  7  5  3  2 1 1
1 27  70  82  73  56  42 30 22 15 11  7  5  3 2 1  1
1 30  84 105  94  76  56 42 30 22 15 11  7  5 3 2  1 1
1 33 101 131 123  99  77 56 42 30 22 15 11  7 5 3  2 1 1
1 37 119 164 157 131 101 77 56 42 30 22 15 11 7 5  3 2 1 1
From _Wolfdieter Lang_, Jul 09 2012 (Start)
T(5,1) = 4 from the four partitions of 11 into 3 parts, all of which are odd: [1,1,9], [1,3,7], [1,5,5] and [3,3,5].
T(5,1) = 4 from the four partitions of 7 = 5+1+1 into 3 parts:
[1,1,5], [1,2,4], [1,3,3] and [2,2,3].
(End)
		

Links

• Alois P. Heinz, Rows n = 0..200, flattened

Crossrefs

Cf. A078408 (row sums), A107379, A152140, A152146, A008284.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0, 1/sqrt(x), `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..degree(p)))(b(2*n+1, 2*n+1)):
seq(T(n), n=0..12);  # Alois P. Heinz, Jun 21 2021

Mathematica

(* p = A008284 *) p[n_, 1] = 1; p[n_, k_] := p[n, k] = If[n >= k, Sum[p[n - i, k - 1], {i, 1, n - 1}] - Sum[p[n - i, k], {i, 1, k - 1}], 0];
T[n_, k_] := p[n + k + 1, 2 k + 1];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 28 2019, after Wolfdieter Lang *)

Formula

T(n,k) = A152140(2n+1,2k+1).

T(n,k) = p(n+k+1,2*k+1), n >= 0, k >= 0, with p(N,M)= A008284(N,M), the number of partitions of N into M parts. See the sketch of the proof given above as a comment. - Wolfdieter Lang, Jul 09 2012

O.g.f. for column k: (x^k)/product(1-x^j,j=1..(2*k+1)), k>=0.

From the o.g.f.s of A008284. - Wolfdieter Lang, Jul 10 2012

Extensions

Indices corrected by R. J. Mathar, Jul 09 2012