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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A152157 #28 Jun 21 2021 12:52:37
%S A152157 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,
%T A152157 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,
%U A152157 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
%N A152157 Triangle read by rows: T(n,k) (n>=0, 0<=k<=n) = number of partitions of 2n+1 into 2k+1 odd parts.
%C A152157 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
%H A152157 Alois P. Heinz, <a href="/A152157/b152157.txt">Rows n = 0..200, flattened</a>
%F A152157 T(n,k) = A152140(2n+1,2k+1).
%F A152157 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
%F A152157 O.g.f. for column k: (x^k)/product(1-x^j,j=1..(2*k+1)), k>=0.
%F A152157   From the o.g.f.s of A008284. - _Wolfdieter Lang_, Jul 10 2012
%e A152157 Triangle begins:
%e A152157 1
%e A152157 1  1
%e A152157 1  1   1
%e A152157 1  2   1   1
%e A152157 1  3   2   1   1
%e A152157 1  4   3   2   1   1
%e A152157 1  5   5   3   2   1   1
%e A152157 1  7   7   5   3   2   1  1
%e A152157 1  8  10   7   5   3   2  1  1
%e A152157 1 10  13  11   7   5   3  2  1  1
%e A152157 1 12  18  15  11   7   5  3  2  1  1
%e A152157 1 14  23  21  15  11   7  5  3  2  1  1
%e A152157 1 16  30  28  22  15  11  7  5  3  2  1  1
%e A152157 1 19  37  38  30  22  15 11  7  5  3  2  1  1
%e A152157 1 21  47  49  41  30  22 15 11  7  5  3  2  1 1
%e A152157 1 24  57  65  54  42  30 22 15 11  7  5  3  2 1 1
%e A152157 1 27  70  82  73  56  42 30 22 15 11  7  5  3 2 1  1
%e A152157 1 30  84 105  94  76  56 42 30 22 15 11  7  5 3 2  1 1
%e A152157 1 33 101 131 123  99  77 56 42 30 22 15 11  7 5 3  2 1 1
%e A152157 1 37 119 164 157 131 101 77 56 42 30 22 15 11 7 5  3 2 1 1
%e A152157 From _Wolfdieter Lang_, Jul 09 2012 (Start)
%e A152157 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].
%e A152157 T(5,1) = 4 from the four partitions of 7 = 5+1+1 into 3 parts:
%e A152157 [1,1,5], [1,2,4], [1,3,3] and [2,2,3].
%e A152157 (End)
%p A152157 b:= proc(n, i) option remember; `if`(n=0, 1/sqrt(x), `if`(i<1, 0,
%p A152157       b(n, i-2)+`if`(i>n, 0, expand(sqrt(x)*b(n-i, i)))))
%p A152157     end:
%p A152157 T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(2*n+1, 2*n+1)):
%p A152157 seq(T(n), n=0..12);  # _Alois P. Heinz_, Jun 21 2021
%t A152157 (* 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 A152157 T[n_, k_] := p[n + k + 1, 2 k + 1];
%t A152157 Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 28 2019, after _Wolfdieter Lang_ *)
%Y A152157 Cf. A078408 (row sums), A107379, A152140, A152146, A008284.
%K A152157 nonn,tabl
%O A152157 0,8
%A A152157 _R. J. Mathar_, Sep 25 2009
%E A152157 Indices corrected by _R. J. Mathar_, Jul 09 2012