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