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%I #13 May 14 2017 12:21:11
%S 1,1,2,1,2,3,1,2,4,5,1,2,4,6,7,1,2,5,8,10,11,1,2,5,9,12,14,15,1,2,6,
%T 11,16,19,21,22,1,2,6,13,19,24,27,29,30,1,2,7,15,24,31,36,39,41,42,1,
%U 2,7,17,28,38,45,50,53,55,56,1,2,8,20,35,48,59,66,71,74,76,77
%N Number triangle with entry T(n,m) giving the position of the first partition of n with m parts in the Abramowitz-Stegun (A-St) partition ordering.
%C For the Abramowitz-Stegun ordering of partitions see A036036 for the reference and a C. F. Hindenburg link.
%C The present triangle is the partial sum triangle of the triangle t(n,k) = 0 if 0 <= n -1 < k , t(n,0) = 1, n >= 1 and t(n,k) = A008284(n,k) if n-1 >= k >= 1. This triangle with offset [1,0] for [n,k] is 1; 1,1; 1,1,1; 1,1,2,1; 1,1,2,2,1; 1,1,3,3,2,1;... (erase the diagonal of A008284 and add instead a column k=0 with only 1's). See the example section.
%F T(n,m) = sum(p(n,k),k=0..m-1) if n >= m >= 1, otherwise 0, with p(n,0) :=1 and p(n,k) = A008284(n,k) for k=1,2,...,n-1.
%e T(n,m) starts with:
%e n\m 1 2 3 4 5 6 7 8 9 10 11 12...
%e 1 1
%e 2 1 2
%e 3 1 2 3
%e 4 1 2 4 5
%e 5 1 2 4 6 7
%e 6 1 2 5 8 10 11
%e 7 1 2 5 9 12 14 15
%e 8 1 2 6 11 16 19 21 22
%e 9 1 2 6 13 19 24 27 29 30
%e 10 1 2 7 15 24 31 36 39 41 42
%e 11 1 2 7 17 28 38 45 50 53 55 56
%e 12 1 2 8 20 35 48 59 66 71 74 76 77
%e ...
%e T(6,4) = 8 because the 11=T(6,6) partitions for n=6 are, in A-St order: [6]; [1,5],[2,4],[3,3]; [1^2,4],[1,2,3],[2^3]; [1^3,3],[1^2,2^2]; [1^4,2]; [1^6] and the first partition with 4 parts, appears at position 8.
%e This triangle is obtained as partial sum triangle from the triangle t(n,k) (see the comment section) which starts with:
%e n\m 0 1 2 3 4 5 6 7 8 9 10 11 ...
%e 1 1
%e 2 1 1
%e 3 1 1 1
%e 4 1 1 2 1
%e 5 1 1 2 2 1
%e 6 1 1 3 3 2 1
%e 7 1 1 3 4 3 2 1
%e 8 1 1 4 5 5 3 2 1
%e 9 1 1 4 7 6 5 3 2 1
%e 10 1 1 5 8 9 7 5 3 2 1
%e 11 1 1 5 10 11 10 7 5 3 2 1
%e 12 1 1 6 12 15 13 11 7 5 3 2 1
%e ...
%Y Cf. A008284.
%K nonn,tabl
%O 1,3
%A _Wolfdieter Lang_, Jul 24 2012
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