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%I #15 Nov 30 2013 21:30:36
%S 1,1,0,2,1,0,3,0,0,0,5,2,1,0,0,7,1,0,0,0,0,11,4,1,1,0,0,0,15,3,2,0,0,
%T 0,0,0,22,8,2,1,1,0,0,0,0,30,7,3,1,0,0,0,0,0,0,42,15,6,3,1,1,0,0,0,0,
%U 0,56,15,6,2,1,0,0,0,0,0,0,0,77,27,10
%N Square array read by antidiagonals: T(n,k) = number of parts of size k in the last section of the set of partitions of n.
%C It appears that in the column k, starting in row n, the sum of k successive terms is equal to A000041(n-1).
%F It appears that A000041(n) = Sum_{j=1..k) T(n+j,k), n >= 0, k >= 1.
%e Array begins:
%e . 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...
%e . 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...
%e . 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,...
%e . 3, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,...
%e . 5, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0,...
%e . 7, 4, 2, 1, 0, 1, 0, 0, 0, 0, 0, 0,...
%e . 11, 3, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0,...
%e . 15, 8, 3, 3, 1, 1, 0, 1, 0, 0, 0, 0,...
%e . 22, 7, 6, 2, 2, 1, 1, 0, 1, 0, 0, 0,...
%e . 30, 15, 6, 5, 3, 2, 1, 1, 0, 1, 0, 0,...
%e . 42, 15, 10, 5, 4, 2, 2, 1, 1, 0, 1, 0,...
%e . 56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1,...
%e ...
%e For n = 7, from the conjecture we have that p(n-1) = p(6) = 11 = 3+8 = 2+3+6 = 1+3+2+5 = 1+1+2+3+4 = 0+1+1+2+2+5, etc. where p(n) = A000041(n).
%Y Columns 1-4: A000041, A182712, A182713, A182714. Main triangle: A182703.
%Y Cf. A066633, A135010, A138121, A138137.
%K nonn,tabl
%O 1,4
%A _Omar E. Pol_, Feb 04 2012
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