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%I #30 Nov 30 2013 21:30:35
%S 42,15,27,10,14,18,5,10,10,17,4,5,8,10,15,2,5,4,8,9,14,2,2,4,5,7,9,13,
%T 1,2,2,4,4,8,8,13,1,1,2,2,4,4,7,9,12,0,1,1,2,2,4,4,7,8,13,1,0,1,1,2,2,
%U 4,4,7,8,12,0,1,0,1,1,2,2,4,4,7,8,12
%N Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (10 + m).
%C Sub-triangle of A182703 and also of A194812. Note that the sum of row k is also the number of partitions of 10. For further information see A182703 and A135010.
%F T(k,m) = A182703(10+m,k), with T(k,m) = 0 if k > 10+m.
%F T(k,m) = A194812(10+m,k).
%F Beginning with row k=11 each row starts with (k-11) 0's and ends with the subsequence 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12, the initial terms of A002865. - Alois P. Heinz, Feb 15 2012
%e Triangle begins:
%e 42;
%e 15, 27;
%e 10, 14, 18;
%e 5, 10, 10, 17;
%e 4, 5, 8, 10, 15;
%e 2, 5, 4, 8, 9, 14;
%e 2, 2, 4, 5, 7, 9, 13;
%e 1, 2, 2, 4, 4, 8, 8, 13;
%e 1, 1, 2, 2, 4, 4, 7, 9, 12;
%e 0, 1, 1, 2, 2, 4, 4, 7, 8, 13;
%e 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12;
%e 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12;
%e 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12;
%e 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12;
%e ....
%e For k = 1 and m = 1; T(1,1) = 42 because there are 42 parts of size 1 in the last section of the set of partitions of 11, since 10 + m = 11, so a(1) = 42. For k = 2 and m = 1; T(2,1) = 15 because there are 15 parts of size 2 in the last section of the set of partitions of 11, since 10 + m = 11, so a(2) = 15.
%Y Always the sum of row k = p(10) = A000041(10) = 42.
%Y The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194709, this sequence.
%Y Cf. A002865, A135010, A138121, A194812.
%K nonn,tabl
%O 1,1
%A _Omar E. Pol_, Feb 05 2012
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