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%I #30 Jan 07 2024 14:14:45
%S 22,7,15,6,6,10,2,5,5,10,2,3,4,5,8,1,2,2,5,4,8,1,1,2,2,4,5,7,0,1,1,2,
%T 2,4,4,8,1,0,1,1,2,2,4,4,7,0,1,0,1,1,2,2,4,4,7,0,0,1,0,1,1,2,2,4,4,7,
%U 0,0,0,1,0,1,1,2,2,4,4,7,0,0,0,0,1,0,1,1,2,2,4,4,7
%N Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (8 + m).
%C Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 8. For further information see A182703 and A135010.
%H Andrew Howroyd, <a href="/A194708/b194708.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%F T(k,m) = A182703(8+m,k), with T(k,m) = 0 if k > 8+m.
%F T(k,m) = A194812(8+m,k).
%e Triangle begins:
%e 22,
%e 7, 15,
%e 6, 6, 10,
%e 2, 5, 5, 10,
%e 2, 3, 4, 5, 8,
%e ...
%e For k = 1 and m = 1: T(1,1) = 22 because there are 22 parts of size 1 in the last section of the set of partitions of 9, since 8 + m = 9, so a(1) = 22.
%e For k = 2 and m = 1: T(2,1) = 7 because there are seven parts of size 2 in the last section of the set of partitions of 9, since 8 + m = 9, so a(2) = 7.
%o (PARI) P(n)={my(M=matrix(n,n), d=8); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
%o { my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ _Andrew Howroyd_, Feb 19 2020
%Y Always the sum of row k = p(8) = A000041(8) = 22.
%Y The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194707, this sequence, A194709, A194710.
%Y Cf. A135010, A138121, A194812.
%K nonn,tabl
%O 1,1
%A _Omar E. Pol_, Feb 05 2012
%E Terms a(11) and beyond from _Andrew Howroyd_, Feb 19 2020