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A194706 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (6 + m).

%I A194706 #33 Jan 07 2024 14:14:16
%S A194706 11,3,8,2,3,6,1,3,2,5,1,1,2,3,4,0,1,1,2,2,5,1,0,1,1,2,2,4,0,1,0,1,1,2,
%T A194706 2,4,0,0,1,0,1,1,2,2,4,0,0,0,1,0,1,1,2,2,4,0,0,0,0,1,0,1,1,2,2,4,0,0,
%U A194706 0,0,0,1,0,1,1,2,2,4,0,0,0,0,0,0,1,0,1,1,2,2,4
%N A194706 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (6 + m).
%C A194706 Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 6. For further information see A182703 and A135010.
%H A194706 Andrew Howroyd, <a href="/A194706/b194706.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%F A194706 T(k,m) = A182703(6+m,k), with T(k,m) = 0 if k > 6+m.
%F A194706 T(k,m) = A194812(6+m,k).
%e A194706 Triangle begins:
%e A194706   11,
%e A194706    3, 8,
%e A194706    2, 3, 6,
%e A194706    1, 3, 2, 5,
%e A194706    1, 1, 2, 3, 4,
%e A194706    0, 1, 1, 2, 2, 5,
%e A194706   ...
%e A194706 For k = 1 and m = 1: T(1,1) = 11 because there are 11 parts of size 1 in the last section of the set of partitions of 7, since 6 + m = 7, so a(1) = 11.
%e A194706 For k = 2 and m = 1: T(2,1) = 3 because there are three parts of size 2 in the last section of the set of partitions of 7, since 6 + m = 7, so a(2) = 3.
%o A194706 (PARI) P(n)={my(M=matrix(n,n), d=6); 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 A194706 { my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ _Andrew Howroyd_, Feb 19 2020
%Y A194706 Always the sum of row k = p(6) = A000041(6) = 11.
%Y A194706 The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194705, this sequence, A194707-A194710.
%Y A194706 Cf. A135010, A138121, A194812.
%K A194706 nonn,tabl
%O A194706 1,1
%A A194706 _Omar E. Pol_, Feb 05 2012
%E A194706 Terms a(22) and beyond from _Andrew Howroyd_, Feb 19 2020