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A194704 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).

%I A194704 #35 Jan 07 2024 14:13:37
%S A194704 5,1,4,1,2,2,0,1,1,3,1,0,1,1,2,0,1,0,1,1,2,0,0,1,0,1,1,2,0,0,0,1,0,1,
%T A194704 1,2,0,0,0,0,1,0,1,1,2,0,0,0,0,0,1,0,1,1,2,0,0,0,0,0,0,1,0,1,1,2,0,0,
%U A194704 0,0,0,0,0,1,0,1,1,2,0,0,0,0,0,0,0,0,1,0,1,1,2
%N A194704 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).
%C A194704 Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 4. For further information see A182703 and A135010.
%H A194704 Andrew Howroyd, <a href="/A194704/b194704.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%F A194704 T(k,m) = A182703(4+m,k), with T(k,m) = 0 if k > 4+m.
%F A194704 T(k,m) = A194812(4+m,k).
%e A194704 Triangle begins:
%e A194704   5,
%e A194704   1, 4,
%e A194704   1, 2, 2,
%e A194704   0, 1, 1, 3,
%e A194704   1, 0, 1, 1, 2,
%e A194704   ...
%e A194704 For k = 1 and m = 1: T(1,1) = 5 because there are five parts of size 1 in the last section of the set of partitions of 5, since 4 + m = 5, so a(1) = 5.
%e A194704 For k = 2 and m = 1: T(2,1) = 1 because there is only one part of size 2 in the last section of the set of partitions of 5, since 4 + m = 5, so a(2) = 1.
%o A194704 (PARI) P(n)={my(M=matrix(n,n), d=4); 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 A194704 { my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ _Andrew Howroyd_, Feb 19 2020
%Y A194704 Always the sum of row k = p(4) = A000041(4) = 5.
%Y A194704 The first (0-10) members of this family of triangles are A023531, A129186, A194702, A194703, this sequence, A194705-A194710.
%Y A194704 Cf. A135010, A138121, A182712-A182714, A194812.
%K A194704 nonn,tabl
%O A194704 1,1
%A A194704 _Omar E. Pol_, Feb 05 2012
%E A194704 Terms a(16) and beyond from _Andrew Howroyd_, Feb 19 2020