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Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).
3

%I #35 Jan 07 2024 14:13:37

%S 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 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 0,0,0,0,0,1,0,1,1,2,0,0,0,0,0,0,0,0,1,0,1,1,2

%N 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 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 Andrew Howroyd, <a href="/A194704/b194704.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)

%F T(k,m) = A182703(4+m,k), with T(k,m) = 0 if k > 4+m.

%F T(k,m) = A194812(4+m,k).

%e Triangle begins:

%e 5,

%e 1, 4,

%e 1, 2, 2,

%e 0, 1, 1, 3,

%e 1, 0, 1, 1, 2,

%e ...

%e 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 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 (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 { 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(4) = A000041(4) = 5.

%Y The first (0-10) members of this family of triangles are A023531, A129186, A194702, A194703, this sequence, A194705-A194710.

%Y Cf. A135010, A138121, A182712-A182714, A194812.

%K nonn,tabl

%O 1,1

%A _Omar E. Pol_, Feb 05 2012

%E Terms a(16) and beyond from _Andrew Howroyd_, Feb 19 2020