A194709 - OEIS

%I #34 Jan 07 2024 14:13:04

%S 30,15,15,6,10,14,5,5,10,10,3,4,5,8,10,2,2,5,4,8,9,1,2,2,4,5,7,9,1,1,

%T 2,2,4,4,8,8,0,1,1,2,2,4,4,7,9,1,0,1,1,2,2,4,4,7,8,0,1,0,1,1,2,2,4,4,

%U 7,8,0,0,1,0,1,1,2,2,4,4,7,8,0,0,0,1,0,1,1,2,2,4,4,7,8

%N Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (9 + m).

%C Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 9. For further information see A182703 and A135010.

%H Andrew Howroyd, <a href="/A194709/b194709.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)

%F T(k,m) = A182703(9+m,k), with T(k,m) = 0 if k > 9+m.

%F T(k,m) = A194812(9+m,k).

%e Triangle begins:

%e   30;

%e   15, 15;

%e    6, 10, 14;

%e    5,  5, 10, 10;

%e    3,  4,  5,  8, 10;

%e    2,  2,  5,  4,  8, 9;

%e   ...

%e For k = 1 and  m = 1; T(1,1) = 30 because there are 30 parts of size 1 in the last section of the set of partitions of 10, since 9 + m = 10, so a(1) = 30. 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 10, since 9 + m = 10, so a(2) = 15.

%o (PARI) P(n)={my(M=matrix(n,n), d=9); 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(9) = A000041(n) = 30.

%Y The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194708, this sequence, A194710.

%Y Cf. A135010, A138121, A194812.

%K nonn,tabl

%O 1,1

%A _Omar E. Pol_, Feb 05 2012

%E Terms a(7) and beyond from _Andrew Howroyd_, Feb 19 2020