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A194708
Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (8 + m).
3
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, 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, 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
OFFSET
1,1
COMMENTS
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.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
FORMULA
T(k,m) = A182703(8+m,k), with T(k,m) = 0 if k > 8+m.
T(k,m) = A194812(8+m,k).
EXAMPLE
Triangle begins:
22,
7, 15,
6, 6, 10,
2, 5, 5, 10,
2, 3, 4, 5, 8,
...
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.
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.
PROG
(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}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020
CROSSREFS
Always the sum of row k = p(8) = A000041(8) = 22.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194707, this sequence, A194709, A194710.
Sequence in context: A317740 A317901 A040467 * A069285 A159990 A243629
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Feb 05 2012
EXTENSIONS
Terms a(11) and beyond from Andrew Howroyd, Feb 19 2020
STATUS
approved