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A194709 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (9 + m). 3
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, 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, 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 (list; table; graph; refs; listen; history; text; internal format)
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 9. 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(9+m,k), with T(k,m) = 0 if k > 9+m.
T(k,m) = A194812(9+m,k).
EXAMPLE
Triangle begins:
30;
15, 15;
6, 10, 14;
5, 5, 10, 10;
3, 4, 5, 8, 10;
2, 2, 5, 4, 8, 9;
...
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.
PROG
(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}
{ 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(9) = A000041(n) = 30.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194708, this sequence, A194710.
Sequence in context: A263927 A033350 A070293 * A070657 A089332 A040873
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Feb 05 2012
EXTENSIONS
Terms a(7) and beyond from Andrew Howroyd, Feb 19 2020
STATUS
approved

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Last modified March 29 08:53 EDT 2024. Contains 371268 sequences. (Running on oeis4.)