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).
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
Offset: 1
Examples
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.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
Crossrefs
Programs
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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
Extensions
Terms a(7) and beyond from Andrew Howroyd, Feb 19 2020
Comments