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A194709
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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 (9 + m).
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3
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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T(k,m) = A182703(9+m,k), with T(k,m) = 0 if k > 9+m.
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EXAMPLE
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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.
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PROG
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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
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CROSSREFS
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Always the sum of row k = p(9) = A000041(n) = 30.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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