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A194705
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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 (5 + m).
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3
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7, 4, 3, 2, 2, 3, 1, 1, 3, 2, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 2, 0, 1, 0, 1, 1, 2, 2, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2
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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 5. 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(5+m,k), with T(k,m) = 0 if k > 5+m.
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EXAMPLE
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Triangle begins:
7,
4, 3,
2, 2, 3,
1, 1, 3, 2,
0, 1, 1, 2, 3,
1, 0, 1, 1, 2, 2,
0, 1, 0, 1, 1, 2, 2,
...
For k = 1 and m = 1: T(1,1) = 7 because there are seven parts of size 1 in the last section of the set of partitions of 6, since 5 + m = 6, so a(1) = 7.
For k = 2 and m = 1: T(2,1) = 4 because there are four parts of size 2 in the last section of the set of partitions of 6, since 5 + m = 6, so a(2) = 4.
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PROG
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(PARI) P(n)={my(M=matrix(n, n), d=5); 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(5) = A000041(5) = 7.
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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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