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A106254
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Partition table in square format.
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1
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1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 1, 3, 4, 3, 2, 1, 1, 4, 5, 5, 3, 2, 1, 1, 4, 7, 6, 5, 3, 2, 1, 1, 5, 8, 9, 7, 5, 3, 2, 1, 1, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1, 6, 12, 15, 13, 11, 7, 5, 3, 2, 1, 1, 6, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 1, 7, 16, 23, 23, 20, 15, 11, 7, 5, 3, 2, 1
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OFFSET
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1,5
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COMMENTS
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The square is the following table:
1 1 1 1 1 1 1...
1 2 2 2 2 2 2...
1 2 3 3 3 3 3...
1 3 4 5 5 5 5...
1 3 5 6 7 7 7...
1 4 7 9 10 11 11...
1 4 8 11 13 14 15...
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REFERENCES
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Ivan Niven, "Mathematics of Choice, How to Count Without Counting", MAA, 1965, pp. 98-99 (table p. 98).
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LINKS
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FORMULA
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Antidiagonals of table of values of p_k(n) (the number of partitions of n with no summand greater than k).
T(n,m) = sum_{i=1..m} A008284(n,i). T(n,m) = A026820(n,m) if m<=n and T(n,m)=T(n,n) if m>=n. G.f. column m: 1/(1-x)/(1-x^2)/.../(1-x^m) = sum_(n=1,2,3..) T(n,m) x^n [Comtet]. - R. J. Mathar, Aug 31 2007
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EXAMPLE
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The partitions of 6 are:
1 + 1 + 1 + 1 + 1 + 1; 2 + 1 + 1 + 1 + 1; 3 + 1 + 1 + 1; 2 + 2 + 1 + 1; 4 + 1 + 1; 3 + 2 + 1; 2 + 2 + 2; 5 + 1, 4 + 2, 3 + 3, 6.
There are 9 partitions of 6 having summands no larger than 4, so p_4(6) = 9.
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MATHEMATICA
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T[n_, k_] := T[n, k] = If[n == 0 || k == 1, 1, T[n, k-1] + If[k > n, 0, T[n-k, k]]];
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CROSSREFS
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Essentially the same as A008284, except for missing one diagonal thereof (which would be zero row of this array).
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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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