A245095 Triangle read by rows: T(n,k) = A006218(k)*A002865(n-k).
1, 0, 3, 1, 0, 5, 1, 3, 0, 8, 2, 3, 5, 0, 10, 2, 6, 5, 8, 0, 14, 4, 6, 10, 8, 10, 0, 16, 4, 12, 10, 16, 10, 14, 0, 20, 7, 12, 20, 16, 20, 14, 16, 0, 23, 8, 21, 20, 32, 20, 28, 16, 20, 0, 27, 12, 24, 35, 32, 40, 28, 32, 20, 23, 0, 29, 14, 36, 40, 56, 40, 56, 32, 40, 23, 27, 0, 35
Offset: 1
Examples
Triangle begins: 1; 0, 3; 1, 0, 5; 1, 3, 0, 8; 2, 3, 5, 0, 10; 2, 6, 5, 8, 0, 14; 4, 6, 10, 8, 10, 0, 16; 4, 12, 10, 16, 10, 14, 0, 20; 7, 12, 20, 16, 20, 14, 16, 0, 23; 8, 21, 20, 32, 20, 28, 16, 20, 0, 27; 12, 24, 35, 32, 40, 28, 32, 20, 23, 0, 29; 14, 36, 40, 56, 40, 56, 32, 40, 23, 27, 0, 35; ... For n = 6: ------------------------- k A006218 T(6,k) ------------------------- 1 1 * 2 = 2 2 3 * 2 = 6 3 5 * 1 = 5 4 8 * 1 = 8 5 10 * 0 = 0 6 14 * 1 = 14 . A002865 ------------------------- So row 6 is [2, 6, 5, 8, 0, 14] and the sum of row 6 is 2+6+5+8+0+14 = 35 equaling A006128(6) = 35.
Links
- Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of the triangle, flattened)
Programs
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Mathematica
A245095row[n_]:=Accumulate[DivisorSigma[0,Range[n]]]Reverse[Differences[PartitionsP[Range[-1,n-1]]]];Array[A245095row,10] (* Paolo Xausa, Sep 04 2023 *)
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PARI
a006218(n) = sum(k=1, n, n\k); a002865(n) = if(n, numbpart(n)-numbpart(n-1), 1); row(n) = vector(n, i, a006218(i)*a002865(n-i)); \\ Michel Marcus, Jul 18 2014
Comments