A210764 Square array T(n,k), n>=0, k>=0, read by antidiagonals in which column k gives the partial sums of column k of A144064.
1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 4, 1, 1, 12, 18, 13, 5, 1, 1, 19, 38, 35, 19, 6, 1, 1, 30, 74, 86, 59, 26, 7, 1, 1, 45, 139, 194, 164, 91, 34, 8, 1, 1, 67, 249, 415, 416, 281, 132, 43, 9, 1, 1, 97, 434, 844, 990, 787, 447, 183, 53, 10, 1
Offset: 0
Examples
Array begins: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 4, 8, 13, 19, 26, 34, 43, 53, 1, 7, 18, 35, 59, 91, 132, 183, 1, 12, 38, 86, 164, 281, 447, 1, 19, 74, 194, 416, 787, 1, 30, 139, 415, 990, 1, 45, 249, 844, 1, 67, 434, 1, 97, 1,
Links
- Alois P. Heinz, Rows n = 0..140, flattened
Crossrefs
Programs
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Maple
with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; `if`(n=0, 1, add(add(d*p(d), d=divisors(j))*b(n-j), j=1..n)/n) end end: A:= (n, k)-> etr(j-> k +`if`(j=1, 1, 0))(n): seq(seq(A(d-k, k), k=0..d), d=0..14); # Alois P. Heinz, May 20 2013
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Mathematica
etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[ j]}]*b[n-j], {j, 1, n}]/n]; b]; A[n_, k_] := etr[Function[{j}, k + If[j == 1, 1, 0]]][n]; Table[Table[A[d-k, k], {k, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Mar 05 2015, after Alois P. Heinz *)
Comments