A258850 A(n,k) = k-th pi-based arithmetic derivative of n; square array A(n,k), n>=0, k>=0, read by antidiagonals.
0, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 2, 4, 0, 0, 0, 1, 4, 5, 0, 0, 0, 0, 4, 3, 6, 0, 0, 0, 0, 4, 2, 7, 7, 0, 0, 0, 0, 4, 1, 4, 4, 8, 0, 0, 0, 0, 4, 0, 4, 4, 12, 9, 0, 0, 0, 0, 4, 0, 4, 4, 20, 12, 10, 0, 0, 0, 0, 4, 0, 4, 4, 32, 20, 11, 11, 0, 0, 0, 0, 4, 0, 4, 4, 80, 32, 5, 5, 12
Offset: 0
Examples
Square array A(n,k) begins: 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, ... 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, ... 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ... 5, 3, 2, 1, 0, 0, 0, 0, 0, 0, ... 6, 7, 4, 4, 4, 4, 4, 4, 4, 4, ... 7, 4, 4, 4, 4, 4, 4, 4, 4, 4, ... 8, 12, 20, 32, 80, 208, 512, 2304, 12288, 81920, ... 9, 12, 20, 32, 80, 208, 512, 2304, 12288, 81920, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..31, flattened
Crossrefs
Programs
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Maple
with(numtheory): d:= n-> n*add(i[2]*pi(i[1])/i[1], i=ifactors(n)[2]): A:= proc(n, k) option remember; `if`(k=0, n, d(A(n, k-1))) end: seq(seq(A(n, h-n), n=0..h), h=0..14);
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Mathematica
d[n_] := n*Total[Last[#]*PrimePi[First[#]]/First[#]& /@ FactorInteger[n]]; d[0] = 0; A[n_, k_] := A[n, k] = If[k == 0, n, d[A[n, k-1]]]; Table[Table[A[n, h-n], {n, 0, h}], {h, 0, 14}] // Flatten (* Jean-François Alcover, Apr 24 2016, adapted from Maple *)