[go: up one dir, main page]

login
A258651
A(n,k) = n^(k) = k-th arithmetic derivative of n; square array A(n,k), n>=0, k>=0, read by antidiagonals.
17
0, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 1, 4, 0, 0, 0, 0, 4, 5, 0, 0, 0, 0, 4, 1, 6, 0, 0, 0, 0, 4, 0, 5, 7, 0, 0, 0, 0, 4, 0, 1, 1, 8, 0, 0, 0, 0, 4, 0, 0, 0, 12, 9, 0, 0, 0, 0, 4, 0, 0, 0, 16, 6, 10, 0, 0, 0, 0, 4, 0, 0, 0, 32, 5, 7, 11, 0, 0, 0, 0, 4, 0, 0, 0, 80, 1, 1, 1, 12
OFFSET
0,6
LINKS
J. Kovič, The Arithmetic Derivative and Antiderivative, Journal of Integer Sequences 15 (2012), Article 12.3.8
FORMULA
A(n,k) = A003415^k(n).
EXAMPLE
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, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ...
5, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
6, 5, 1, 0, 0, 0, 0, 0, 0, 0, ...
7, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
8, 12, 16, 32, 80, 176, 368, 752, 1520, 3424, ...
9, 6, 5, 1, 0, 0, 0, 0, 0, 0, ...
MAPLE
d:= n-> n*add(i[2]/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);
MATHEMATICA
d[n_] := n*Sum[i[[2]]/i[[1]], {i, FactorInteger[n]}]; d[0] = d[1] = 0;
A[n_, k_] := A[n, k] = If[k == 0, n, d[A[n, k-1]]];
Table[A[n, h-n], {h, 0, 14}, {n, 0, h}] // Flatten (* Jean-François Alcover, Apr 27 2017, translated from Maple *)
CROSSREFS
Rows n=0,1,4,8 give: A000004, A000007, A010709, A129150.
Row 15: A090636, Row 28: A090637, Row 63: A090635, Row 81: A129151, Row 128: A369638, Row 1024: A214800, Row 15625: A129152.
Main diagonal gives A185232.
Antidiagonal sums give A258652.
Cf. also A328383.
Sequence in context: A325201 A260019 A153036 * A350530 A258850 A182114
KEYWORD
nonn,tabl,look
AUTHOR
Alois P. Heinz, Jun 06 2015
STATUS
approved