A341996 a(n) = 1 if there is at least one such prime p that p^p divides the arithmetic derivative of n, A003415(n); a(0) = a(1) = 0 by convention.

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0

Offset

0

Comments

Question: What is the asymptotic mean of this sequence and its complement A368915? See also A360111. - Antti Karttunen, Jan 11 2024

Links

Antti Karttunen, Table of n, a(n) for n = 0..65537

Index entries for characteristic functions

Formula

a(n) = [A327928(n)>0], where [ ] is the Iverson bracket.

For all n > 1, a(n) >= [A129251(n)>0], i.e., if A129251(n) is nonzero, then certainly a(n) = 1.

For all n >= 0, a(n) <= A341999(n).

For n > 0, a(n) = 1 - A368915(n). - Antti Karttunen, Jan 11 2024

Prog

(PARI)
A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A129251(n) = { my(f = factor(n)); sum(k=1, #f~, (f[k, 2]>=f[k, 1])); };
A327928(n) = if(n<=1, 0, A129251(A003415(n)));
A341996(n) = (A327928(n)>0);

Crossrefs

Characteristic function of A327929.

Positions of zeros is given by {0, 1} U A358215.

Cf. A003415, A129251, A327928, A341994, A341995, A341997, A341999, A360111, A368915 (one's complement).

Differs from A327928 for the first time at n=81, where a(81)=1.

Sequence in context: A353814 A144596 A188187 * A341999 A118685 A244063

Adjacent sequences: A341993 A341994 A341995 * A341997 A341998 A341999

Keyword

nonn

Author

Antti Karttunen, Feb 28 2021

Status

approved