A341996 - OEIS

%I #18 Jan 11 2024 09:18:28

%S 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,

%T 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,

%U 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

%N 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.

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

%H Antti Karttunen, <a href="/A341996/b341996.txt">Table of n, a(n) for n = 0..65537</a>

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

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

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

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

%F For n > 0, a(n) = 1 - A368915(n). - _Antti Karttunen_, Jan 11 2024

%o (PARI)

%o 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

%o A129251(n) = { my(f = factor(n)); sum(k=1, #f~, (f[k, 2]>=f[k, 1])); };

%o A327928(n) = if(n<=1,0,A129251(A003415(n)));

%o A341996(n) = (A327928(n)>0);

%Y Characteristic function of A327929.

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

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

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

%K nonn

%O 0

%A _Antti Karttunen_, Feb 28 2021