%I #21 Jan 06 2023 15:19:30

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,

%T 1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,

%U 1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,1,0,1,0,0,1,1,0,1

%N a(n) = 1 if there is no factor of the form p^p in n, but for some k-th arithmetic derivative (k >= 1) of n such a factor exists; otherwise 0.

%C Question: What can be said about the distribution of 0's and 1's in this sequence? Compare also to A328308, A341996 and A359543.

%H Antti Karttunen, <a href="/A359546/b359546.txt">Table of n, a(n) for n = 1..100000</a>

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

%F a(n) = A341999(n) - A342023(n).

%F a(n) = A359550(n) * A341999(n).

%F a(n) = [A256750(n) < 0], where [ ] is the Iverson bracket.

%e a(15) = 1, because although 15 itself is not in A100716, its arithmetic derivative 15' = 8 is there.

%e a(26) = 1, as although neither 26 nor 26' = 15 are in A100716, the second derivative of 26, 26'' = 15' = 8 is there.

%o (PARI)

%o A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));

%o A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k, 2] = (f[k, 2]>=f[k, 1])); factorback(f); };

%o A341999(n) = if(!n,n,while(n>1, n = A003415checked(n)); (!n));

%o A359546(n) = ((1==A327936(n))&&A341999(n));

%Y Characteristic function of A359547.

%Y Cf. A003415, A048103, A256750, A327936, A328308, A341996, A341999, A342023, A359543, A359550.

%K nonn

%O 1

%A _Antti Karttunen_, Jan 05 2023