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a(n) = 1 if there is no prime p such that p^p divides A342001(n), otherwise 0.
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%I #22 Jan 14 2024 20:34:55

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

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

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

%N a(n) = 1 if there is no prime p such that p^p divides A342001(n), otherwise 0.

%C Question: What is the asymptotic mean of this sequence?

%C Answer: Apparently about 0.77... See A368920. - _Antti Karttunen_, Jan 14 2024

%H Antti Karttunen, <a href="/A368914/b368914.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(1) = 0, and for n > 1, a(n) = A359550(A342001(n)).

%F a(n) = A368913(n) + A368915(n).

%F For all n >= 1, a(n) >= A368912(n).

%o (PARI)

%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));

%o A003557(n) = (n/factorback(factorint(n)[, 1]));

%o A342001(n) = (A003415(n) / A003557(n));

%o A359550(n) = { my(f = factor(n)); prod(k=1, #f~, (f[k, 2]<f[k, 1])); };

%o A368914(n) = ((n>1)&&A359550(A342001(n)));

%Y Characteristic function of A368904, whose complement A368996 gives the positions of 0's.

%Y Cf. A003415, A342001, A359550, A368912, A368913, A368915, A368916 [= a(A276086(n))], A368905 [= A005940(1+n)], A368920.

%K nonn

%O 1

%A _Antti Karttunen_, Jan 09 2024