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A342023
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a(n) = 1 if there is a prime p such that p^p divides n, otherwise 0.
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8
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0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 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, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 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, 0, 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, 0, 0, 0, 1, 0, 0, 0, 1
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OFFSET
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1
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LINKS
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FORMULA
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a(n) = min(1, A129251(n)) = [A129251(n) > 0], where [ ] is the Iverson bracket.
For all n >= 1,
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 - Product_{p prime} (1 - 1/p^p) = 0.2780097655... . - Amiram Eldar, Jul 24 2022
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MATHEMATICA
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Array[Function[{D, q}, Boole[Total@ Table[Count[D, _?(IntegerExponent[#, p] == p &)], {p, Prime@ Range@ q}] > 0]] @@ {Divisors[#], PrimePi@ Floor[Sqrt[#]]} &, 120] (* Michael De Vlieger, Mar 11 2021 *)
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PROG
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(PARI) A342023(n) = if(1==n, 0, my(f = factor(n)); for(k=1, #f~, if(f[k, 2]>=f[k, 1], return(1))); (0));
(Python)
from sympy import factorint
f = factorint(n)
for p in f:
if p <= f[p]:
return 1
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CROSSREFS
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Characteristic function of A100716.
Cf. A003415, A008966, A048103 (positions of zeros), A107078, A276086, A341996, A341999, A327936, A341999, A342004, A342007, A359546, A359550 (one's complement).
Differs from A129251 and A276077 for the first time at n=108, as here a(108) = 1.
Differs from A342024 for the first time at n=625, where a(625)=0, while A342024(625)=1.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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