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A376362
The number of unitary divisors that are squares of primes applied to the powerful numbers.
5
0, 1, 0, 1, 0, 1, 0, 0, 2, 1, 0, 1, 0, 2, 1, 1, 0, 0, 1, 1, 2, 1, 0, 2, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 2, 2, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 1, 1, 0, 3, 1, 1, 1, 0, 0, 2, 1, 1, 2, 2, 0, 1, 0, 1, 1, 1, 2, 2, 1, 1, 1, 0, 3, 2, 1, 1, 0, 0, 1, 0, 2, 0, 0, 1, 1, 1, 0, 1, 0, 2, 2, 1, 0, 1, 1, 1, 2, 1, 0, 1, 1, 2, 1, 2, 0
OFFSET
1,9
LINKS
Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, On the number of prime factors with a given multiplicity over h-free and h-full numbers, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.3.
FORMULA
a(n) = A369427(A001694(n)).
Sum_{A001694(k) <= x} a(k) = c * sqrt(x) * (log(log(x)) + B - log(2) - L(2, 4)) + O(sqrt(x)/log(x)), where c = zeta(3/2)/zeta(3) (A090699), B is Mertens's constant (A077761), L(h, r) = Sum_{p prime} 1/(p^(r/h - 1) * (p - p^(1 - 1/h) + 1)), and L(2, 4) = 0.57937575954505652569... (Das et al., 2024).
MATHEMATICA
f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # > 1 &], Count[e, 2], Nothing]]; Array[f, 3500]
PROG
(PARI) lista(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] == 1, is = 0; break)); if(is, print1(#select(x -> x == 2, e), ", "))); }
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Sep 21 2024
STATUS
approved