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A183093
a(n) = number of divisors d of n such that d > 1 and if d = Product_(i) (p_i^e_i) then e_i = 1 for at least one i.
6
0, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 3, 1, 1, 4, 1, 4, 3, 3, 1, 5, 1, 3, 1, 4, 1, 7, 1, 1, 3, 3, 3, 5, 1, 3, 3, 5, 1, 7, 1, 4, 4, 3, 1, 6, 1, 4, 3, 4, 1, 5, 3, 5, 3, 3, 1, 10, 1, 3, 4, 1, 3, 7, 1, 4, 3, 7, 1, 6, 1, 3, 4, 4, 3, 7, 1, 6, 1, 3, 1, 10, 3, 3, 3, 5, 1, 10, 3, 4, 3, 3, 3, 7, 1, 4, 4, 5
OFFSET
1,6
COMMENTS
a(n) = number of non-powerful divisors d of n where powerful numbers are numbers of the form A001694(m) for m >= 2.
Sequence is not the same as A183096: a(72) = 6, A183096(72) = 7.
FORMULA
a(n) = A000005(n) - A005361(n) = A183095(n) - 1.
a(1) = 0, a(p) = 1, a(pq) = 3, a(pq...z) = 2^k - 1, a(p^k) = 1, for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.
Sum_{k=1..n} a(k) ~ n*(log(n) + 2*gamma - zeta(2)*zeta(3)/zeta(6) - 1), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 30 2025
EXAMPLE
For n = 12, set of such divisors is {2, 3, 6, 12}; a(12) = 4.
MATHEMATICA
nonpowQ[n_] := Min[FactorInteger[n][[;; , 2]]] == 1; a[n_] := DivisorSum[n, 1 &, # > 1 && nonpowQ[#] &]; Array[a, 100] (* Amiram Eldar, Jan 30 2025 *)
PROG
(Scheme) (define (A183093 n) (- (A000005 n) (A005361 n))) ;; Antti Karttunen, Nov 23 2017
(PARI) a(n) = sumdiv(n, d, d > 1 && vecmin(factor(d)[, 2]) == 1); \\ Amiram Eldar, Jan 30 2025
KEYWORD
nonn,changed
AUTHOR
Jaroslav Krizek, Dec 25 2010
EXTENSIONS
Name corrected by Amiram Eldar, Jan 30 2025
STATUS
approved