A112624 If p^b(p,n) is the highest power of the prime p dividing n, then a(n) = Product_{p|n} b(p,n)!.
1, 1, 1, 2, 1, 1, 1, 6, 2, 1, 1, 2, 1, 1, 1, 24, 1, 2, 1, 2, 1, 1, 1, 6, 2, 1, 6, 2, 1, 1, 1, 120, 1, 1, 1, 4, 1, 1, 1, 6, 1, 1, 1, 2, 2, 1, 1, 24, 2, 2, 1, 2, 1, 6, 1, 6, 1, 1, 1, 2, 1, 1, 2, 720, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 2, 2, 1, 1, 1, 24, 24, 1, 1, 2, 1, 1, 1, 6, 1, 2, 1, 2, 1, 1, 1, 120, 1, 2, 2, 4, 1
Offset: 1
Examples
45 = 3^2 * 5^1. So a(45) = 2! * 1! = 2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Rafael Jakimczuk, Generalizations of Mertens's Formula and k-Free and s-Full Numbers with Prime Divisors in Arithmetic Progression, ResearchGate, 2024.
- Index entries for sequences computed from exponents in factorization of n.
Crossrefs
Programs
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Maple
w := n -> op(2, ifactors(n)): a := n -> mul(factorial(w(n)[j][2]), j = 1..nops(w(n))): seq(a(n), n = 1..101); # Emeric Deutsch, May 17 2012
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Mathematica
f[n_] := Block[{fi = Last@Transpose@FactorInteger@n}, Times @@ (fi!)]; Array[f, 101] (* Robert G. Wilson v, Dec 27 2005 *)
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PARI
A112624(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= f[k, 2]!; ); m; } \\ Antti Karttunen, May 28 2017
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Sage
def A112624(n): return mul(factorial(s[1]) for s in factor(n)) [A112624(i) for i in (1..101)] # Peter Luschny, Jun 15 2013
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Scheme
(define (A112624 n) (if (= 1 n) n (* (A000142 (A067029 n)) (A112624 (A028234 n))))) ;; Antti Karttunen, May 29 2017
Formula
From Antti Karttunen, May 29 2017: (Start)
From Mats Granvik, Mar 05 2019: (Start)
log(a(n)) = inverse Möbius transform of log(A306694(n)).
From Amiram Eldar, Mar 08 2024: (Start)
Let f(n) = 1/a(n). Formulas from Jakimczuk (2024, pp. 12-15):
Dirichlet g.f. of f(n): Sum_{n>=1} f(n)/n^s = exp(P(s)), where P(s) is the prime zeta function.
Sum_{k=1..n} f(k) = c * n + o(n), where c = A240953.
Sum_{k=1..n} f(k)/k = c * log(n) + o(log(n)), where c = A240953. (End)
Extensions
More terms from Robert G. Wilson v, Dec 27 2005
Comments