OFFSET
1,4
COMMENTS
a(n) depends only on prime signature of n (cf. A025487, A052306). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3,1).
There was a comment here that said "a(n) is the number of nilpotents elements in the ring Z/nZ", but this is false, see A003557.
a(n) is the number of square-full divisors of n. a(n) is also the number of divisors d of n such that d and n have the same prime factors, i.e., A007947(d) = A007947(n). - Laszlo Toth, May 22 2009
Number of divisors u of n such that u|(u^n/n). Row lengths in triangle of A284318. - Juri-Stepan Gerasimov, Apr 05 2017
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from T. D. Noe)
Imanuel Chen and Michael Z. Spivey, Integral Generalized Binomial Coefficients of Multiplicative Functions, Preprint 2015; Summer Research Paper 238, Univ. Puget Sound.
P. Erdős and T. Motzkin, Problem 5735, Amer. Math. Monthly, 78 (1971), 680-681. (Incorrect solution!)
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
J. Knopfmacher, A prime-divisor function, Proc. Amer. Math. Soc., 40 (1973), 373-377.
MathStackExchange, Dirichlet series for zeta(s)*zeta(2*s)*zeta(3*s)*zeta(6*s)^(-1).
H. N. Shapiro, Problem 5735, Amer. Math. Monthly, 97 (1990), 937.
D. Suryanarayana and R. Sitaramachandra Rao, The number of square-full divisors of an integer, Proc. Amer. Math. Soc., Vol. 34, No. 1 (1972), pp. 79-80.
FORMULA
n = Product (p_j^k_j) -> a(n) = Product (k_j).
Dirichlet g.f.: zeta(s)*zeta(2s)*zeta(3s)/zeta(6s).
Multiplicative with a(p^e) = e. - David W. Wilson, Aug 01 2001
a(n) = Sum_{d dividing n} floor(rad(d)/rad(n)) where rad(n) is A007947. - Enrique Pérez Herrero, Nov 06 2009
a(n) = Sum_{k=1..n}(floor(cos^2(Pi*k^n/n))*floor(cos^2(Pi*n/k))). - Anthony Browne, May 11 2016
From Antti Karttunen, Mar 06 2017: (Start)
(End)
Let (b(n)) be multiplicative with b(p^e) = -1 + ( (floor((e-1)/3)+floor(e/3)) mod 4 ) for p prime and e > 0, then b(n) is the Dirichlet inverse of (a(n)). - Werner Schulte, Feb 23 2018
Sum_{i=1..k} a(i) ~ (zeta(2)*zeta(3)/zeta(6)) * k (Suryanarayana and Sitaramachandra Rao, 1972). - Amiram Eldar, Apr 13 2020
More precise asymptotics: Sum_{k=1..n} a(k) ~ 315*zeta(3)*n / (2*Pi^4) + zeta(1/2)*zeta(3/2)*sqrt(n) / zeta(3) + 6*zeta(1/3)*zeta(2/3)*n^(1/3) / Pi^2 [Knopfmacher, 1973]. - Vaclav Kotesovec, Jun 13 2020
MAPLE
A005361 := proc(n)
local a, p ;
a := 1 ;
for p in ifactors(n)[2] do
a := a*op(2, p) ;
end do:
a ;
end proc:
seq(A005361(n), n=1..30) ; # R. J. Mathar, Nov 20 2012
# second Maple program:
a:= n-> mul(i[2], i=ifactors(n)[2]):
seq(a(n), n=1..80); # Alois P. Heinz, Feb 18 2020
MATHEMATICA
Prepend[ Array[ Times @@ Last[ Transpose[ FactorInteger[ # ] ] ]&, 100, 2 ], 1 ]
Array[Times@@Transpose[FactorInteger[#]][[2]]&, 80] (* Harvey P. Dale, Aug 15 2012 *)
PROG
(PARI) for(n=1, 100, f=factor(n); print1(prod(i=1, omega(f), f[i, 2]), ", ")) \\ edited by M. F. Hasler, Feb 18 2020
(PARI) a(n)=factorback(factor(n)[, 2]) \\ Charles R Greathouse IV, Nov 07 2014
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - X + X^2)/(1 - X)^2)[n], ", ")) \\ Vaclav Kotesovec, Jun 14 2020
(Haskell)
a005361 = product . a124010_row -- Reinhard Zumkeller, Jan 09 2012
(Scheme) (define (A005361 n) (if (= 1 n) 1 (* (A067029 n) (A005361 (A028234 n))))) ;; Antti Karttunen, Mar 06 2017
(Python)
from math import prod
from sympy import factorint
def a(n): return prod(factorint(n).values())
print([a(n) for n in range(1, 91)]) # Michael S. Branicky, Jul 04 2022
CROSSREFS
KEYWORD
nonn,easy,nice,mult
AUTHOR
STATUS
approved