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A005361
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Product of exponents of prime factorization of n.
(Formerly M0063)
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130
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1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1
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OFFSET
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1,4
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COMMENTS
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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
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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P. Erdős and T. Motzkin, Problem 5735, Amer. Math. Monthly, 78 (1971), 680-681. (Incorrect solution!)
H. N. Shapiro, Problem 5735, Amer. Math. Monthly, 97 (1990), 937.
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FORMULA
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n = Product (p_j^k_j) -> a(n) = Product (k_j).
Dirichlet g.f.: zeta(s)*zeta(2s)*zeta(3s)/zeta(6s).
a(n) = Sum_{k=1..n}(floor(cos^2(Pi*k^n/n))*floor(cos^2(Pi*n/k))). - Anthony Browne, May 11 2016
For all n >= 1, a(prime^n) = n, a(A002110(n)) = a(A005117(n)) = 1. [From Crossrefs section.]
(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
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MAPLE
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local a, p ;
a := 1 ;
for p in ifactors(n)[2] do
a := a*op(2, p) ;
end do:
a ;
end proc:
# second Maple program:
a:= n-> mul(i[2], i=ifactors(n)[2]):
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MATHEMATICA
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Prepend[ Array[ Times @@ Last[ Transpose[ FactorInteger[ # ] ] ]&, 100, 2 ], 1 ]
Array[Times@@Transpose[FactorInteger[#]][[2]]&, 80] (* Harvey P. Dale, Aug 15 2012 *)
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PROG
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(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) for(n=1, 100, print1(direuler(p=2, n, (1 - X + X^2)/(1 - X)^2)[n], ", ")) \\ Vaclav Kotesovec, Jun 14 2020
(Haskell)
(Python)
from math import prod
from sympy import factorint
def a(n): return prod(factorint(n).values())
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CROSSREFS
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KEYWORD
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nonn,easy,nice,mult
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AUTHOR
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STATUS
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approved
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