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A005361
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Product of exponents of prime factorization of n.
(Formerly M0063)
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57
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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 an old 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 u such that u^n|n|u. Row lengths in triangle of A284318. - Juri-Stepan Gerasimov, Apr 05 2017
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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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T. D. Noe and 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, T. Motzkin, Problem 5735, Amer. Math. Monthly, 78 (1971), 680-681. (Incorrect solution!)
J. Knopfmacher, A prime-divisor function, Proc. Amer. Math. Soc., 40 (1973), 373-377.
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., 34 (1972), 79-80.
Index entries for sequences computed from exponents in factorization of n
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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).
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
For n > 1: a(n) = Product_{k=1..A001221(n)} A124010(n,k). - Reinhard Zumkeller, Aug 27 2011
a(n) = tau(n/rad(n)), where tau is A000005 and rad is A007947. - Anthony Browne, May 11 2016
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)
For all n >= 1, a(prime^n) = n, a(A002110(n)) = a(A005117(n)) = 1. [From Crossrefs section.]
a(1) = 1; for n > 1, a(n) = A067029(n) * a(A028234(n)).
(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
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MAPLE
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A005361 := proc(n)
local a ;
if n = 1 then
1 ;
else
a := 1 ;
for p in ifactors(n)[2] do
a := a*op(2, p) ;
end do:
end if;
end proc:
seq(A005361(n), n=1..30) ; # R. J. Mathar, Nov 20 2012
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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, print1(prod(i=1, omega(n), component(component(factor(n), 2), i)), ", "))
(PARI) a(n)=factorback(factor(n)[, 2]) \\ Charles R Greathouse IV, Nov 07 2014
(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
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CROSSREFS
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Cf. A000005, A002110, A005117 (indices of ones), A028234, A052306, A067029, A072411, A284318.
Sequence in context: A290107 A212180 A091050 * A303915 A322885 A292582
Adjacent sequences: A005358 A005359 A005360 * A005362 A005363 A005364
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KEYWORD
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nonn,easy,nice,mult
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AUTHOR
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Jeffrey Shallit, Olivier Gérard
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STATUS
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approved
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