A064549 a(n) = n * Product_{primes p|n} p.
1, 4, 9, 8, 25, 36, 49, 16, 27, 100, 121, 72, 169, 196, 225, 32, 289, 108, 361, 200, 441, 484, 529, 144, 125, 676, 81, 392, 841, 900, 961, 64, 1089, 1156, 1225, 216, 1369, 1444, 1521, 400, 1681, 1764, 1849, 968, 675, 2116, 2209, 288, 343, 500, 2601, 1352
Offset: 1
Examples
a(12) = 72 since 12 = 2^2*3 and 12*2*3 = 72.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
- Nadia Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, Vol. 113, No. 8 (2006), pp. 1732-1745; arXiv preprint, arXiv:math/0509316 [math.NT], 2005-2006.
Crossrefs
Programs
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Haskell
a064549 n = a007947 n * n -- Reinhard Zumkeller, Jul 23 2013
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Magma
[n^2/( (&+[Floor(k^n/n)-Floor((k^n - 1)/n) : k in [1..n]]) ): n in [1..50]]; // G. C. Greubel, Nov 02 2018
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Maple
a:= n -> n * convert(numtheory:-factorset(n), `*`): seq(a(n),n=1..100); # Robert Israel, Jul 25 2014
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Mathematica
a[n_] := n * Times @@ FactorInteger[n][[All, 1]]; Array[a, 100] (* Jean-François Alcover, Feb 17 2017 *) Table[n*Product[If[PrimeQ[d], d, 1], {d, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Jun 15 2019 *)
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PARI
popf(n)= { local(f,p=1); f=factor(n); for(i=1, matsize(f)[1], p*=f[i, 1]); return(p) } { for (n=1, 1000, write("b064549.txt", n, " ", n*popf(n)) ) } \\ Harry J. Smith, Sep 18 2009
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PARI
A064549(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 2]++); factorback(f); }; \\ Antti Karttunen, Aug 30 2018
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PARI
for(n=1, 100, print1(direuler(p=2, n, (1 - p*X + p^2*X)/(1 - p*X))[n], ", ")) \\ Vaclav Kotesovec, Jun 24 2020
Formula
Multiplicative with a(p^k)=p^(k+1) when k>0.
a(n) = A078310(n) - 1. - Reinhard Zumkeller, Jul 23 2013
G.f.: Sum_{k>=1} mu(k)^2*phi(k)*k*x^k/(1 - x^k)^2. - Ilya Gutkovskiy, Nov 02 2018
From Vaclav Kotesovec, Jun 24 2020: (Start)
Dirichlet g.f.: zeta(s-2) * zeta(s-1) * Product_{primes p} (1 + p^(3-2*s) - p^(4-2*s) - p^(1-s)).
(End)
Sum_{k>=1} 1/a(k) = zeta(2)*zeta(3)/zeta(6) = A082695. - Vaclav Kotesovec, Sep 19 2020
Sum_{k>=1} (-1)^(k+1)/a(k) = zeta(2)*zeta(3)/(3*zeta(6)) = (1/3) * A082695. - Amiram Eldar, Nov 18 2020
Comments