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A003557
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n divided by largest squarefree divisor of n; if n = Product p(k)^e(k) then a(n) = Product p(k)^(e(k)-1), with a(1) = 1.
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313
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1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 8, 1, 3, 1, 2, 1, 1, 1, 4, 5, 1, 9, 2, 1, 1, 1, 16, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 8, 7, 5, 1, 2, 1, 9, 1, 4, 1, 1, 1, 2, 1, 1, 3, 32, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 5, 2, 1, 1, 1, 8, 27, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 16, 1, 7
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OFFSET
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1,4
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COMMENTS
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a(n) is the size of the Frattini subgroup of the cyclic group C_n - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 07 2001.
Also of the Frattini subgroup of the dihedral group with 2*n elements. - Sharon Sela (sharonsela(AT)hotmail.com), Jan 01 2002
Number of solutions to x^m==0 (mod n) provided that n < 2^(m+1), i.e. the sequence of sequences A000188, A000189, A000190, etc. converges to this sequence. - Henry Bottomley, Sep 18 2001
a(n) is the number of nilpotent elements in the ring Z/nZ. - Laszlo Toth, May 22 2009
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LINKS
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FORMULA
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a(n) = Sum_{k=1..n}(floor(k^n/n)-floor((k^n-1)/n)). - Anthony Browne, May 11 2016
a(n) = e^[Sum_{k=2..n} (floor(n/k)-floor((n-1)/k))*(1-A010051(k))*Mangoldt(k)] where Mangoldt is the Mangoldt function. - Anthony Browne, Jun 16 2016
a(n) = Sum_{d|n} mu(d) * phi(d) * (n/d), where mu(d) is the Moebius function and phi(d) is the Euler totient function (rephrases formula of Dec 2011). - Daniel Suteu, Jun 19 2018
G.f.: Sum_{k>=1} mu(k)*phi(k)*x^k/(1 - x^k)^2. - Ilya Gutkovskiy, Nov 02 2018
Dirichlet g.f.: Product_{primes p} (1 + 1/(p^s - p)). - Vaclav Kotesovec, Jun 24 2020
a(n) = Sum_{k=1..n} mu(n/gcd(n,k))*gcd(n,k).
a(n) = Sum_{k=1..n} mu(gcd(n,k))*(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). (End)
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MAPLE
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A003557 := n -> n/ilcm(op(numtheory[factorset](n))):
seq(n / NumberTheory:-Radical(n), n = 1..98); # Peter Luschny, Jul 20 2021
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MATHEMATICA
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Prepend[ Array[ #/Times@@(First[ Transpose[ FactorInteger[ # ] ] ])&, 100, 2 ], 1 ] (* Olivier Gérard, Apr 10 1997 *)
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PROG
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(Sage) def A003557(n) : return n*mul(1/p for p in prime_divisors(n))
(Haskell)
a003557 n = product $ zipWith (^)
(a027748_row n) (map (subtract 1) $ a124010_row n)
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - p*X + X)/(1 - p*X))[n], ", ")) \\ Vaclav Kotesovec, Jun 20 2020
(Python)
from sympy.ntheory.factor_ import core
from sympy import divisors
def a(n): return n / max(i for i in divisors(n) if core(i) == i)
(Python)
from math import prod
from sympy import primefactors
(Magma) [(&+[(Floor(k^n/n)-Floor((k^n-1)/n)): k in [1..n]]): n in [1..100]]; // G. C. Greubel, Nov 02 2018
(Julia)
using Nemo
n < 4 && return 1
q = prod([p for (p, e) ∈ Nemo.factor(fmpz(n))])
return n == q ? 1 : div(n, q)
end
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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