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A000189
Number of solutions to x^3 == 0 (mod n).
16
1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 4, 5, 1, 9, 2, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 4, 7, 5, 1, 2, 1, 9, 1, 4, 1, 1, 1, 2, 1, 1, 3, 16, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 5, 2, 1, 1, 1, 4, 9, 1, 1, 2, 1, 1, 1, 4, 1, 3
OFFSET
1,4
COMMENTS
Shadow transform of the cubes A000578. - Michel Marcus, Jun 06 2013
LINKS
Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
Lorenz Halbeisen and Norbert Hungerbuehler, Number theoretic aspects of a combinatorial function, Notes on Number Theory and Discrete Mathematics 5(4) (1999), 138-150. (ps, pdf); see Definition 7 for the shadow transform.
OEIS Wiki, Shadow transform.
N. J. A. Sloane, Transforms.
FORMULA
Multiplicative with a(p^e) = p^[2e/3]. - David W. Wilson, Aug 01 2001
a(n) = n/A019555(n). - Petros Hadjicostas, Sep 15 2019
Dirichlet g.f.: zeta(3*s-2) * Product_{p prime} (1 + 1/p^s + 1/p^(2*s-1)). - Amiram Eldar, Sep 09 2023
From Vaclav Kotesovec, Sep 09 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-1) * zeta(3*s-2) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s-1) - 1/p^(4*s-2) + 1/p^(4*s-1) + 1/p^(5*s-2)).
Let f(s) = Product_{primes p} (1 - 1/p^(2*s) - 1/p^(3*s-1) - 1/p^(4*s-2) + 1/p^(4*s-1) + 1/p^(5*s-2)).
Sum_{k=1..n} a(k) ~ (f(1)*n/6) * (log(n)^2/2 + (6*gamma - 1 + f'(1)/f(1))*log(n) + 1 - 6*gamma + 11*gamma^2 - 14*sg1 + (6*gamma - 1)*f'(1)/f(1) + f''(1)/(2*f(1))), where
f(1) = Product_{primes p} (1 - 3/p^2 + 2/p^3) = A065473 = 0.2867474284344787341078927127898384464343318440970569956414778593366522431...,
f'(1) = f(1) * Sum_{primes p} 9*log(p) / (p^2 + p - 2) = f(1) * 4.1970213428422788650375569145777616746065054412058004220013841318980729375...,
f''(1) = f'(1)^2/f(1) + f(1) * Sum_{primes p} (-29*p^2 - 17*p + 1) * log(p)^2 / (p^2 + p - 2)^2 = f'(1)^2/f(1) + f(1) * (-21.3646716550082193262514333696570765444176783899223644201265894338042468...),
gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). (End)
EXAMPLE
a(4) = 2 because 0^3 == 0, 1^3 == 1, 2^3 == 0, and 3^3 == 3 (mod 4); also, a(9) = 3 because 0^3 = 0, 3^3 == 0, and 6^3 = 0 (mod 9), while x^3 =/= 0 (mod 9) for x = 1, 2, 4, 5, 7, 8. - Petros Hadjicostas, Sep 16 2019
MATHEMATICA
Array[ Function[ n, Count[ Array[ PowerMod[ #, 3, n ]&, n, 0 ], 0 ] ], 100 ]
f[p_, e_] := p^Floor[2*e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 19 2020 *)
PROG
(PARI) a(n)=my(f=factor(n)); prod(i=1, #f[, 1], f[i, 1]^(2*f[i, 2]\3)) \\ Charles R Greathouse IV, Jun 06 2013
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X + p*X^2)/(1 - p^2*X^3))[n], ", ")) \\ Vaclav Kotesovec, Aug 30 2021
CROSSREFS
Sequence in context: A125653 A104445 A359762 * A000190 A348037 A003557
KEYWORD
nonn,mult,easy
STATUS
approved