A070474 a(n) = n^3 mod 12, n^5 mod 12.

0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3

Offset

0,3

Comments

n^5 - n^3 == 0 (mod 12) is shown explicitly for n = 0 to 11, then the induction n -> n+12 for the 5th-order polynomial followed by binomial expansion of (n+12)^k concludes that the zero (mod 12) is periodically extended to the other integers. - R. J. Mathar, Jul 23 2009

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Formula

From R. J. Mathar, Jul 23 2009: (Start)

a(n) = a(n-12).

G.f.: -x*(1 + 8*x + 3*x^2 + 4*x^3 + 5*x^4 + 7*x^6 + 8*x^7 + 9*x^8 + 4*x^9 + 11*x^10)/ ((x-1) *(1+x+x ^2) *(1+x) *(1-x+x^2) *(1+x^2) *(x^4-x^2+1)). (End)

Mathematica

Table[Mod[n^3, 12], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)
PowerMod[Range[0, 100], 3, 12] (* Harvey P. Dale, Oct 29 2014 *)

Prog

(Sage) [power_mod(n, 7, 12)for n in range(0, 100)] # Zerinvary Lajos, Oct 28 2009
(Magma) [Modexp(n, 3, 12 ): n in [0..100]]; // Vincenzo Librandi, Mar 27 2016
(PARI) a(n)=n^3%12 \\ Charles R Greathouse IV, Apr 06 2016

Crossrefs

Cf. A167176.

Sequence in context: A021849 A338462 A201754 * A070597 A222232 A091895

Adjacent sequences: A070471 A070472 A070473 * A070475 A070476 A070477

Keyword

nonn,easy

Author

N. J. A. Sloane, May 12 2002

Status

approved