A070477 a(n) = n^3 mod 15.

0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14

Offset

0,3

Comments

It is also true that a(n) = n^(4k+3) mod 15 holds any k. - Gary Detlefs, Dec 15 2021

Links

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

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

Formula

From G. C. Greubel, Mar 28 2016: (Start)

a(n) = a(n-15).

G.f.: (-x -8*x^2 -12*x^3 -4*x^4 -5*x^5 -6*x^6 -13*x^7 -2*x^8 -9*x^9 - 10*x^10 -11*x^11 -3*x^12 -7*x^13 -14*x^14)/(-1 + x^15). (End)

Mathematica

PowerMod[Range[0, 90], 3, 15] (* or *) PadRight[{}, 90, {0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14}] (* Harvey P. Dale, Jan 27 2014 *)
Table[Mod[n^3, 15], {n, 0, 100}] (* Vincenzo Librandi, Jun 19 2014 *)

Prog

(Sage) [power_mod(n, 3, 15) for n in range(0, 90)] # Zerinvary Lajos, Oct 29 2009
(Magma) [Modexp(n, 3, 15): n in [0..100]]; // Vincenzo Librandi, Mar 28 2016
(PARI) a(n)=n^3%15 \\ Charles R Greathouse IV, Apr 06 2016

Crossrefs

Sequence in context: A172391 A227239 A037449 * A070697 A004473 A114404

Adjacent sequences: A070474 A070475 A070476 * A070478 A070479 A070480

Keyword

nonn,easy

Author

N. J. A. Sloane, May 12 2002

Status

approved