login
A070474
a(n) = n^3 mod 12, n^5 mod 12.
4
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
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
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, May 12 2002
STATUS
approved