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A070511 a(n) = n^4 mod 6. 1
0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

a(n) = A070431(n). Proof: n^4-n^2 == 0 (mod 6) is shown explicitly for n = 0..5, then the induction n -> n+6 for the 4th order polynomial followed by binomial expansion of (n+6)^k concludes that the zero (mod 6) is periodically extended to the other integers. -  R. J. Mathar, Jul 23 2009

Equivalently n^6 mod 6. - Zerinvary Lajos, Nov 06 2009

Equivalently: n^(2*m + 4) mod 6; n^(2*m + 2) mod 6. - G. C. Greubel, Apr 01 2016

LINKS

Table of n, a(n) for n=0..100.

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

FORMULA

G.f. -x*(1+4*x+3*x^2+4*x^3+x^4) / ( (x-1)*(1+x)*(1+x+x^2)*(x^2-x+1) ). - R. J. Mathar, Mar 14 2011

From G. C. Greubel, Apr 01 2016: (Start)

a(6*m) = 0.

a(2*n) = 4*A011655(n).

a(n) = (1/6)*(13 + 3*(-1)^n - 12*cos(n*Pi/3) - 4*cos(2*n*Pi/3)).

G.f.: (x +4*x^2 +3*x^3 + 4*x^4 +x^5)/(1 - x^6). (End)

MATHEMATICA

Table[Mod[n^4, 6], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)

PowerMod[Range[0, 100], 4, 6] (* Bruno Berselli, Mar 31 2016 *)

PROG

(Sage) [power_mod(n, 4, 6) for n in xrange(0, 101)] # Zerinvary Lajos, Oct 30 2009

(MAGMA) [Modexp(n, 4, 6): n in [0..100]]; // Bruno Berselli, Mar 31 2016

(PARI) a(n)=n^4%6 \\ Charles R Greathouse IV, Apr 06 2016

CROSSREFS

Cf. A070430, A070431.

Sequence in context: A243149 A048156 A070431 * A066340 A195597 A143505

Adjacent sequences:  A070508 A070509 A070510 * A070512 A070513 A070514

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane, May 13 2002

STATUS

approved

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Last modified October 18 02:23 EDT 2019. Contains 328135 sequences. (Running on oeis4.)