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 A036117 a(n) = 2^n mod 11. 12
 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(k) = k has only one solution, namely k=7. - Jon Perry, Oct 30 2014 As 2 is a primitive root of 11, all integers 1 through 10 are present. - Jon Perry, Oct 30 2014 REFERENCES H. Cohn, A Second Course in Number Theory, Wiley, NY, 1962, p. 256. I. M. Vinogradov, Elements of Number Theory, pp. 220 ff. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Wikipedia, Primitive roots Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,-1,1). FORMULA a(n) = a(n-1) - a(n-5) + a(n-6). - R. J. Mathar, Apr 13 2010 G.f.: (1+x+2*x^2+4*x^3-3*x^4+6*x^5)/ ((1-x) * (1+x) * (x^4-x^3+x^2-x+1)). - R. J. Mathar, Apr 13 2010 a(n+10) = a(n). - Jon Perry, Oct 30 2014 a(n+5) = 11 - a(n) for all n in Z. - Michael Somos, Oct 17 2018 EXAMPLE 2^6 = 64 = 66 - 2 == -2 mod 11 == 9 mod 11, so a(6) = 9. MAPLE i := 5: [ seq(numtheory[primroot](ithprime(i))^j mod ithprime(i), j=0..100) ]; MATHEMATICA Table[Mod[2^n, 11], {n, 0, 6!}] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2010 *) PROG (Sage) [power_mod(2, n, 11) for n in range(0, 78)] # Zerinvary Lajos, Nov 03 2009 (PARI) a(n)=lift(Mod(2, 11)^n) \\ Charles R Greathouse IV, Jul 02 2013 (Magma) [Modexp(2, n, 11): n in [0..100]]; // G. C. Greubel, Oct 16 2018 (GAP) List([0..70], n->PowerMod(2, n, 11)); # Muniru A Asiru, Oct 18 2018 CROSSREFS Cf. A000079 (2^n), A008830, A168429. Sequence in context: A227818 A300890 A021893 * A307357 A116624 A361191 Adjacent sequences: A036114 A036115 A036116 * A036118 A036119 A036120 KEYWORD nonn,easy AUTHOR N. J. A. Sloane STATUS approved

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Last modified June 1 17:30 EDT 2023. Contains 363076 sequences. (Running on oeis4.)