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 A245480 Numbers n such that the n-th cyclotomic polynomial has a root mod 11. 6
 1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605, 1210, 1331, 2662, 6655, 13310, 14641, 29282, 73205, 146410, 161051, 322102, 805255, 1610510, 1771561, 3543122, 8857805, 17715610, 19487171, 38974342, 97435855, 194871710, 214358881, 428717762, 1071794405, 2143588810 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Numbers of the form d*11^j for d=1,2,5,10. REFERENCES Trygve Nagell, Introduction to Number Theory. New York: Wiley, 1951, pp. 164-168. LINKS Eric M. Schmidt, Table of n, a(n) for n = 1..500 Eric Weisstein, Cyclotomic Polynomial. Index entries for linear recurrences with constant coefficients, signature (0,0,0,11). FORMULA From Benedict W. J. Irwin, Jul 29 2016: (Start) a(n) = 11*a(n-4). G.f.: x*(1 + 2*x)*(1 + 5*x^2)/(1 - 11*x^4). a(n) appears to satisfy x*Prod_{n>=0} (1 + a(2^n+1)x^(2^n)) = Sum_{n>=1} a(n)*x^n. Then a(n+1) = a(2^x+1)*a(2^y+1)*a(2^z+1)..., where n=2^x+2^y+2^z+... . For example, n=31=2^0+2^1+2^2+2^3+2^4, then a(31+1)=a(2)*a(3)*a(5)*a(9)*a(17) i.e. 194871710=2*5*11*121*14641. (End) EXAMPLE The 5th cyclotomic polynomial x^4 + x^3 + x^2 + x + 1 considered modulo 11 has a root x = 3, so 5 is in the sequence. MATHEMATICA CoefficientList[Series[x(2x+1)(5x^2+1)/(1-11x^4), {x, 0, 20}], x] (* Benedict W. J. Irwin, Jul 24 2016 *) PROG (Sage) def A245480(n) : return [10, 1, 2, 5][n%4]*11^((n-1)//4) (PARI) for(n=1, 10^6, if(#polrootsmod(polcyclo(n), 11), print1(n, ", "))) /* by definition; rather inefficient. - Joerg Arndt, Jul 28 2014 */ (PARI) a(n)=11^((n-1)\4)*[10, 1, 2, 5][n%4+1] \\ Charles R Greathouse IV, Jun 11 2015 CROSSREFS Cf. A000079, A038754, A245478, A245479, A245481. Sequence in context: A196168 A018514 A018288 * A080792 A182656 A090050 Adjacent sequences:  A245477 A245478 A245479 * A245481 A245482 A245483 KEYWORD nonn,easy AUTHOR Eric M. Schmidt, Jul 23 2014 STATUS approved

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Last modified July 30 04:06 EDT 2021. Contains 346348 sequences. (Running on oeis4.)