A245480 Numbers n such that the n-th cyclotomic polynomial has a root mod 11.

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

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 *)
LinearRecurrence[{0, 0, 0, 11}, {1, 2, 5, 10}, 40] (* Harvey P. Dale, Aug 04 2021 *)

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 A351062 A182656

Adjacent sequences: A245477 A245478 A245479 * A245481 A245482 A245483

Keyword

nonn,easy

Author

Eric M. Schmidt, Jul 23 2014

Status

approved