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

Author

Eric M. Schmidt, Jul 23 2014

Keywords

• nonn
• easy

Comments

Numbers of the form d*11^j for d=1,2,5,10.

Examples

			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.
		

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).

Crossrefs

Cf. A000079, A038754, A245478, A245479, A245481.

Programs

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 *)

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

Sage

def A245480(n) : return [10,1,2,5][n%4]*11^((n-1)//4)

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)