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
KEYWORD
nonn,easy
AUTHOR
Eric M. Schmidt, Jul 23 2014
STATUS
approved