%I #22 Feb 16 2025 08:33:23
%S 1,2,5,10,11,22,55,110,121,242,605,1210,1331,2662,6655,13310,14641,
%T 29282,73205,146410,161051,322102,805255,1610510,1771561,3543122,
%U 8857805,17715610,19487171,38974342,97435855,194871710,214358881,428717762,1071794405,2143588810
%N Numbers n such that the n-th cyclotomic polynomial has a root mod 11.
%C Numbers of the form d*11^j for d=1,2,5,10.
%D Trygve Nagell, Introduction to Number Theory. New York: Wiley, 1951, pp. 164-168.
%H Eric M. Schmidt, <a href="/A245480/b245480.txt">Table of n, a(n) for n = 1..500</a>
%H Eric Weisstein, <a href="https://mathworld.wolfram.com/CyclotomicPolynomial.html">Cyclotomic Polynomial</a>.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,11).
%F From _Benedict W. J. Irwin_, Jul 29 2016: (Start)
%F a(n) = 11*a(n-4).
%F G.f.: x*(1 + 2*x)*(1 + 5*x^2)/(1 - 11*x^4).
%F a(n) appears to satisfy x*Prod_{n>=0} (1 + a(2^n+1)x^(2^n)) = Sum_{n>=1} a(n)*x^n.
%F Then a(n+1) = a(2^x+1)*a(2^y+1)*a(2^z+1)..., where n=2^x+2^y+2^z+... .
%F 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.
%F (End)
%e 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.
%t CoefficientList[Series[x(2x+1)(5x^2+1)/(1-11x^4), {x, 0, 20}], x] (* _Benedict W. J. Irwin_, Jul 24 2016 *)
%t LinearRecurrence[{0,0,0,11},{1,2,5,10},40] (* _Harvey P. Dale_, Aug 04 2021 *)
%o (Sage) def A245480(n) : return [10,1,2,5][n%4]*11^((n-1)//4)
%o (PARI) for(n=1,10^6,if(#polrootsmod(polcyclo(n),11),print1(n,", "))) /* by definition; rather inefficient. - _Joerg Arndt_, Jul 28 2014 */
%o (PARI) a(n)=11^((n-1)\4)*[10,1,2,5][n%4+1] \\ _Charles R Greathouse IV_, Jun 11 2015
%Y Cf. A000079, A038754, A245478, A245479, A245481.
%K nonn,easy,changed
%O 1,2
%A _Eric M. Schmidt_, Jul 23 2014