A105838 Nonnegative integers n such that 11*n^2 + 11*n + 1 is a square.
0, 39, 159, 15720, 63480, 6256719, 25265079, 2490158640, 10055438160, 991076882199, 4002039122799, 394446108956760, 1592801515436040, 156988560287908479, 633931001104421319, 62481052548478618080, 252302945638044249120, 24867301925734202087559
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (1,398,-398,-1,1).
Crossrefs
Cf. A105837 (square roots of 11*a(n)^2+11*a(n)+1).
Cf. similar sequences indexed in A222390. [Bruno Berselli, Feb 20 2013]
Programs
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Magma
m:=17; R
:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(3*(13+40*x+13*x^2)/((1-x)*(1-20*x+x^2)*(1+20*x+x^2)))); // Bruno Berselli, Feb 20 2013 -
Mathematica
LinearRecurrence[{1, 398, -398, -1, 1}, {0, 39, 159, 15720, 63480}, 18] (* Bruno Berselli, Feb 20 2013 *) -
Maxima
makelist(expand(-1/2+((11+2*sqrt(11)*(-1)^n)*(10-3*sqrt(11))^(2*floor(n/2))+(11-2*sqrt(11)*(-1)^n)*(10+3*sqrt(11))^(2*floor(n/2)))/44), n, 1, 18); /* Bruno Berselli, Feb 20 2013 */
Formula
a(1)=0, a(2)=39, a(3)=398*a(1)+198-a(2), a(4)=398*a(2)+198-a(1), a(n) = 398*a(n-2) + 198 - a(n-4) for n>4.
From Bruno Berselli, Feb 20 2013: (Start)
G.f.: 3*x*(13+40*x+13*x^2)/((1-x)*(1-20*x+x^2)*(1+20*x+x^2)).
a(n) = a(-n+1) = -1/2+((11+2*t*(-1)^n)*(10-3*t)^(2*floor(n/2))+(11-2*t*(-1)^n)*(10+3*t)^(2*floor(n/2)))/44, where t=sqrt(11). (End)
Extensions
More terms from Bruno Berselli, Feb 20 2013