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A198947 x values in the solution to 11*x^2 - 10 = y^2.

%I A198947 #42 Nov 01 2024 20:48:57
%S A198947 1,7,13,139,259,2773,5167,55321,103081,1103647,2056453,22017619,
%T A198947 41025979,439248733,818463127,8762957041,16328236561,174819892087,
%U A198947 325746268093,3487634884699,6498597125299,69577877801893,129646196237887,1388069921153161,2586425327632441
%N A198947 x values in the solution to 11*x^2 - 10 = y^2.
%C A198947 When are n and 11*n+1 perfect squares? This problem gives rise to the Diophantine equation 11*x^2 - 10 = y^2.
%C A198947 Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 90 = 0. - _Colin Barker_, Feb 18 2014
%H A198947 Vincenzo Librandi, <a href="/A198947/b198947.txt">Table of n, a(n) for n = 1..1000</a>
%H A198947 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 20, 0, -1).
%F A198947 a(n+4) = 20*a(n+2) - a(n) with a(1)=1, a(2)=7, a(3)=13, a(4)=139.
%F A198947 From _Bruno Berselli_, Nov 06 2011:  (Start)
%F A198947 G.f.: x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4).
%F A198947 a(n) = ((11+(-1)^n*t)*(10-3*t)^floor(n/2)+(11-(-1)^n*t)*(10+3*t)^floor(n/2))/22 with t=sqrt(11).  (End).
%t A198947 LinearRecurrence[{0,20,0,-1},{1,7,13,139},30] (* _Vincenzo Librandi_, Feb 06 2012 *)
%o A198947 (PARI) v=vector(25); v[1]=1; v[2]=7; v[3]=13; v[4]=139; for(i=5, #v, v[i]=20*v[i-2]-v[i-4]); v \\ _Bruno Berselli_, Nov 07 2011
%o A198947 (Magma) m:=26; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4))); // _Bruno Berselli_, Nov 07 2011
%o A198947 (Maxima) makelist(expand(((11+(-1)^n*sqrt(11))*(10-3*sqrt(11))^floor(n/2)+(11-(-1)^n*sqrt(11))*(10+3*sqrt(11))^floor(n/2))/22), n, 1, 25); /* _Bruno Berselli_, Nov 07 2011 */
%Y A198947 Cf. A198949, A221762.
%K A198947 nonn,easy
%O A198947 1,2
%A A198947 _Sture Sjöstedt_, Oct 31 2011
%E A198947 Terms a(1)-a(7) confirmed, a(8)-a(15) added by _John W. Layman_, Nov 04 2011
%E A198947 a(16)-a(25) from _Bruno Berselli_, Nov 06 2011