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%I #34 Sep 08 2022 08:44:50
%S 1,35,1190,40426,1373295,46651605,1584781276,53835911780,
%T 1828836219245,62126595542551,2110475412227490,71694037420192110,
%U 2435486796874304251,82734857056306152425,2810549653117534878200,95475953348939879706376,3243371864210838375138585
%N Expansion of 1/( (1-x)*(1-34*x+x^2) ).
%C Numbers m such that r = 24*m+1 and 2*r-1 are both squares. - _Bruno Berselli_, Jul 17 2014
%H Vincenzo Librandi, <a href="/A029546/b029546.txt">Table of n, a(n) for n = 0..200</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F a(n) = A029549(n+1)/6 = A075528(n+1)/3.
%F From _Colin Barker_, Mar 02 2016: (Start)
%F a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n > 2.
%F a(n) = ( (99-70*sqrt(2))*(17-12*sqrt(2))^n - 6 + (99+70*sqrt(2))*(17+12*sqrt(2))^n )/192. (End)
%F a(n) = ( Pell(2*n+3)^2 - 1)/24 = (Q(4*n+6) - 6)/192, where Q(n) = Pell-Lucas numbers. - _G. C. Greubel_, Jan 13 2020
%p seq(coeff(series(1/( (1-x)*(1-34*x+x^2) ), x, n+1), x, n), n = 0..20); # _G. C. Greubel_, Jan 13 2020
%t LinearRecurrence[{35,-35,1},{1,35,1190},20] (* _Vincenzo Librandi_, Nov 22 2011 *)
%t Table[(Fibonacci[2*n+3, 2]^2 -1)/24, {n,0,20}] (* _G. C. Greubel_, Jan 13 2020 *)
%o (Magma) I:=[1,35,1190]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2) +Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Nov 22 2011
%o (PARI) Vec(1/(1-35*x+35*x^2-x^3)+O(x^20)) \\ _Charles R Greathouse IV_, Sep 23 2012
%o (Sage) [(lucas_number2(4*n+6, 2,-1) -6)/192 for n in (0..20)] # _G. C. Greubel_, Jan 13 2020
%o (GAP) List([0..20], n-> (Lucas(2,-1, 4*n+6)[2] -6)/192 ); # _G. C. Greubel_, Jan 13 2020
%Y Cf. A029547 (first differences), A245031 (see Comments line).
%Y Cf. A000129, A002203.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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