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%I #44 Feb 11 2024 04:36:31
%S 1,3,9,19,53,111,309,647,1801,3771,10497,21979,61181,128103,356589,
%T 746639,2078353,4351731,12113529,25363747,70602821,147830751,
%U 411503397,861620759,2398417561,5021893803,13979001969,29269742059,81475594253
%N 2*a(n)^2 + 7 is a square.
%C Lim. n -> Inf. a(n)/a(n-2) = 3 + 2*Sqrt(2) = R1*R2. Lim. k -> Inf. a(2*k-1)/a(2*k) = (9 + 4*Sqrt(2))/7 = R1 (ratio #1). Lim. k -> Inf. a(2*k)/a(2*k-1) = (11 + 6*Sqrt(2))/7 = R2 (ratio #2).
%C a(n) gives for n >= 0 all positive y-values solving the (generalized) Pell equation x^2 - 2*y^2 = 7. A077443(n+1) gives the corresponding x-values. See, e.g., the Nagell reference on how to find all solutions. - _Wolfdieter Lang_, Feb 05 2015
%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
%D A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
%D Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
%D T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.
%H Colin Barker, <a href="/A077442/b077442.txt">Table of n, a(n) for n = 0..1000</a>
%H J. J. O'Connor and E. F. Robertson, <a href="https://web.archive.org/web/20170729132724/http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Pell.html">History of Pell's Equation</a>
%H J. P. Robertson, <a href="https://web.archive.org/web/20180831180333/http://www.jpr2718.org/pell.pdf">Solving the Generalized Pell Equation</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PellEquation.html">Pell Equation.</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).
%F For n>0, a(2n) = A046090(n) + A001653(n) + A001652(n-1); a(2n+1) = A001652(n+1) - A001652(n-1) - A001653(n-1); e.g. 53=21+29+3; 111=119-3-5. - _Charlie Marion_, Aug 14 2003
%F The same recurrences hold for the odd and even indices respectively : a(n+2) = 6*a(n+1) - a(n), a(n+1) = 3*a(n) + 2*(2*a(n)^2+7)^0.5. - _Richard Choulet_, Oct 11 2007
%F G.f.: (x+1)^3/(x^2+2*x-1)/(x^2-2*x-1). a(n)= [ -A077985(n)-3*A077985(n-1)+3*A000129(n+1)+A000129(n)]/2. - _R. J. Mathar_, Nov 16 2007
%F a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=3, a(3)=9, a(4)=19. - _Sture Sjöstedt_, Oct 08 2012
%F a(n) = ((-(-1 - sqrt(2))^n*(-2+sqrt(2)) - (-1+sqrt(2))^n*(2+sqrt(2)) + (1-sqrt(2))^n*(-4+3*sqrt(2)) + (1+sqrt(2))^n*(4+3*sqrt(2))))/(4*sqrt(2)). - _Colin Barker_, Mar 27 2016
%e a(4)^2 - 2*a(3)^2 = 27^2 - 2*19^2 = +7. - _Wolfdieter Lang_, Feb 05 2015
%t CoefficientList[Series[(1+3 x+3 x^2+x^3)/ (1-6 x^2+x^4),{x,0,50}],x] (* _Harvey P. Dale_, Mar 12 2011 *)
%t LinearRecurrence[{0, 6, 0, -1},{1,3,9,19},50] (* _Sture Sjöstedt_, Oct 08 2012 *)
%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;3;9;19])[1,1] \\ _Charles R Greathouse IV_, Jun 20 2015
%o (PARI) Vec((x+1)^3/(x^2+2*x-1)/(x^2-2*x-1) + O(x^50)) \\ _Colin Barker_, Mar 27 2016
%Y Cf. A077443, A038762, A038761, A101386, A253811.
%K nonn,easy
%O 0,2
%A _Gregory V. Richardson_, Nov 06 2002
%E Edited: n in Name replaced by a(n). Pell comments moved to comment section. - _Wolfdieter Lang_, Feb 05 2015
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