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%I #56 Sep 26 2021 00:54:16
%S 1,4,23,134,781,4552,26531,154634,901273,5253004,30616751,178447502,
%T 1040068261,6061962064,35331704123,205928262674,1200237871921,
%U 6995498968852,40772755941191,237641036678294,1385073464128573
%N a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.
%C This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - _Wolfdieter Lang_, Feb 26 2015
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
%H I. Adler, <a href="http://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), pp. 181-193.
%H Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, <a href="https://arxiv.org/abs/2004.04535">Symmetries of biplanes</a>, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
%H E. I. Emerson, <a href="http://www.fq.math.ca/Scanned/7-3/emerson.pdf">Recurrent Sequences in the Equation DQ^2=R^2+N</a>, Fib. Quart., 7 (1969), pp. 231-242.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - _Antonio Alberto Olivares_, Mar 29 2008
%F a(n) = A001653(n+1) - A001109(n). - _Antonio Alberto Olivares_, Mar 29 2008
%F Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - _Michael Somos_, Sep 28 2008
%F G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - _Michael Somos_, Sep 28 2008
%F a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - _Philippe Deléham_, Mar 05 2014
%F a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - _Wolfdieter Lang_, Feb 26 2015
%F a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - _Wolfdieter Lang_, Mar 19 2015
%F E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - _Stefano Spezia_, Apr 30 2020
%e n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
%e 65^2 - 2*(2*23)^2 = -7,
%e (4*a(2))^2 - 2*A054490(2)^2 =
%e (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
%e a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015
%p a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # _Zerinvary Lajos_, Jul 26 2006
%t LinearRecurrence[{6,-1},{1,4},30] (* _Harvey P. Dale_, Aug 06 2020 *)
%o (PARI) {a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* _Michael Somos_, Sep 28 2008 */
%o (PARI) {a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* _Michael Somos_, Sep 28 2008 */
%Y Cf. A001109, A001541, A001653, A054490.
%Y A038725(n) = a(-n).
%K easy,nonn
%O 0,2
%A _Barry E. Williams_, May 02 2000
%E More terms from _James A. Sellers_, May 03 2000
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