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A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

%I A001091 M3637 N1479 #103 Dec 12 2023 07:47:00
%S A001091 1,4,31,244,1921,15124,119071,937444,7380481,58106404,457470751,
%T A001091 3601659604,28355806081,223244789044,1757602506271,13837575261124,
%U A001091 108942999582721,857706421400644,6752708371622431,53163960551578804
%N A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.
%C A001091 a(15+30k)-1 and a(15+30k)+1 are consecutive odd powerful numbers. The first pair is 13837575261124 +- 1. See A076445. - _T. D. Noe_, May 04 2006
%C A001091 This sequence gives the values of x in solutions of the Diophantine equation x^2 - 15*y^2 = 1. The corresponding y values are in A001090. - _Vincenzo Librandi_, Nov 12 2010 [edited by _Jon E. Schoenfield_, May 04 2014]
%C A001091 The square root of 15*(n^2-1) at those numbers = 5 * A136325. - _Richard R. Forberg_, Nov 22 2013
%C A001091 For the above Diophantine equation x^2-15*y^2=1, x + y = A105426. - _Richard R. Forberg_, Nov 22 2013
%C A001091 a(n) solves for x in the Diophantine equation floor(3*x^2/5)= y^2. The corresponding y solutions are provided by A136325(n).  x + y = A070997(n). - _Richard R. Forberg_, Nov 22 2013
%C A001091 Except for the first term, values of x (or y) in the solutions to x^2 - 8xy + y^2 + 15 = 0. - _Colin Barker_, Feb 05 2014
%D A001091 Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From _N. J. A. Sloane_, May 30 2012
%D A001091 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A001091 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A001091 T. D. Noe, <a href="/A001091/b001091.txt">Table of n, a(n) for n=0..200</a>
%H A001091 H. Brocard, <a href="http://resolver.sub.uni-goettingen.de/purl?PPN598948236_0004/DMDLOG_0053">Notes élémentaires sur le problème de Peel [sic]</a>, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
%H A001091 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A001091 Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H A001091 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H A001091 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H A001091 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-1).
%F A001091 G.f.: A(x) = (1-4*x)/(1-8*x+x^2). - _Simon Plouffe_ in his 1992 dissertation
%F A001091 For all elements x of the sequence, 15*(x^2 -1) is a square. Limit_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(15). - _Gregory V. Richardson_, Oct 11 2002
%F A001091 a(n) = sqrt(15*((A001090(n))^2)+1).
%F A001091 a(n) = ((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
%F A001091 a(n) = 4*S(n-1, 8) - S(n-2, 8) = (S(n, 8) - S(n-2, 8))/2, n>=1; S(n, x) := U(n, x/2) with Chebyshev's polynomials of the 2nd kind, A049310, with S(-1, x) := 0 and S(-2, x) := -1.
%F A001091 a(n) = T(n, 4) with Chebyshev's polynomials of the first kind; see A053120.
%F A001091 a(n)=a(-n). - _Ralf Stephan_, Jun 06 2005
%F A001091 a(n)*a(n+3) - a(n+1)*a(n+2) = 120. - _Ralf Stephan_, Jun 06 2005
%F A001091 From _Peter Bala_, Feb 19 2022: (Start)
%F A001091 a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*15^k*binomial(n,2*k).
%F A001091 a(n) = [x^n] (4*x + sqrt(1 + 15*x^2))^n.
%F A001091 The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 16*x + 4*x^2) is the g.f. of A098269.
%F A001091 The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k. (End)
%F A001091 From _Peter Bala_, Aug 17 2022: (Start)
%F A001091 Sum_{n >= 1} 1/(a(n) - (5/2)/a(n)) = 1/3.
%F A001091 Sum_{n >= 1} (-1)^(n+1)/(a(n) + (3/2)/a(n)) = 1/5.
%F A001091 Sum_{n >= 1} 1/(a(n)^2 - 5/2) =  1/3 - 1/sqrt(15). (End)
%F A001091 a(n) = A001090(n+1)-4*A001090(n). - _R. J. Mathar_, Dec 12 2023
%t A001091 LinearRecurrence[{8,-1},{1,4},20] (* _Harvey P. Dale_, May 01 2014 *)
%o A001091 (PARI) a(n)=subst(poltchebi(n),x,4)
%o A001091 (PARI) a(n)=n=abs(n); polcoeff((1-4*x)/(1-8*x+x^2)+x*O(x^n),n) /* _Michael Somos_, Jun 07 2005 */
%o A001091 (Magma) R<x>:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1-4*x)/(1-8*x+x^2) )); // _G. C. Greubel_, Aug 26 2019
%o A001091 (Sage)
%o A001091 def A001091_list(prec):
%o A001091     P.<x> = PowerSeriesRing(ZZ, prec)
%o A001091     return P( (1-4*x)/(1-8*x+x^2) ).list()
%o A001091 A001091_list(20) # _G. C. Greubel_, Aug 26 2019
%o A001091 (GAP) a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Aug 26 2019
%Y A001091 Cf. A001090, A090965, A098269, A322836 (column 4).
%K A001091 nonn,easy
%O A001091 0,2
%A A001091 _N. J. A. Sloane_
%E A001091 More terms from Larry Reeves (larryr(AT)acm.org), Aug 25 2000
%E A001091 Chebyshev comments from _Wolfdieter Lang_, Oct 31 2002