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%I #114 Mar 17 2023 09:55:52
%S 2,7,47,322,2207,15127,103682,710647,4870847,33385282,228826127,
%T 1568397607,10749957122,73681302247,505019158607,3461452808002,
%U 23725150497407,162614600673847,1114577054219522,7639424778862807,52361396397820127,358890350005878082,2459871053643326447
%N a(n) = Lucas(4*n).
%C a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - _Wolfdieter Lang_, Jun 26 2013
%C Lucas numbers of the form n^2-2. - _Michel Lagneau_, Aug 11 2014
%D R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
%H G. C. Greubel, <a href="/A056854/b056854.txt">Table of n, a(n) for n = 0..1190</a>
%H Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%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 Dale Gerdemann, <a href="http://www.fq.math.ca/Papers1/46_47-3/Gerdemann2.pdf">Combinatorial proofs of Zeckendorf family identities</a>, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
%H A. F. Horadam, <a href="http://www.fq.math.ca/Scanned/5-5/horadam.pdf">Special Properties of the Sequence W(n){a,b; p,q}</a>, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A. V. Zarelua, <a href="https://doi.org/10.1007/s11006-006-0090-y">On Matrix Analogs of Fermat's Little Theorem</a>, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-1).
%F a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
%F a(n) = A000032(4*n), where A000032 = Lucas numbers.
%F a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
%F a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
%F G.f.: (2-7x)/(1-7x+x^2).
%F a(n) = A005248(2*n); bisection of A005248.
%F a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - _Gary Detlefs_, Dec 26 2010
%F a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - _Wolfdieter Lang_, Aug 27 2012
%F From _Peter Bala_, Jan 06 2013: (Start)
%F Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
%F Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
%F F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
%F Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
%F a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - _Bruno Berselli_, May 25 2015
%F a(n) = sqrt((45*(A004187(n))^2)+4.
%F From _Peter Bala_, Oct 13 2019: (Start)
%F a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
%F a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
%F Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) (mod p^k) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
%F 5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
%F 9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
%F Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
%F Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
%F x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
%F E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - _Stefano Spezia_, Oct 18 2019
%F a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - _Greg Dresden_ and _Tracy Z. Wu_, Sep 10 2020
%e Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper); n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013
%t a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* _Robert G. Wilson v_, Jan 30 2004 *)
%t LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* _Harvey P. Dale_, Aug 08 2011 *)
%o (PARI) a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])
%o (PARI) a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))
%o (Sage) [lucas_number2(n,7,1) for n in range(27)] #_Zerinvary Lajos_, Jun 25 2008
%o (Magma) [Lucas(4*n): n in [0..100]]; // _Vincenzo Librandi_, Apr 14 2011
%Y Cf. A004187, A005248.
%Y Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
%Y Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).
%K nonn,easy
%O 0,1
%A _Barry E. Williams_, Aug 29 2000
%E More terms from _James A. Sellers_, Aug 31 2000
%E Chebyshev comments from _Wolfdieter Lang_, Oct 31 2002
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