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 A056854 a(n) = Lucas(4*n). 21
 2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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 Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1190 E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242. [ Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [N. J. A. Sloane, Dec 05 2009] A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434. Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (7,-1) FORMULA a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7. a(n) = A000032(4*n), where A000032 = Lucas numbers. 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. a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n. G.f.: (2-7x)/(1-7x+x^2). a(n) = A005248(2*n); bisection of A005248. a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs Dec 26 2010 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 From Peter Bala, Jan 06 2013: (Start) 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 + ...))). 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(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 + ...))))))). (End) a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. Bruno Berselli, May 25 2015 a(n) = sqrt((45*(A004187(n))^2)+4. EXAMPLE 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 MATHEMATICA 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 *) LinearRecurrence[{7, -1}, {2, 7}, 25] (* or *) LucasL[4*Range[0, 25]] (* Harvey P. Dale, Aug 08 2011 *) PROG (PARI) a(n)=if(n<0, 0, polsym(1-7*x+x^2, n)[n+1]) (PARI) a(n)=if(n<0, 0, 2*subst(poltchebi(n), x, 7/2)) (Sage) [lucas_number2(n, 7, 1) for n in range(27)] - Zerinvary Lajos, Jun 25 2008 (MAGMA) [Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011 CROSSREFS Cf. A004187, A005248. Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth). Sequence in context: A290488 A276649 A091117 * A117141 A305533 A125813 Adjacent sequences:  A056851 A056852 A056853 * A056855 A056856 A056857 KEYWORD nonn,easy AUTHOR Barry E. Williams, Aug 29 2000 EXTENSIONS More terms from James A. Sellers, Aug 31 2000 Chebyshev comments from Wolfdieter Lang, Oct 31 2002 STATUS approved

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Last modified October 21 17:10 EDT 2018. Contains 316427 sequences. (Running on oeis4.)