A005319 a(n) = 6*a(n-1) - a(n-2).

0, 4, 24, 140, 816, 4756, 27720, 161564, 941664, 5488420, 31988856, 186444716, 1086679440, 6333631924, 36915112104, 215157040700, 1254027132096, 7309005751876, 42600007379160, 248291038523084, 1447146223759344, 8434586304032980

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Solutions y of the equation 2x^2-y^2=2; the corresponding x values are given by A001541. - N-E. Fahssi, Feb 25 2008

The lower intermediate convergents to 2^(1/2) beginning with 4/3, 24/17, 140/99, 816/577, form a strictly increasing sequence; essentially, numerators=A005319 and denominators=A001541. - Clark Kimberling, Aug 26 2008

Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 1 + n*n/2. - Ctibor O. Zizka, Nov 09 2009

All nonnegative solutions of the indefinite binary quadratic form X^2 + 4*X*Y -4*Y^2 of discriminant 32, representing -4 are (X(n), Y(n)) = (a(n), A001653(n+1)), for n >= 0. - Wolfdieter Lang, Jun 13 2018

Also the number of edge covers in the n-triangular snake graph. - Eric W. Weisstein, Jun 08 2019

All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

a(n) is the sum of 4*n consecutive powers of the silver ratio 1+sqrt(2), starting at (1+sqrt(2))^(-2*n) and ending at (1+sqrt(2))^(2*n-1). - Greg Dresden and Ruxin Sheng, Jul 25 2024

Examples

			G.f. = 4*x + 24*x^2 + 140*x^3 + 816*x^4 + 4756*x^5 + ... - _Michael Somos_, Jun 26 2022
		

References

• P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 160, middle display.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Marius A. Burtea, Table of n, a(n) for n = 0..100
• K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
• Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
• John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
• Tanya Khovanova, Recursive Sequences
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
• Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
• Eric Weisstein's World of Mathematics, Edge Cover
• Eric Weisstein's World of Mathematics, Triangular Snake Graph
• Index entries for linear recurrences with constant coefficients, signature (6,-1).

Crossrefs

Cf. A000129, A001109, A001541, A001542, A001652, A001653, A046090, A046729, A090390.

Programs

Magma

a:=[0,4]; [n le 2 select a[n] else 6*Self(n-1) - Self(n-2):n in [1..22]]; // Marius A. Burtea, Sep 19 2019

Mathematica

LinearRecurrence[{6, -1}, {0, 4}, 22] (* Jean-François Alcover, Sep 26 2017 *)
Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/Sqrt[2], {n, 20}] // Expand (* Eric W. Weisstein, Jun 08 2019 *)
CoefficientList[Series[(4 x)/(1 - 6 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 08 2019 *)
a[ n_] := 4*ChebyshevU[n-1, 3]; (* Michael Somos, Jun 26 2022 *)

PARI

{a(n) = 4*polchebyshev(n-1, 2, 3)}; /* Michael Somos, Jun 26 2022 */

Formula

G.f.: 4*x / ( 1-6*x+x^2 ). - Simon Plouffe in his 1992 dissertation.

G.f. for signed version beginning with 1: (1+2*x+x^2)/(1+6*x+x^2).

For any term n of the sequence, 2*n^2 + 4 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002

a(n) = ((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) / sqrt(2). - Gregory V. Richardson, Oct 06 2002

(-1)^(n+1) = A090390(n+1) + A001542(n+1) + A046729(n) - a(n) (conjectured). - Creighton Dement, Nov 17 2004

For n > 0, a(n) = A000129(n+1)^2 - A000129(n-1)^2; a(n) = A046090(n-1) + A001652(n); e.g., 816 = 120 + 696; a(n) = A001653(n) - A001653(n-1); e.g., 816 = 985 - 169. - Charlie Marion Jul 22 2005

a(n) = 4*A001109(n). - M. F. Hasler, Mar 2009

For n > 1, a(n) is the denominator of continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the numerators, see A001653. - Greg Dresden, Sep 10 2019

1/a(n) - 1/a(n+1) = 1/(Pell(2*n+1) - 1/Pell(2*n+1)) for n >= 1, where Pell(n) = A000129(n). - Peter Bala, Aug 21 2022

E.g.f.: sqrt(2)*exp(3*x)*sinh(2*sqrt(2)*x). - Stefano Spezia, Nov 25 2022

a(n) = 2*A000129(2*n). - Tanya Khovanova and MIT PRIMES STEP senior group, Apr 17 2024