A108412 Expansion of (1 + x + x^2)/(1 - 4x^2 + x^4).

1, 1, 5, 4, 19, 15, 71, 56, 265, 209, 989, 780, 3691, 2911, 13775, 10864, 51409, 40545, 191861, 151316, 716035, 564719, 2672279, 2107560, 9973081, 7865521, 37220045, 29354524, 138907099, 109552575, 518408351, 408855776, 1934726305

Offset: 0

Author

Ralf Stephan, Jun 05 2005

Keywords

• nonn
• easy

Comments

This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 6 and Q = 1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence satisfies a linear recurrence of order four. - Peter Bala, Apr 18 2014

The sequence of convergents of the 2-periodic continued fraction [0; 1, -6, 1, -6, ...] = 1/(1 - 1/(6 - 1/(1 - 1/(6 - ...)))) = 3 - sqrt(3) begins [0/1, 1/1, 6/5, 5/4, 24/19, 19/15, 90/71,...]. The present sequence is the sequence of denominators; the sequence of numerators of the continued fraction convergents [1, 6, 5, 24, 19, 90,...] is also a strong divisibility sequence. Cf. A005013 and A203976. - Peter Bala, May 19 2014

From Peter Bala, Mar 25 2018: (Start)

The following remarks assume an offset of 1.

Define a binary operation o on the real numbers by x o y = x*sqrt(1 + (1/2)*y^2) + y*sqrt(1 + (1/2)*x^2). The operation o is commutative and associative with identity 0. We have a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and sqrt(6)*a(2*n) = (1 o 1 o ... o 1) (2*n terms). Cf. A005013 and A084068. For example, 1 o 1 = sqrt(6) and 1 o 1 o 1 = sqrt(6) o 1 = 5 = a(3).

From the obvious identity ( 1 o 1 o ... o 1 (2*n terms) ) o ( 1 o 1 o ... o 1 (2*m terms) ) = 1 o 1 o ... o 1 (2*n + 2*m terms) we find the relation a(2*n+2*m) = a(2*n)*sqrt(1 + 3*a(2*m)^2) + a(2*m)*sqrt(1 + 3*a(2*n)^2).

Similarly, from a(2*n+1) o a(2*m+1) = sqrt(6)*a(2*n+2*m+2) we find sqrt(6)*a(2*n+2*m+2) = a(2*n+1)*sqrt(1 + (1/2)*a(2*m+1)^2) + a(2*m+1)*sqrt(1 + (1/2)*a(2*n+1)^2). (End)

Examples

			G.f. = 1 + x + 5*x^2 + 4*x^3 + 19*x^4 + 15*x^5 + 71*x^6 + 56*x^7 + ...
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• P. Bala, Notes on 2-periodic continued fractions and Lehmer sequences
• Seong Ju Kim, R. Stees, L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
• Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
• E. W. Weisstein, MathWorld: Lehmer Number
• Index to divisibility sequences
• Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Crossrefs

Cf. A001353, A001834.

Cf. A026741, A005013, A084068.

Programs

Maple

a := proc (n) if `mod`(n, 2) = 1 then 1/sqrt(2)*( ((sqrt(6) + sqrt(2))/2 )^n - ( (sqrt(6) - sqrt(2))/2 )^n) else 1/sqrt(12)*( ((sqrt(6) + sqrt(2))/2 )^n - ( (sqrt(6) - sqrt(2))/2 )^n) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

Mathematica

CoefficientList[Series[(1+x+x^2)/(1-4x^2+x^4),{x,0,40}],x] (* or *) LinearRecurrence[{0,4,0,-1},{1,1,5,4},40] (* Harvey P. Dale, Nov 15 2012 *)

PARI

{a(n) = my( w = quadgen(24)); simplify( polchebyshev( n, 2, w/2) / if( n%2, w, 1))}; /* Michael Somos, Feb 10 2015 */

Formula

a(0)=a(1)=1, a(2)=5, a(n)a(n+3) - a(n+1)a(n+2) = -1.

a(0)=1, a(1)=1, a(2)=5, a(3)=4, a(n) = 4*a(n-2)-a(n-4). - Harvey P. Dale, Nov 15 2012

a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = (1/2)*(sqrt(6) + sqrt(2)) (A188887) and beta = (1/2)*(sqrt(6) - sqrt(2)) (A101263). Equivalently, a(n) = U(n-1,sqrt(6)/2) for n odd and a(n) = (1/sqrt(6))*U(n-1,sqrt(6)/2) for n even, where U(n,x) is the Chebyshev polynomial of the second kind. - Peter Bala, Apr 18 2014

a(2*n) = A001834(n). a(2*n + 1) = A001353(n+1). - Michael Somos, Feb 10 2015

a(n) = -a(-2-n) for all n in Z. - Michael Somos, Feb 10 2015