OFFSET
0,3
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)
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Seong Ju Kim, R. Stees, and 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.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
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(n) = -a(-2-n) for all n in Z. - Michael Somos, Feb 10 2015
EXAMPLE
G.f. = 1 + x + 5*x^2 + 4*x^3 + 19*x^4 + 15*x^5 + 71*x^6 + 56*x^7 + ...
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 *)
PROG
(PARI) {a(n) = my( w = quadgen(24)); simplify( polchebyshev( n, 2, w/2) / if( n%2, w, 1))}; /* Michael Somos, Feb 10 2015 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ralf Stephan, Jun 05 2005
STATUS
approved