OFFSET
1,2
COMMENTS
The bisection {1,5,19,265,...} appears to be A001834 and to satisfy the recurrence a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 6 is a square. The other bisection {2,8,30,112,...} appears to be A052530 and one-half of this bisection, {1,4,15,56,...}, appears to be A001353 and to satisfy a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 1 is a square.
Conjecturally, a(n) = x + y, where these values solve x^2 - floor(y^2/3) = 1, see related sequences and formula below. - Richard R. Forberg, Sep 08 2013
Let alpha be an algebraic integer and define a sequence of integers a(alpha,n) by the condition a(alpha,n) = max { integer d : alpha^n = = 1 (mod d)}. Silverman shows that a(alpha,n) is a strong-divisibility sequence, that is gcd(a(n), a(m)) = a(gcd(n, m)) for all n and m in N; in particular, if n divides m then a(n) divides a(m). This sequence appears to be the strong divisibility sequence a(2 + sqrt(3),n) (Silverman, Example 4). - Peter Bala, Jan 10 2014
This sequence appears as the coefficients of the defining inequalities of a polyhedral realization of the B(infinity) crystal of the Kac-Moody Lie algebra with Cartan matrix [2,-2;-3,2] (see Nakashima-Zelevinsky reference). - Paul E. Gunnells, May 05 2019
LINKS
Toshiki Nakashima and Andrei Zelevinsky, Polyhedral realizations of crystal bases for quantized Kac-Moody algebras, Adv. Math. 131 (1997), no. 1, 253-278.
J. H. Silverman, Divisibility sequences and powers of algebraic integers, Documenta Mathematica, Extra Volume: John H. Coates' Sixtieth Birthday (2006) 711-727.
FORMULA
The sequence appears to satisfy a(n) = 4*a(n-2) - a(n-4).
Apparently a(n)*a(n+3) = -2 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
Conjecturally, a(n) = A143643(n-1) + A005246(n), for n => 2, as derived from comment above. - Richard R. Forberg, Sep 08 2013
If the above conjectures are true, then a(n) = A001353(n)/A005246(n+1). - Andrey Zabolotskiy, Sep 26 2024
EXAMPLE
Let S(0) = (1,1). Since 7*1 <= 11*1 we obtain S(1) = (1,2,1). Then since 7*2 > 11*1 and 7*1 <= 11*2, we obtain S(2) = (1,2,3,1). Continuing, we get S(3) = (1,2,5,3,4,1), S(4) = (1,2,5,8,3,7,4,5,1), S(5) = (1,2,5,8,11,3,...), S(6) =(1,2,5,8,19,11,...), etc.
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Layman, May 23 2003
EXTENSIONS
Edited by M. F. Hasler, Nov 06 2018
STATUS
approved