A082630 Limit of the sequence obtained from S(0) = (1,1) and, for n > 0, S(n) = I(S(n-1)), where I consists of inserting, for i = 1, 2, 3..., the term a(i) + a(i+1) between any two terms for which 7*a(i+1) <= 11*a(i).
1, 2, 5, 8, 19, 30, 71, 112, 265, 418, 989, 1560, 3691, 5822, 13775, 21728, 51409, 81090, 191861, 302632, 716035, 1129438, 2672279, 4215120, 9973081, 15731042, 37220045, 58709048, 138907099, 219105150, 518408351, 817711552, 1934726305, 3051741058, 7220496869
Offset: 1
Keywords
Examples
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.
Links
- Zhuorui He, Table of n, a(n) for n = 1..1000
- John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types
- John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types [local copy, corrected]
- 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.
Programs
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Python
def A082630_list(n): a = [] x = y = 1 while len(a) < n: a.append(x) while 7*(x+y) <= 11*x: y += x x += y return a # Zhuorui He, Jul 16 2025
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
Extensions
Edited by M. F. Hasler, Nov 06 2018
Comments