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

Author

John W. Layman, May 23 2003

Keywords

• nonn

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

From Zhuorui He, Jul 16 2025: (Start)

This sequence is Ratio-determined insertion sequence I(7/11) (see the Layman link below).

If S(0) in the definition is (1,1,a,b,c...) (all numbers >= 0) instead of (1,1), the resulting sequence is still the same.

For a finite sequence S, let k be the least i such that 7*S(i+1) <= 11*S(i). If k didn't exist then I(S)=S. Else, let k' be the least i such that 7*I(S)(i+1) <= 11*I(S)(i). Then k <= k' <= k+1.

This sequence can be generated by this process:

Step 1: Let X=1 and Y=1.

Step 2: If 7*(X+Y)<=11*X, then Y:=X+Y, repeat this step. Else go to step 3.

Step 3: Append X to the sequence. Let X:=X+Y, go back to step 2. (End)

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.

Crossrefs

Cf. A001353, A005246, A001834, A001835, A052530.

Programs

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