cp's OEIS Frontend

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).

%I A082630 #68 Jul 23 2025 21:09:01
%S A082630 1,2,5,8,19,30,71,112,265,418,989,1560,3691,5822,13775,21728,51409,
%T A082630 81090,191861,302632,716035,1129438,2672279,4215120,9973081,15731042,
%U A082630 37220045,58709048,138907099,219105150,518408351,817711552,1934726305,3051741058,7220496869
%N 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).
%C A082630 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.
%C A082630 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
%C A082630 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
%C A082630 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
%C A082630 From _Zhuorui He_, Jul 16 2025: (Start)
%C A082630 This sequence is Ratio-determined insertion sequence I(7/11) (see the Layman link below).
%C A082630 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.
%C A082630 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.
%C A082630 This sequence can be generated by this process:
%C A082630 Step 1: Let X=1 and Y=1.
%C A082630 Step 2: If 7*(X+Y)<=11*X, then Y:=X+Y, repeat this step. Else go to step 3.
%C A082630 Step 3: Append X to the sequence. Let X:=X+Y, go back to step 2. (End)
%H A082630 Zhuorui He, <a href="/A082630/b082630.txt">Table of n, a(n) for n = 1..1000</a>
%H A082630 John W. Layman, <a href="https://personal.math.vt.edu/layman/sequences/ins_seq.htm">Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types</a>
%H A082630 John W. Layman, <a href="/A085376/a085376.txt">Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types</a> [local copy, corrected]
%H A082630 Toshiki Nakashima and Andrei Zelevinsky, <a href="https://doi.org/10.1006/aima.1997.1670">Polyhedral realizations of crystal bases for quantized Kac-Moody algebras</a>, Adv. Math. 131 (1997), no. 1, 253-278.
%H A082630 J. H. Silverman, <a href="https://doi.org/10.4171/dms/4/21">Divisibility sequences and powers of algebraic integers</a>, Documenta Mathematica, Extra Volume: John H. Coates' Sixtieth Birthday (2006) 711-727.
%F A082630 The sequence appears to satisfy a(n) = 4*a(n-2) - a(n-4).
%F A082630 Apparently a(n)*a(n+3) = -2 + a(n+1)*a(n+2). - _Ralf Stephan_, May 29 2004
%F A082630 Conjecturally, a(n) = A143643(n-1) + A005246(n), for n => 2, as derived from comment above. - _Richard R. Forberg_, Sep 08 2013
%F A082630 If the above conjectures are true, then a(n) = A001353(n)/A005246(n+1). - _Andrey Zabolotskiy_, Sep 26 2024
%e A082630 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.
%o A082630 (Python)
%o A082630 def A082630_list(n):
%o A082630   a = []
%o A082630   x = y = 1
%o A082630   while len(a) < n:
%o A082630     a.append(x)
%o A082630     while 7*(x+y) <= 11*x:
%o A082630       y += x
%o A082630     x += y
%o A082630   return a # _Zhuorui He_, Jul 16 2025
%Y A082630 Cf. A001353, A005246, A001834, A001835, A052530.
%K A082630 nonn
%O A082630 1,2
%A A082630 _John W. Layman_, May 23 2003
%E A082630 Edited by _M. F. Hasler_, Nov 06 2018