A208207 a(n)=(a(n-1)^3*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.

1, 1, 1, 2, 9, 1459, 13975855106, 442535332406378982945622818194705

Offset

0,4

Comments

This is the case a=1, b=3, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term has 105 digits. - Harvey P. Dale, Jul 04 2022

Links

Seiichi Manyama, Table of n, a(n) for n = 0..9

Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Formula

From Vaclav Kotesovec, May 20 2015: (Start)

a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where

d1 = -0.675130870566646070889621798150060480808032527677372732612153869841...

d2 = 0.4608111271891108834741240973014799919001128904578732982807715533323...

d3 = 3.2143197433775351874154977008485804889079196372194994343313823165091...

are the roots of the equation d^3 + 1 = 3*d^2 + d and

c1 = 0.8399660110229591295951614867364338523629139731316529610703364786466...

c2 = 0.5166029105674572719002224224720428001985297645051505025129589573676...

c3 = 1.0214282112585594227681235564690028577352359049566082298453239674712...

(End)

Maple

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^3*a(n-2)+1)/a(n-3): end: seq(a(i), i=0..10);

Mathematica

RecurrenceTable[{a[n] == (a[n - 1]^3*a[n - 2] + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 8}] (* Michael De Vlieger, Mar 19 2017 *)
nxt[{a_, b_, c_}] := {b, c, (c^3 b + 1)/a}; NestList[nxt, {1, 1, 1}, 10][[All, 1]] (* Harvey P. Dale, Jul 04 2022 *)

Crossrefs

Cf. A005246, A208206, A208208, A208213.

Sequence in context: A024226 A252584 A305851 * A229050 A221177 A181865

Adjacent sequences: A208204 A208205 A208206 * A208208 A208209 A208210

Keyword

nonn

Author

Matthew C. Russell, Apr 23 2012

Status

approved