A208205 a(n)=(a(n-1)*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1.

1, 1, 1, 2, 3, 97, 11786, 33736797423001, 79097781524295318019203322936641

Offset

0,4

Comments

This is the case a=5, b=1, 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 (a(9)) has 96 digits. - _Harvey P. Dale_, Sep 13 2022

Links

Seiichi Manyama, Table of n, a(n) for n = 0..11

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 = -1.903211925911553287485216224057094233775314050044332659604216582082...

d2 = 0.1939365664746304482560845569332033002552873106788960042162607290276...

d3 = 2.7092753594369228392291316671238909335200267393654366553879558530545...

are the roots of the equation d^3 + 1 = d^2 + 5*d and

c1 = 0.9741074409555962981370572554321352591111177638556227590517984272608...

c2 = 0.0499759123576461468686480770163694779918691526759585723897652462761...

c3 = 1.0272217210627198315132544386598971884129462517962425299212701250318...

(End)

Maple

a:=proc(n) if n<3 then return 1: fi: return (a(n-1)*a(n-2)^5+1)/a(n-3): end: seq(a(i), i=1..10);

Mathematica

RecurrenceTable[{a[n] == (a[n - 1] a[n - 2]^5 + 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*b^5+1)/a}; NestList[nxt, {1, 1, 1}, 10][[All, 1]] (* _Harvey P. Dale_, Sep 13 2022 *)

Crossrefs

Cf. A005246, A208204.

Sequence in context: A247652 A132499 A199541 * A128931 A323463 A212556

Adjacent sequences: A208202 A208203 A208204 * A208206 A208207 A208208

Keyword

nonn

Author

_Matthew C. Russell_, Apr 23 2012

Status

approved