A271330 Rank 2 cluster algebra recurrence with parameters b=c=3 and initial conditions a(0)=2 and a(1)=3.

2, 3, 14, 915, 54718634, 179054255520688074387, 104910436622395373760090792140594526518830577847960206

Offset

0,1

Comments

This recurrence comes from taking a(0)=2, a(1)=3 with parameters b=c=3 in the well known (rank 2 cluster algebra) recurrence:

a(n) = (1+a(n-1)^b)/a(n-2) if n is even,

a(n) = (1+a(n-1)^c)/a(n-2) if n is odd.

Links

Table of n, a(n) for n=0..6.

Formula

a(n) = (1+a(n-1)^3)/a(n-2), n>=2, a(0)=2, a(1)=3.

a(n) ~ c^(((1+sqrt(5))/2)^(2*n)), where c = 1.46103748126346957816556... . - Vaclav Kotesovec, Apr 08 2016

Maple

# Here we set x=a(0) and y=a(1) to be appropriate initial conditions to produce an integer sequence.
# In general we do not obtain an integer sequence.
# The values x=2, y=3 are appropriate initial conditions when b=c=3.
a:=proc(n) local resp, j; option remember;
  if n=0 then return x fi:
   if n=1 then return y fi:
   if type(n, even)=true then
      resp:=simplify((1+a(n-1)^b)/a(n-2)):
      else resp:=simplify((1+a(n-1)^c)/a(n-2)):
     fi:
  return simplify(resp)
  end proc:

Mathematica

a[0] = 2; a[1] = 3; a[n_] := a[n] = (1 + a[n - 1]^3)/a[n - 2]; Array[a, 8, 0] (* Michael De Vlieger, Apr 08 2016 *)

Prog

(PARI) a(n) = if (n==0, 2, if (n==1, 3, if (n%2, (1+a(n-1)^3)/a(n-2), (1+a(n-1)^3)/a(n-2)))); \\ Michel Marcus, Apr 04 2016

Crossrefs

Cf. A271331.

Sequence in context: A042071 A042817 A224848 * A262462 A180698 A266618

Adjacent sequences: A271327 A271328 A271329 * A271331 A271332 A271333

Keyword

nonn

Author

Hector J. Blandin N., Apr 04 2016

Status

approved