login
A271330
Rank 2 cluster algebra recurrence with parameters b=c=3 and initial conditions a(0)=2 and a(1)=3.
1
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.
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
KEYWORD
nonn
AUTHOR
STATUS
approved