A076737 Let u(1)=u(2)=u(3)=2, u(n)=(1+u(n-1)u(n-2))/u(n-3); then a(n) is the numerator of u(n).

2, 2, 2, 5, 3, 17, 11, 65, 43, 257, 171, 1025, 683, 4097, 2731, 16385, 10923, 65537, 43691, 262145, 174763, 1048577, 699051, 4194305, 2796203, 16777217, 11184811, 67108865, 44739243, 268435457, 178956971, 1073741825, 715827883, 4294967297

Offset: 1

Author

Benoit Cloitre, Nov 24 2002

Keywords

• nonn
• frac

Links

• Robert Israel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,5,0,-4).

Crossrefs

Cf. A005246, A076736 (denominator of u(n)).

Programs

Maple

2,2,2,seq(2/3+(1/6)*2^k+(1/12)*(-1)^k*2^k+(1/3)*(-1)^k,k=4..50); # Robert Israel, Aug 10 2015
H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -8):
a := n -> `if`(n < 5, [2, 2, 2, 5][n], H(n-2, irem(n, 2), 1/2)):
seq(simplify(a(n)), n=1..34); # Peter Luschny, Sep 03 2019

Mathematica

nxt[{a_,b_,c_}]:={b,c,(1+b c)/a}; NestList[nxt,{2,2,2},40][[All,1]]// Numerator (* Harvey P. Dale, Oct 31 2021 *)

Formula

For n>4, a(n) = 2^A028242(n-4)*u(n); u(2n) = 2^(n-1)+1/2^n hence a(2n) = 4^(n-1)+1.

From Michael Somos (via Benoit Cloitre), Nov 29 2002: (Start)

a(1)=a(2)=a(3)=2, a(n+2) = (1+2^n)/(1+2*(n mod 2)).

For k>=2, a(2k+1) = A001045(2k-1). (End)

Empirical g.f.: x*(4*x^6+x^4-5*x^3-8*x^2+2*x+2) / ((x-1)*(x+1)*(2*x-1)*(2*x+1)). - Colin Barker, Oct 14 2014

This follows from the Somos formula for a(n+2). - Robert Israel, Aug 10 2015

a(1)=a(2)=a(3)=2 and, for n>3, a(n) = denominator(1/2+6/(4+2^n)). - Gerry Martens, Aug 10 2015

a(n) = H(n - 2, n mod 2, 1/2) for n >= 5 where H(n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -8). - Peter Luschny, Sep 03 2019