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

1,1

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).

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

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 *)

Crossrefs

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

Sequence in context: A333388 A174577 A194684 * A246119 A210562 A208512

Adjacent sequences: A076734 A076735 A076736 * A076738 A076739 A076740

Keyword

nonn,frac

Author

Benoit Cloitre, Nov 24 2002

Status

approved