A097110 Expansion of (1 + 2x - 2x^3) / (1 - 3x^2 + 2x^4).

1, 2, 3, 4, 7, 8, 15, 16, 31, 32, 63, 64, 127, 128, 255, 256, 511, 512, 1023, 1024, 2047, 2048, 4095, 4096, 8191, 8192, 16383, 16384, 32767, 32768, 65535, 65536, 131071, 131072, 262143, 262144, 524287, 524288, 1048575, 1048576, 2097151, 2097152

Offset

0,2

Comments

Union of A000079 and A000225 without 0 = 2^0 - 1. - Reinhard Zumkeller, Jan 18 2005

Let f(0)=1, f(1)=1, and f(n) = f(n - 1 - (1 + (-1)^n)/2) + f(n-2); then a(n-1) = f(n). - John M. Campbell, May 22 2011

The same sequence is obtained iteratively by setting a(0)=1, a(1)=2, and a(n) = a(n-1) + a(n-2) - (GCD(1+a(n-1), a(n-2))-1), showing a kind of affinity to Fibonacci numbers. - Stanislav Sykora, Oct 16 2015

Links

Table of n, a(n) for n=0..41.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-2).

Formula

G.f.: 2*(1+x)/(1-2*x^2)-1/(1-x^2);

a(n) = 3*a(n-2) - 2*a(n-4);

a(n) = (1+sqrt(2)/2)*(sqrt(2))^n + (1/2-sqrt(2)/2)*(-sqrt(2))^n - (1+(-1)^n)/2;

a(n) = Sum_{k=0..n} binomial(floor(n/2), floor(k/2)).

a(n) = 2^floor((n+2)/2) - 1 + (n mod 2). - Reinhard Zumkeller, Jan 18 2005

Maple

seq(op([2^n-1, 2^n]), n=1..100); # Robert Israel, Oct 16 2015

Mathematica

t={1}; Do[AppendTo[t, t[[-1]]+1]; AppendTo[t, t[[-1]]+t[[-2]]], {n, 10}]; t (* Vladimir Joseph Stephan Orlovsky, Jan 27 2012 *)
CoefficientList[Series[(1 + 2*x - 2*x^3)/(1 - 3*x^2 + 2*x^4), {x, 0, 40}], x] (* T. D. Noe, Jan 27 2012 *)

Prog

(PARI) a=vector(1000); a[0]=1; a[1]=2; for(n=3, #a, a[n]=a[n-1]+a[n-2]+1-gcd(1+a[n-1], a[n-2])) \\ Stanislav Sykora, Oct 16 2015
(PARI) Vec(2*(1+x)/(1-2*x^2)-1/(1-x^2) + O(x^100)) \\ Altug Alkan, Oct 16 2015

Crossrefs

Cf. A000045, A000079, A000225.

Sequence in context: A278180 A368448 A015927 * A116961 A300486 A120611

Adjacent sequences: A097107 A097108 A097109 * A097111 A097112 A097113

Keyword

easy,nonn

Author

Paul Barry, Jul 25 2004, corrected Sep 05 2006

Status

approved