A079289 For even n, a(n) = a(n-2) + a(n-1) + 2^(n/2-2), n>2. For odd n, a(n) = a(n-2) + a(n-1).

1, 1, 2, 3, 6, 9, 17, 26, 47, 73, 128, 201, 345, 546, 923, 1469, 2456, 3925, 6509, 10434, 17199, 27633, 45344, 72977, 119345, 192322, 313715, 506037, 823848, 1329885, 2161925, 3491810, 5670119, 9161929, 14864816, 24026745, 38957097, 62983842

Offset: 0

Author

Paul Barry, Feb 08 2003

Keywords

• easy
• nonn

Comments

Generalized Fibonacci sequence: a(n) = a(n-2) + a(n-1), and for even n a row sum of Pascal's triangle (a power of two) is added.

Call a multiset of nonzero integers good if the sum of the cubes is the square of the sum. The number of ascending chains of good multisets starting from the empty set by adding one element at a time is a(n). - Michael Somos, Apr 14 2005

a(n) is the number of compositions of n which consist of an initial (possibly empty) subsequence of even summands and a remaining (possibly empty) sequence of odd summands.

Examples

			a(4) = 6 from the good multisets {-1,-1,1,1}, {-1,1,1,2}, {-2,-1,1,2}, {-2,1,2,2}, {-3,1,2,3}, {1,2,3,4}.
a(4) = 6 because there are six compositions of four, in which the initial parts are all even and the final parts are all odd: 4, 3+1, 1+3, 2+2, 2+1+1, 1+1+1+1.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-2).

Crossrefs

Cf. A000045, A005674, A007318, A011782, A061667 (bisection).

Programs

Magma

I:=[1,1,2,3,6]; [n le 5 select I[n] else Self(n-1)+3*Self(n-2) -2*Self(n-3)-2*Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 05 2013

Mathematica

CoefficientList[Series[(1-x^2)^2/(1-x-x^2)/(1-2x^2),{x,0,37}],x]
LinearRecurrence[{1,3,-2,-2}, {1,1,2,3,6}, 25] (* G. C. Greubel, Aug 16 2016; corrected by Georg Fischer, Apr 02 2019 *)
nxt[{n_,a_,b_}]:={n+1,b,If[EvenQ[n],a+b,a+b+2^((n+1)/2-2)]}; Join[{1}, NestList[ nxt,{2,1,2},40][[All,2]]] (* Harvey P. Dale, Jul 13 2019 *)

PARI

{a(n)=local(A); if(n<3,(n>=0)+(n>1), A=vector(n,i,i); for(i=3,n,A[i]=A[i-1]+A[i-2]+ if(i%2==0,2^(i/2-2))); A[n])} /* Michael Somos, Apr 14 2005 */

Formula

a(n) = a(n-2) + a(n-1) + floor(2^(n/2-2))*(1-(-1)^(n+1))/2 for n>1.

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

a(n) = -A016116(n+1)/2 +A000045(n+2), n>0. - R. J. Mathar, Sep 27 2012

From Gregory L. Simay, Jul 25 2016: (Start)

If n = 2k+1, a(n) = the convolution Sum_{j=0,..k} c(j)*F(n-2j), where c(j) = A011782(j) = 2^(j-1) and f(j)= A000045(j).

If n = 2k, a(n) = c(k) + the convolution Sum_{j=0,..(k-1)} c(j)*F(n-2j), where c(j)=A011782(j)=2^(j-1) and f(j)= A000045(j). (End)