A152891 a(1) = b(1) = 0; for n > 1, b(n) = b(n-1) + n-1 + a(n-1) and a(n) = a(n-1) + n-1 + b(n).

0, 2, 9, 29, 83, 226, 602, 1588, 4171, 10935, 28645, 75012, 196404, 514214, 1346253, 3524561, 9227447, 24157798, 63245966, 165580120, 433494415, 1134903147, 2971215049, 7778742024, 20365011048, 53316291146, 139583862417

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Dec 14 2008

Keywords

• nonn

Comments

Partial sums of A035508. - R. J. Mathar, Dec 15 2008

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Index entries for linear recurrences with constant coefficients, signature (5,-8,5,-1).

Crossrefs

Cf. A000045, A001906, A035508, A054452 (b sequence).

Programs

Maple

with(combinat): seq(fibonacci(2*n+1)-n-1, n = 1 .. 27); # Emeric Deutsch, Jun 01 2009

Mathematica

lst={};a=b=0;Do[b+=n+a;a+=n+b;AppendTo[lst,a],{n,0,2*4!}];lst
Table[Fibonacci[2n+1]-n-1,{n,30}] (* or *) LinearRecurrence[{5,-8,5,-1},{0,2,9,29},30] (* Harvey P. Dale, Sep 24 2013 *)

Formula

From R. J. Mathar, Dec 15 2008: (Start)

G.f.: x^2*(2 - x)/((1 - 3*x + x^2)*(1 - x)^2).

a(n) = A001906(n+1) - A001906(n) - n - 1. (End)

a(n) = Fibonacci(2*n+1) - n - 1. - Emeric Deutsch, Jun 01 2009

Extensions

Name corrected by Jon E. Schoenfield, Feb 19 2019