A168084 Fibonacci 13-step numbers.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16381, 32760, 65516, 131024, 262032, 524032, 1048000, 2095872, 4191488, 8382464, 16763904, 33525760, 67047424, 134086657, 268156933, 536281106, 1072496696
Offset: 1
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
- M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
- Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).
Programs
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Maple
k:=13:a:=taylor((z^(k-1)-z^(k))/(1-2*z+z^(k+1)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); k:=13:for n from 0 to 50 do l(n):=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))):od:seq(l(n),n=0..50); # Richard Choulet, Feb 22 2010
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Mathematica
a={1,0,0,0,0,0,0,0,0,0,0,0,0};Flatten[Prepend[Table[s=Plus@@a;a=RotateLeft[a];a[[ -1]]=s,{n,60}],Table[0,{m,Length[a]-1}]]] LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50] With[{nn=13},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)
Formula
Another form of the g.f. f: f(z)=(z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=13. then a(n)=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))) with k=13 and convention sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010