A168083 Fibonacci 12-step numbers.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095, 8189, 16376, 32748, 65488, 130960, 261888, 523712, 1047296, 2094336, 4188160, 8375296, 16748544, 33492993, 66977797, 133939218, 267845688, 535625888, 1071120816

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

Keywords

• nonn

Comments

From Ruediger Jehn, Nov 30 2020: (Start)

a(n+12) is the number of compositions of n with no part greater than 12.

a(n+12) is the number of ways of throwing n with an unstated number of dodecahedra (here dice with numbers from 1 to 12).

(End)

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.
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1,1,1).

Programs

Maple

k:=12: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); 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); # Richard Choulet, Feb 22 2010

Mathematica

a={1,0,0,0,0,0,0,0,0,0,0,0};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}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50]
With[{nn=12},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=12. 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=12 and convention sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010