OFFSET
1,15
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).
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
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
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Nov 18 2009
STATUS
approved