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A122265 10th-order Fibonacci numbers: a(n+1) = a(n)+...+a(n-9) with a(0) = ... = a(8) = 0, a(9) = 1. 5
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2045, 4088, 8172, 16336, 32656, 65280, 130496, 260864, 521472, 1042432, 2083841, 4165637, 8327186, 16646200, 33276064, 66519472, 132973664, 265816832, 531372800, 1062224128 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,12
COMMENTS
The (1,10)-entry of the matrix M^n, where M is the 10 X 10 matrix {{0,1,0,0,0, 0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0},{0,0,0,1,0,0,0,0,0,0},{0,0,0,0,1,0,0,0,0,0}, {0,0,0,0,0,1,0,0,0,0},{0,0,0,0,0,0,1,0,0,0},{0,0,0,0,0,0,0,1,0,0},{0,0,0,0,0, 0,0,0,1,0},{0,0,0,0,0,0,0,0,0,1},{1,1,1,1,1,1,1,1,1,1}}.
LINKS
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).
FORMULA
a(n) = Sum_{j=1..10} a(n-j) for n>=10; a(n) = 0 for 0<=n<=8, a(9) = 1 (follows from the minimal polynomial of M; a Maple program based on this recurrence relation is much slower than the given Maple program, based on the definition).
G.f.: -x^9/(-1+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
Another form of the g.f. f: f(z)=(z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=10. 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=10 and sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010
MAPLE
with(linalg): p:=-1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9+x^10: M[1]:=transpose(companion(p, x)): for n from 2 to 40 do M[n]:=multiply(M[n-1], M[1]) od: seq(M[n][1, 10], n=1..40);
k:=10: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); k:=10: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
M = {{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}; v[1] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[Floor[v[n][[1]]], {n, 1, 50}]
a={1, 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}]]] (* Vladimir Joseph Stephan Orlovsky, Nov 18 2009 *)
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
With[{nn=10}, LinearRecurrence[Table[1, {nn}], Join[Table[0, {nn-1}], {1}], 50]] (* Harvey P. Dale, Aug 17 2013 *)
CROSSREFS
Cf. A257227, A257228 for primes in this sequence.
Sequence in context: A145116 A172319 A234591 * A339073 A194633 A243088
KEYWORD
nonn
AUTHOR
EXTENSIONS
Edited by N. J. A. Sloane, Oct 29 2006 and Mar 05 2011
STATUS
approved

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)