cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A122265 10th-order Fibonacci numbers: a(n+1) = a(n)+...+a(n-9) with a(0) = ... = a(8) = 0, a(9) = 1.

Original entry on oeis.org

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
Offset: 0

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Author

Roger L. Bagula and Gary W. Adamson, Oct 18 2006

Keywords

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}}.

Crossrefs

Cf. A257227, A257228 for primes in this sequence.

Programs

  • 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 *)

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

Extensions

Edited by N. J. A. Sloane, Oct 29 2006 and Mar 05 2011