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A122265
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10th-order Fibonacci numbers: a(n+1) = a(n)+...+a(n-9) with a(0) = ... = a(8) = 0, a(9) = 1.
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5
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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
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OFFSET
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0,12
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COMMENTS
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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}}.
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LINKS
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Robert Price, Table of n, a(n) for n = 0..1000
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).
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1).
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FORMULA
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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
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MAPLE
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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Cf. A000322, A001591, A001592, A079262.
Cf. A001591, A001592, A122189, A079262, A104144, A168082, A168083, A168084. - Richard Choulet, Feb 22 2010
Cf. A257227, A257228 for primes in this sequence.
Sequence in context: A145116 A172319 A234591 * A339073 A194633 A243088
Adjacent sequences: A122262 A122263 A122264 * A122266 A122267 A122268
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KEYWORD
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nonn
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AUTHOR
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Roger L. Bagula and Gary W. Adamson, Oct 18 2006
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EXTENSIONS
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Edited by N. J. A. Sloane, Oct 29 2006 and Mar 05 2011
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STATUS
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approved
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