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A015440
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Generalized Fibonacci numbers.
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16
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1, 1, 6, 11, 41, 96, 301, 781, 2286, 6191, 17621, 48576, 136681, 379561, 1062966, 2960771, 8275601, 23079456, 64457461, 179854741, 502142046, 1401415751, 3912125981, 10919204736, 30479834641, 85075858321, 237475031526
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 6*a(n-2) equals the number of 6-colored compositions of n with all parts >=2, such that no adjacent parts have the same color.-Milan Janjic, Nov 26 2011
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LINKS
| Joerg Arndt, Fxtbook
Index entries for sequences related to linear recurrences with constant coefficients, signature 1,5.
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FORMULA
| a(n) = a(n-1) + 5 a(n-2).
a(n)={[ (1+sqrt(21))/2 ]^(n+1) - [ (1-sqrt(21))/2 ]^(n+1)}/sqrt(21).
a(n)=sum(k=0, ceil(n/2), 5^k*binomial(n-k, k)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 06 2004
G.f.: 1/(1-x-5x^2). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 03 2008]
a(n)=Sum_{k, 0<=k<=n} A109466(n,k)*(-5)^(n-k). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 26 2008]
A special case of a more general class of Lucas sequences given by
U(n) = U(n-1) + (4^(m-1)-1)/3 U(n-2).
U(n)={[ (1+sqrt((4^(m)-1)/3))/2 ]^(n+1) - [ (1-sqrt((4^(m)-1)/3))/2 ]^(n+1)}/sqrt((4^(m)-1)/3). Fix m = 2 to get the formula for the Fibonacci sequence, fix m = 3 to get the formula for a(n). [From Jeffrey R. Goodwin (jeff.r.goodwin(AT)gmail.com), May 28 2011]
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MATHEMATICA
| a[n_]:=(MatrixPower[{{1, 3}, {1, -2}}, n].{{1}, {1}})[[2, 1]]; Table[Abs[a[n]], {n, -1, 40}] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Feb 19 2010]
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PROG
| (Other) sage: [lucas_number1(n, 1, -5) for n in xrange(1, 28)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 22 2009]
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CROSSREFS
| Cf. A006130, A006131, A015441.
Sequence in context: A108698 A002570 A038265 * A193488 A094555 A099437
Adjacent sequences: A015437 A015438 A015439 * A015441 A015442 A015443
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KEYWORD
| nonn,easy
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AUTHOR
| Olivier Gerard (olivier.gerard(AT)gmail.com)
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