A015449 - OEIS

%I #83 Feb 27 2024 21:02:36

%S 1,1,6,31,161,836,4341,22541,117046,607771,3155901,16387276,85092281,

%T 441848681,2294335686,11913527111,61861971241,321223383316,

%U 1667978887821,8661117822421,44973567999926,233528957822051

%N Expansion of (1-4*x)/(1-5*x-x^2).

%C Row m=5 of A135597.

%C Binomial transform of A152187. - _Johannes W. Meijer_, Aug 01 2010

%C For n>=1, row sums of triangle

%C m/k.|..0.....1.....2.....3.....4.....5.....6.....7

%C ==================================================

%C .0..|..1

%C .1..|..1.....5

%C .2..|..1.....5....25

%C .3..|..1....10....25.....125

%C .4..|..1....10....75.....125....625

%C .5..|..1....15....75.....500....625....3125

%C .6..|..1....15...150.....500...3125....3125...15625

%C .7..|..1....20...150....1250...3125...18750...15625...78125

%C which is triangle for numbers 5^k*C(m,k) with duplicated diagonals. - _Vladimir Shevelev_, Apr 12 2012

%C a(n+1) is (for n>=0) the number of length-n strings of 6 letters {0,1,2,3,4,5} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - _Joerg Arndt_, Oct 11 2012

%C With offset 1, the sequence is the INVERT transform (1, 5, 5*4, 5*4^2, 5*4^3, ...); i.e., of A003947. The sequence can also be obtained by taking powers of the matrix [(1,5); (1,4)] and extracting the upper left terms. - _Gary W. Adamson_, Jul 31 2016

%H Vincenzo Librandi, <a href="/A015449/b015449.txt">Table of n, a(n) for n = 0..1000</a>

%H Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>

%H Taras Goy and Mark Shattuck, <a href="https://doi.org/10.2478/amsil-2023-0027">Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries</a>, Ann. Math. Silesianae (2023). See p. 16.

%H Milan Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Janjic/janjic63.html">On Linear Recurrence Equations Arising from Compositions of Positive Integers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,1).

%F a(n) = 5*a(n-1) + a(n-2).

%F a(n) = Sum_{k=0..n} 4^k*A055830(n,k). - _Philippe Deléham_, Oct 18 2006

%F G.f.: (1-4*x)/(1-5*x-x^2). - _Philippe Deléham_, Nov 20 2008

%F For n >= 2, a(n) = F_n(5) + F_(n+1)(5), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - _Vladimir Shevelev_, Apr 13 2012

%F a(n) = Sum_{k=0..n} A046854(n-1,k)*5^k. - _R. J. Mathar_, Feb 10 2024

%p a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..21); # _Zerinvary Lajos_, Jul 26 2006

%t Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{1,5},#]}]&, {1,1},40]][[1]]  (* _Harvey P. Dale_, Mar 23 2011 *)

%t LinearRecurrence[{5,1}, {1,1}, 30] (* _Vincenzo Librandi_, Nov 06 2012 *)

%t CoefficientList[Series[(1-4*x)/(1-5*x-x^2), {x,0,30}], x] (* _G. C. Greubel_, Dec 19 2017 *)

%t Sum[Fibonacci[Range[30] +k-2, 5], {k,0,1}] (* _G. C. Greubel_, Oct 23 2019 *)

%o (Magma) [n le 2 select 1 else 5*Self(n-1)+Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Nov 06 2012

%o (PARI) Vec((1-4*x)/(1-5*x-x^2) +O('x^30)) \\ _G. C. Greubel, Dec 19 2017

%o (Sage)

%o def A015449_list(prec):

%o     P.<x> = PowerSeriesRing(ZZ, prec)

%o     return P((1-4*x)/(1-5*x-x^2)).list()

%o A015449_list(30) # _G. C. Greubel_, Oct 23 2019

%o (GAP) a:=[1,1];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # _G. C. Greubel_, Oct 23 2019

%Y Cf. A084057, A108306, A164549.

%K nonn,easy

%O 0,3

%A _Olivier Gérard_