A017898 Expansion of (1-x)/(1-x-x^4).
1, 0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272, 8657, 11949, 16493, 22765, 31422, 43371, 59864, 82629, 114051, 157422, 217286, 299915, 413966, 571388, 788674
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Christian Ballot, On Functions Expressible as Words on a Pair of Beatty Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.2.
- I. M. Gessel, Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5
- J. Hermes, Anzahl der Zerlegungen einer ganzen rationalen Zahl in Summanden, Math. Ann., 45 (1894), 371-380.
- Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3
- T. Mansour, M. Shattuck, A monotonicity property for generalized Fibonacci sequences, arXiv:1410.6943 [math.CO], 2014.
- Augustine O. Munagi, Integer Compositions and Higher-Order Conjugation, J. Int. Seq., Vol. 21 (2018), Article 18.8.5.
- J. D. Opdyke, A unified approach to algorithms generating unrestricted.., J. Math. Model. Algor. 9 (2010) 53-97
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,1).
Crossrefs
Programs
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Maple
f := proc(r) local t1,i; t1 := []; for i from 1 to r do t1 := [op(t1),0]; od: for i from 1 to r+1 do t1 := [op(t1),1]; od: for i from 2*r+2 to 50 do t1 := [op(t1),t1[i-1]+t1[i-1-r]]; od: t1; end; # set r = order a:= n-> (Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 0$2, 1][i] else 0 fi)^n)[4,4]: seq(a(n), n=0..42); # Alois P. Heinz, Aug 04 2008
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Mathematica
LinearRecurrence[{1, 0, 0, 1}, {1, 0, 0, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 11 2012 *) CoefficientList[Series[(1-x)/(1-x-x^4),{x,0,50}],x] (* Harvey P. Dale, Sep 12 2019 *)
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PARI
a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,0,0,1]^n*[1;0;0;0])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
Formula
a(n) = a(n-1) + a(n-4). - R. J. Mathar, Mar 06 2008
G.f.: 1/(1-sum(k>=4, x^k)). - Joerg Arndt, Aug 13 2012
Apparently a(n) = hypergeometric([1-1/4*n, 5/4-1/4*n, 3/2-1/4*n, 7/4-1/4*n],[4/3-1/3*n, 5/3-1/3*n, 2-1/3*n], -4^4/3^3) for n>=13. - Peter Luschny, Sep 18 2014
Comments