OFFSET
0,13
COMMENTS
A Lamé sequence of higher order.
Number of compositions of n into parts >= 6. - Milan Janjic, Jun 28 2010
a(n+6) equals the number of n-length binary words such that 0 appears only in a run which length is a multiple of 6. - Milan Janjic, Feb 17 2015
Same as sequence A005708 with 1, 0, 0, 0, 0, 0 prepended. - Linas Vepstas, Feb 06 2024
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..1000
I. M. Gessel and 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.
Augustine O. Munagi, Integer Compositions and Higher-Order Conjugation, J. Int. Seq., Vol. 21 (2018), Article 18.8.5.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1).
FORMULA
G.f.: 1/(1-Sum_{k>=6} x^k).
G.f.: (1-x)/(1-x-x^6). - Alois P. Heinz, Aug 04 2008
For positive integers n and k such that k <= n <= 6*k, and 5 divides n-k, define c(n,k) = binomial(k,(n-k)/5), and c(n,k) = 0 otherwise. Then, for n >= 1, a(n+6) = Sum_{k=1..n} c(n,k). - Milan Janjic, Dec 09 2011
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(6, (i, j)-> `if`(i=j-1, 1, `if`(j=1, [1, 0$4, 1][i], 0)))^n)[6, 6]: seq(a(n), n=0..80); # Alois P. Heinz, Aug 04 2008
MATHEMATICA
f[n_] := If[n < 1, 1, Sum[ Binomial[ n - 5 k - 5, k], {k, 0, (n - 5)/6}]]; Array[f, 49, 0] (* Adi Dani, Robert G. Wilson v, Jul 04 2011 *)
LinearRecurrence[{1, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0}, 60] (* Jean-François Alcover, Feb 13 2016 *)
PROG
(PARI) Vec((1-x)/(1-x-x^6)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved