OFFSET
0,11
COMMENTS
a(n) is the number of compositions of n into parts >=5. - Joerg Arndt, Jun 22 2011
a(n+5) equals the number of binary words such that 0 appears only in runs whose lengths are a multiple of 5. - Milan Janjic, Feb 17 2015
a(n-5) equals the number of circular arrangements of the first n positive integers such that adjacent terms have absolute difference 2 or 3. - Ethan Patrick White, Jun 24 2020
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..1000
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.
J. D. Opdyke, A unified approach to algorithms generating unrestricted.., J. Math. Model. Algor. 9 (2010) 53-97.
Ethan P. White, Richard K. Guy, Renate Scheidler, Difference Necklaces, arXiv:2006.15250 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1).
FORMULA
G.f.: (1-x)/(1-x-x^5) = 1/(1-Sum_{k>=5} x^k).
For positive integers n and k such that k <= n <= 5*k, and 4 divides n-k, define c(n,k) = binomial(k,(n-k)/4), and c(n,k) = 0, otherwise. Then, for n>=1, a(n+5) = 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(5, (i, j)-> if (i=j-1) then 1 elif j=1 then [1, 0$3, 1][i] else 0 fi)^n)[5, 5]: seq(a(n), n=0..50); # Alois P. Heinz, Aug 04 2008
MATHEMATICA
CoefficientList[ Series[(1 - x)/(1 - x - x^5), {x, 0, 50}], x] (* Adi Dani, Jun 25 2011 *)
LinearRecurrence[{1, 0, 0, 0, 1}, {1, 0, 0, 0, 0}, 60] (* Harvey P. Dale, Jun 07 2015 *)
PROG
(PARI) Vec((1-x)/(1-x-x^5)+O(x^99)) \\ Charles R Greathouse IV, Jun 21 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved