A017904 Expansion of 1/(1 - x^10 - x^11 - ...).

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 16, 20, 25, 31, 38, 46, 55, 65, 76, 89, 105, 125, 150, 181, 219, 265, 320, 385, 461, 550, 655, 780, 930, 1111, 1330, 1595, 1915, 2300, 2761, 3311, 3966, 4746, 5676

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

A Lamé sequence of higher order.

a(n) = number of compositions of n in which each part is >=10. - Milan Janjic, Jun 28 2010

a(n+19) equals the number of binary words of length n having at least 9 zeros between every two successive ones. - Milan Janjic, Feb 09 2015

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.
• Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Crossrefs

For Lamé sequences of orders 1 through 9 see A000045, A000930, A017898-A017903, and this one.

Programs

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(10, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 0$8, 1][i] else 0 fi)^n)[10,10]: seq(a(n), n=0..80); # Alois P. Heinz, Aug 04 2008

Mathematica

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2012 *)

PARI

a(n)=([0,1,0,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0,0; 0,0,0,0,1,0,0,0,0,0; 0,0,0,0,0,1,0,0,0,0; 0,0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,0,1; 1,0,0,0,0,0,0,0,0,1]^n)[1,1] \\ Charles R Greathouse IV, Oct 03 2016

Formula

G.f.: (x-1)/(x-1+x^10). - Alois P. Heinz, Aug 04 2008

For positive integers n and k such that k <= n <= 10*k, and 9 divides n-k, define c(n,k) = binomial(k,(n-k)/9), and c(n,k) = 0, otherwise. Then, for n>= 1, a(n+10) = sum(c(n,k), k=1..n). - Milan Janjic, Dec 09 2011