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a(n) = a(n-7) + a(n-11), starting a(0)=a(1)=...= a(10) = 1.
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%I #27 Oct 09 2018 20:27:07

%S 1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,4,4,4,5,5,5,5,7,7,7,8,9,

%T 9,9,12,12,12,13,16,16,16,20,21,21,22,28,28,28,33,37,37,38,48,49,49,

%U 55,65,65,66,81,86,86,93,113,114,115,136,151,151,159,194,200,201,229,264,265,274

%N a(n) = a(n-7) + a(n-11), starting a(0)=a(1)=...= a(10) = 1.

%D Kenneth H. Rosen, Discrete Mathematics and its Applications, McGraw-Hill, 2012, 501-503.

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,1,0,0,0,1).

%F G.f.: (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)/(1 - x^7 - x^11). - _R. J. Mathar_ and _N. J. A. Sloane_, Nov 10 2017

%t LinearRecurrence[{0,0,0,0,0,0,1,0,0,0,1},{1,1,1,1,1,1,1,1,1,1,1},80] (* _Harvey P. Dale_, Oct 09 2018 *)

%o (Java)

%o import java.util.Arrays;

%o public class IntegerSequences

%o {

%o public static void main(String[] args)

%o {

%o int j = 7;

%o int k = 11;

%o // Set N to the number of terms you would like to generate.

%o int N = 200;

%o long[] G = new long[N];

%o for(int i=0; i<k; i++)

%o {

%o G[i] = 1;

%o }

%o for(int i=k; i<N; i++)

%o {

%o G[i] = G[i-j]+G[i-k];

%o }

%o System.out.println(Arrays.toString(G));

%o }

%o }

%Y Cf. A000930, A003269, A003520, A005708, A005709.

%K nonn,easy

%O 0,12

%A _Jason Bruce_, Sep 07 2017