A212404 Number of binary arrays of length 2*n+2 with no more than n ones in any length 2n subsequence (=50% duty cycle).

8, 33, 132, 527, 2104, 8402, 33560, 134075, 535728, 2140910, 8556568, 34201078, 136713872, 546528612, 2184925808, 8735357267, 34925461088, 139642914902, 558353310488, 2232601256162, 8927375430608, 35698163696252, 142750104755408

Offset: 1

Author

R. H. Hardin May 14 2012

Keywords

• nonn

Comments

Row 3 of A212402

Examples

			Some solutions for n=3
..0....0....0....1....0....1....0....0....0....1....0....0....1....1....1....0
..0....0....1....0....0....0....1....1....0....0....1....1....0....0....0....0
..1....1....0....1....1....0....1....0....0....0....1....0....0....0....1....1
..1....0....1....0....1....0....0....0....0....1....0....0....0....1....0....0
..1....0....1....1....0....1....0....1....1....1....0....1....0....0....0....1
..0....1....0....0....0....1....1....0....0....0....0....0....0....1....0....0
..0....1....0....0....1....1....0....0....0....0....0....1....1....0....0....0
..0....0....1....0....0....0....1....1....1....1....1....0....0....1....1....1
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210

Programs

Mathematica

Flatten[{8,33,RecurrenceTable[{(n-4)*n*a[n]==2*(n-1)*(4*n-15)*a[n-1]-8*(n-3)*(2*n-5)*a[n-2],a[3]==132,a[4]==527},a,{n,3,20}]}] (* Vaclav Kotesovec, Oct 19 2012 *)
Table[2^(2*n+1)+Binomial[2*n-2,n],{n,1,20}] (* Vaclav Kotesovec, Oct 28 2012 *)

Formula

Recurrence (for n>4): (n-4)*n*a(n) = 2*(n-1)*(4*n-15)*a(n-1) - 8*(n-3)*(2*n-5)*a(n-2). - Vaclav Kotesovec, Oct 19 2012

a(n) = 2^(2*n+1) + C(2*n-2,n). - Vaclav Kotesovec, Oct 28 2012