OFFSET
1,1
COMMENTS
Row 3 of A212402
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
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
EXAMPLE
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
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin May 14 2012
STATUS
approved