A329718 The number of open tours by a biased rook on a specific f(n) X 1 board, where f(n) = A070941(n) and cells are colored white or black according to the binary representation of 2n.
1, 2, 4, 4, 8, 6, 14, 8, 16, 10, 24, 10, 46, 24, 46, 16, 32, 18, 44, 14, 84, 34, 68, 18, 146, 68, 138, 44, 230, 84, 146, 32, 64, 34, 84, 22, 160, 54, 112, 22, 276, 106, 224, 54, 376, 106, 192, 34, 454, 192, 406, 112, 690, 224, 406, 84, 1066, 376, 690, 160
Offset: 0
Keywords
Examples
a(1) = 2 because the binary expansion of 2 is 10 and there are 2 open biased rook's tours, namely 12 and 21. a(2) = 4 because the binary expansion of 4 is 100 and there are 4 open biased rook's tours, namely 132, 213, 231 and 321. a(3) = 4 because the binary expansion of 6 is 110 and there are 4 open biased rook's tours, namely 123, 132, 231 and 312.
Links
- Mikhail Kurkov, Comments on A329718 [verification needed]
Crossrefs
Formula
a(n) = f(n) + f(A059894(n)) = f(n) + f(2*A053645(n)) for n > 0 with a(0) = 1 where f(n) = A329369(n).
Sum_{k=0..2^n-1} a(k) = 2*(n+1)! - 1 for n >= 0.
a((4^n-1)/3) = 2*A110501(n+1) for n > 0.
a(2^1*(2^n-1)) = A027649(n),
a(2^2*(2^n-1)) = A027650(n),
a(2^3*(2^n-1)) = A027651(n),
a(2^4*(2^n-1)) = A283811(n),
and more generally, a(2^m*(2^n-1)) = T(n,m+1) for n >= 0, m >= 0 where T(n,m) = Sum_{k=0..n} k!*(k+1)^m*Stirling2(n,k)*(-1)^(n-k).
Comments