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

0,2

Comments

A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right.

Links

Table of n, a(n) for n=0..59.

Mikhail Kurkov, Comments on A329718 [verification needed]

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).

Example

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.

Crossrefs

Cf. A027649, A027650, A027651, A059894, A110501, A283811, A284005, A329369, A344902, A344947.

Sequence in context: A180444 A065608 A184396 * A077764 A366621 A110794

Adjacent sequences: A329715 A329716 A329717 * A329719 A329720 A329721

Keyword

nonn

Author

Mikhail Kurkov, Nov 19 2019 [verification needed]

Status

approved