A143654 Array T(n,k) read by rows: number of binary bracelets with n beads, k of them 0, with 00 prohibited, (n >= 2, 0 <= k <= floor(n/2)).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 4, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 8, 3, 1, 1, 1, 5, 8, 10, 5, 1, 1, 1, 6, 10, 16, 10, 4, 1, 1, 1, 6, 12, 20, 16, 7, 1, 1, 1, 7, 14, 29, 26, 16, 4, 1, 1, 1, 7, 16, 35, 38, 26, 8, 1, 1, 1, 8, 19, 47, 57, 50

Offset

2,13

Comments

The initial columns give A057427, A057427, A004526, A069905, A005232, A032279, A005513, A032280, A005514, A032281, A005515, A032282, A005516. Row sums give A129526.

A binary bracelet with n beads, k of them 0, with 00 prohibited has from 0 to floor(n/2) beads 0, i.e., 0 <= k <= floor(n/2). If n is even, the bracelet 0101...01 with n/2 beads of each kind does not have 00 and we cannot change any 1 of it to a 0. If n is odd we cannot change a 1 to a 0 in the bracelet 0101...011 with (n-1)/2 beads 0.

The number of binary bracelets with n beads, 0 <= k <= floor(n/2) of them 0 with 00 prohibited, is equal to the number of binary bracelets with n-k beads, k of them 0. See below.

Let B be a binary bracelet with n-k beads, k of them 0. If we insert one 1 (circularly) after a 0 of B, we obtain a bracelet with n-k+1 beads, k of them 0.

If we do this insertion k times, each time after a distinct 0 of B, we obtain a bracelet with n = n-k+k beads, k of them 0, with 00 prohibited.

On the contrary, Let B be a binary bracelet with n beads, k of them 0, with 00 prohibited. If we remove from B one 1 that is after a 0, we obtain a bracelet of n-1 beads, k of them 0. (If not and we undo the removal, the configuration obtained cannot be a bracelet and this is absurd.) If we repeat this removal k times, after each distinct bead 0, we obtain a bracelet with n-k beads, k of them 0.

Links

Table of n, a(n) for n=2..96.

Example

Array begins
1 1
1 1
1 1 1
1 1 1
1 1 2 1
1 1 2 1
1 1 3 2 1
1 1 3 3 1
1 1 4 4 3 1
...
A129526(10) = A057427(10) + A057427(9) + A004526(8) + A069905(7) + A005232(6) +
A032279(5) = 1+1+4+4+3+1 = 14.

Crossrefs

Cf. A057427, A004526, A069905, A005232, A032279, A005513, A032280, A005514, A032281, A005515, A032282, A005516. Row sums of array give A129526.

Sequence in context: A327857 A363850 A227481 * A170981 A161096 A214247

Adjacent sequences: A143651 A143652 A143653 * A143655 A143656 A143657

Keyword

nonn,tabf

Author

Washington Bomfim, Aug 28 2008

Status

approved