A185376 Number of binary necklaces of 2n beads for which a cut exists producing a palindrome.

2, 3, 6, 9, 20, 34, 72, 129, 272, 516, 1056, 2050, 4160, 8200, 16512, 32769, 65792, 131088, 262656, 524292, 1049600, 2097184, 4196352, 8388610, 16781312, 33554496, 67117056, 134217736, 268451840, 536871040, 1073774592, 2147483649

Offset: 1

Author

Tony Bartoletti, Feb 20 2011

Keywords

• nonn

Comments

These are the values of A185333 for even n.

Conjecture: a(n) = 2^(n-1) + 2^((n-2^t)/(2^(t+1))), where t = number of factors of 2 in n.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A185333.

Programs

Mathematica

f[n_] := Block[{k = IntegerExponent[n, 2]}, 2^n/2 + 2^((n - 2^k)/(2^(k + 1)))]; Array[f, 32] (* Robert G. Wilson v, Aug 08 2011 *)

Python

def a185333(n):
    if n%2: return 2**((n + 1)//2)
    k=bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
    return 2**(n//2 - 1) + 2**((n//2 - 2**k)//(2**(k + 1)))
def a(n): return a185333(2*n)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 29 2017, after the formula

Formula

a(n) = A185333(2n).