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

%I #31 May 08 2021 06:29:55

%S 2,3,6,9,20,34,72,129,272,516,1056,2050,4160,8200,16512,32769,65792,

%T 131088,262656,524292,1049600,2097184,4196352,8388610,16781312,

%U 33554496,67117056,134217736,268451840,536871040,1073774592,2147483649

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

%C These are the values of A185333 for even n.

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

%H G. C. Greubel, <a href="/A185376/b185376.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A185333(2n).

%t 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 *)

%o (Python)

%o def a185333(n):

%o if n%2: return 2**((n + 1)//2)

%o k=bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')

%o return 2**(n//2 - 1) + 2**((n//2 - 2**k)//(2**(k + 1)))

%o def a(n): return a185333(2*n)

%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jun 29 2017, after the formula

%Y Cf. A185333.

%K nonn

%O 1,1

%A _Tony Bartoletti_, Feb 20 2011