Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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