|
|
A185376
|
|
Number of binary necklaces of 2n beads for which a cut exists producing a palindrome.
|
|
2
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
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
|
|
|
FORMULA
|
|
|
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 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|