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A022341
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a(n) = 4*A003714(n) + 1; the odd Fibbinary numbers.
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6
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1, 5, 9, 17, 21, 33, 37, 41, 65, 69, 73, 81, 85, 129, 133, 137, 145, 149, 161, 165, 169, 257, 261, 265, 273, 277, 289, 293, 297, 321, 325, 329, 337, 341, 513, 517, 521, 529, 533, 545, 549, 553, 577, 581, 585, 593, 597, 641, 645, 649, 657, 661, 673, 677, 681
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OFFSET
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0,2
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COMMENTS
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Numbers k such that (k+1) does not divide C(3k, k) - C(2k, k). - Benoit Cloitre, May 23 2004
Each term is the unique odd number a(n) = Sum_{i in S} 2^i such that n = Sum_{i in S} F_i, where F_i is the i-th Fibonacci number, A000045(i), and S is a set of nonnegative integers of which no two are adjacent. Note that this corresponds to adding F_0 to the Zeckendorf representation of n, which does not change the number being represented, because F_0 = 0. - Peter Munn, Sep 02 2022
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LINKS
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Estelle Basor, Brian Conrey, and Kent E. Morrison, Knots and ones, arXiv:1703.00990 [math.GT], 2017. See page 2.
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MAPLE
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F:= combinat[fibonacci]:
b:= proc(n) local j;
if n=0 then 0
else for j from 2 while F(j+1)<=n do od;
b(n-F(j))+2^(j-2)
fi
end:
a:= n-> 4*b(n)+1:
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MATHEMATICA
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Select[Range[1, 511, 2], BitAnd[#, 2#] == 0 &] (* Alonso del Arte, Jun 18 2012 *)
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PROG
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(Python)
for n in range(1, 700, 2):
if n*2 & n == 0:
print(n, end=', ')
(Scala) (1 to 511 by 2).filter(n => (n & 2 * n) == 0) // Alonso del Arte, Apr 12 2020
(C#)
public static bool IsOddFibbinaryNum(this int n) => ((n & (n >> 1)) == 0) && (n % 2 == 1) ? true : false; // Frank Hollstein, Jul 07 2021
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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