A000846 -id:A000846 - cp's OEIS frontend

A022341 a(n) = 4*A003714(n) + 1; the odd Fibbinary numbers.

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

Offset: 0

Marc LeBrun

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

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Estelle Basor, Brian Conrey, and Kent E. Morrison, Knots and ones, arXiv:1703.00990 [math.GT], 2017. See page 2.
Linus Lindroos, Andrew Sills, and Hua Wang, Odd fibbinary numbers and the golden ration, Fib. Q., 52 (2014), 61-65.
Linus Lindroos, Andrew Sills, and Hua Wang, Odd fibbinary numbers and the golden ration, Fib. Q., 52 (2014), 61-65.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See page 6.
D. M. McKenna, Fibbinary Zippers in a Monoid of Toroidal Hamiltonian Cycles that Generate Hilbert-style Square-filling Curves, Enumerative Combinatorics and Applications, 2:2 #S2R13 (2021).

Cf. A000045, A003714, A000846, A022340.

First column of A356875.

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:
seq(a(n), n=0..70);  # Alois P. Heinz, May 15 2016

Select[Range[1, 511, 2], BitAnd[#, 2#] == 0 &] (* Alonso del Arte, Jun 18 2012 *)

for n in range(1, 700, 2):
    if n*2 & n == 0:
        print(n, end=',')

def A022341(n):
    tlist, s = [1,2], 0
    while tlist[-1]+tlist[-2] <= n: tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= n:
            s += 1
            n -= d
        s <<= 1
    return (s<<1)|1 # Chai Wah Wu, Apr 24 2025

(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

More terms from Benoit Cloitre, May 23 2004 and Alonso del Arte, Jun 18 2012

Name edited by Peter Munn, Sep 02 2022

A060774 a(n) = number of lattice paths from (0,0,0) to (n,n,n) along the cracks on the surface of a Rubik-ized n X n X n cube so that no step increases distance from goal.

Original entry on oeis.org

1, 6, 54, 384, 2550, 16506, 105840, 677088, 4335606, 27829230, 179161554, 1156987728, 7493841264, 48672149064, 316920674880, 2068273848384, 13525486999542, 88612412883030, 581503640659830, 3821691744347400, 25150239955660050, 165713382866931570

Offset: 0

Len Smiley, Apr 25 2001

3-dimensional version of block-walking (0,0) to (n,n) in binomial(2n,n) ways.

			a(1)=6: XYZ, XZY, YXZ, YZX, ZXY, ZYX.

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (terms n = 1..200 from Harry J. Smith)
W. Li and E. T. H. Wang, A bug's shortest path on a cube, Mathematics Magazine 58:4 (Sept. 1985), pp. 219-221.

Column k=3 of A225094.

A060774 := proc(n)
        `if`(n=0, 1,
        6*(binomial(3*n,n)-binomial(2*n,n)) ) ;
end proc: # R. J. Mathar, Oct 31 2015

Rest[CoefficientList[Series[-(6/Sqrt[1-4z])+(12Cos[ArcCos[1-27z/2]/6])/Sqrt[4-27z], {z, 0, 20}], z]] (* Benedict W. J. Irwin, Jul 12 2016 *)

j=[]; for(n=1,50,j=concat(j,6*(binomial(3*n,n)-binomial(2*n,n)))); j

{ for (n=1, 200, write("b060774.txt", n, " ", 6*(binomial(3*n, n) - binomial(2*n, n))); ) } \\ Harry J. Smith, Jul 11 2009

a(n) = 6*binomial(3n, n) - 6*binomial(2n, n).
a(n) = 6*A000846(n) for n>0. - R. J. Mathar, Oct 31 2015
Conjecture: 2*n*(2*n-1)*(n-1)*a(n) + (n-1)*(13*n^2-209*n+258)*a(n-1) + 2*(-259*n^3+1785*n^2-3728*n+2460)*a(n-2) + 6*(295*n^3-2068*n^2+4833*n-3780)*a(n-3) - 36*(3*n-10)*(2*n-7)*(3*n-11)*a(n-4) = 0. - R. J. Mathar, Oct 31 2015
Conjecture: 2*n*(n-1)*(2*n-1)*(11*n^2-33*n+24)*a(n) - (n-1)*(473*n^4-1892*n^3+2561*n^2-1338*n+216)*a(n-1) + 6*(3*n-5)*(3*n-4)*(2*n-3)*(11*n^2-11*n+2)*a(n-2) = 0. - R. J. Mathar, Oct 31 2015
From Benedict W. J. Irwin, Jul 12 2016: (Start)
G.f.: -6/sqrt(1-4*x) + 12*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x).
E.g.f: -6*E^(2*x)*BesselI(0,2*x) + 6*2F2(1/3,2/3;1/2,1;27*x/4).
(End)
a(n) ~ 4^(-n)*(3^(3*n+3/2))/sqrt(Pi*n). - Ilya Gutkovskiy, Jul 12 2016

Corrected by Franklin T. Adams-Watters and T. D. Noe, Oct 25 2006

a(0)=1 prepended by Alois P. Heinz, Sep 09 2016

cp's OEIS Frontend

A022341 a(n) = 4*A003714(n) + 1; the odd Fibbinary numbers.

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