A131323 Odd numbers whose binary expansion ends in an even number of 1's.

3, 11, 15, 19, 27, 35, 43, 47, 51, 59, 63, 67, 75, 79, 83, 91, 99, 107, 111, 115, 123, 131, 139, 143, 147, 155, 163, 171, 175, 179, 187, 191, 195, 203, 207, 211, 219, 227, 235, 239, 243, 251, 255, 259, 267, 271, 275, 283, 291, 299, 303, 307, 315, 319, 323, 331

Offset: 1

Author

Nadia Heninger and N. J. A. Sloane, Dec 16 2007

Keywords

• nonn
• easy

Comments

Also numbers of the form (4^a)*b - 1 with positive integer a and odd integer b. The sequence has linear growth and the limit of a(n)/n is 6. - Stefan Steinerberger, Dec 18 2007

Evil and odious terms alternate. - Vladimir Shevelev, Jun 22 2009

Also odd numbers of the form m = (A079523(k)-1)/2. - Vladimir Shevelev, Jul 06 2009

As a set, this is the complement of A079523 in the odd numbers. - Michel Dekking, Feb 13 2019

From Ctibor O. Zizka, Dec 28 2024: (Start)

Numbers k >= 1 such that (k + 1)*(k + 2*r)/2 is not a square for any r >= 1.

Numbers k such that A076114(k + 1) = 0. (End)

Examples

			11 in binary is 1011, which ends with two 1's.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Thomas Zaslavsky, Anti-Fibonacci Numbers: A Formula, Sep 26 2016.

Crossrefs

Cf. A076114, A079523, A121539.

Programs

Maple

N:= 1000: # to get all terms up to N
Odds:= [seq(2*i+1,i=0..floor((N-1)/2)]:
f:= proc(n) local L,x;
   L:= convert(n,base,2);
   x:= ListTools:-Search(0,L);
   if x = 0 then type(nops(L),even) else type(x,odd) fi
end proc:
A131323:= select(f,Odds); # Robert Israel, Apr 02 2014

Mathematica

Select[Range[500], OddQ[ # ] && EvenQ[FactorInteger[ # + 1][[1, 2]]] &] (* Stefan Steinerberger, Dec 18 2007 *)
en1Q[n_]:=Module[{ll=Last[Split[IntegerDigits[n,2]]]},Union[ll] =={1} &&EvenQ[Length[ll]]]; Select[Range[1,501,2],en1Q] (* Harvey P. Dale, May 18 2011 *)

PARI

is(n)=n%2 && valuation(n+1,2)%2==0 \\ Charles R Greathouse IV, Aug 20 2013

Python

from itertools import count, islice
def A131323_gen(startvalue=3): # generator of terms >= startvalue
    return map(lambda n:(n<<1)+1,filter(lambda n:(~(n+1)&n).bit_length()&1,count(max(startvalue>>1,1))))
A131323_list = list(islice(A131323_gen(),30)) # Chai Wah Wu, Sep 11 2024

Python

def A131323(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, s = n+x, bin(x+1)[2:]
        l = len(s)
        for i in range(l&1,l,2):
            c -= int(s[i])+int('0'+s[:i],2)
        return c
    return bisection(f,n,n)<<1|1 # Chai Wah Wu, Jan 29 2025

Formula

a(n) = 2*A079523(n) + 1. - Michel Dekking, Feb 13 2019

Extensions

More terms from Stefan Steinerberger, Dec 18 2007