A257999 Numbers of the form, 2^i*3^j, i+j odd.

2, 3, 8, 12, 18, 27, 32, 48, 72, 108, 128, 162, 192, 243, 288, 432, 512, 648, 768, 972, 1152, 1458, 1728, 2048, 2187, 2592, 3072, 3888, 4608, 5832, 6912, 8192, 8748, 10368, 12288, 13122, 15552, 18432, 19683, 23328, 27648, 32768, 34992, 41472, 49152, 52488

Offset: 1

Author

Reinhard Zumkeller, May 16 2015

Keywords

• nonn

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Complement of A036667 with respect to A003586.

Intersection of A026424 and A003586.

Cf. A069352, A022328, A022329.

Programs

Haskell

a257999 n = a257999_list !! (n-1)
a257999_list = filter (odd . flip mod 2 . a001222) a003586_list

Mathematica

max = 53000; Reap[Do[k = 2^i*3^j; If[k <= max && OddQ[i + j], Sow[k]], {i, 0, Log[2, max] // Ceiling}, {j, 0, Log[3, max] // Ceiling}]][[2, 1]] // Union (* Amiram Eldar, Feb 18 2021 after Jean-François Alcover at A036667 *)

Python

from sympy import integer_log
def A257999(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): return n+x-sum((x//3**i).bit_length()+(i&1)>>1 for i in range(integer_log(x, 3)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Jan 30 2025

Formula

A069352(a(n)) mod 2 = 1.

Sum_{n>=1} 1/a(n) = 5/4. - Amiram Eldar, Feb 18 2021