A225858 Numbers of the form 2^i*3^j*(12k+7) or 2^i*3^j*(12k+11), i, j, k >= 0.

7, 11, 14, 19, 21, 22, 23, 28, 31, 33, 35, 38, 42, 43, 44, 46, 47, 55, 56, 57, 59, 62, 63, 66, 67, 69, 70, 71, 76, 79, 83, 84, 86, 88, 91, 92, 93, 94, 95, 99, 103, 105, 107, 110, 112, 114, 115, 118, 119, 124, 126, 127, 129, 131, 132, 134, 138, 139, 140

Offset: 1

Author

Ralf Stephan, May 18 2013

Keywords

• nonn
• easy

Comments

The asymptotic density of this sequence is 1/2. - Amiram Eldar, Nov 14 2023

Examples

			From _David A. Corneth_, Nov 11 2023: (Start)
28 = 2^2 * 7 = 2^2 * 3^0 * (12*0 + 7) is in the sequence as it meets the first form.
76 = 2^2 * 19 = 2^2 * 3^0 * (12*1 + 7) is in the sequence as it meets the first form.
15 = 3 * 5 = 2^0 * 3^1 * (12*0 + 5) is not in the sequence as it does not match any of the desired forms. (End)
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Complement of A225857. Cf. A225838.

Programs

Magma

[n: n in [1..200] | d mod 4 eq 3 where d is n div (2^Valuation(n,2)*3^Valuation(n,3))]; // Bruno Berselli, May 16 2013

Mathematica

Select[Range[140], Mod[#/Times @@ ({2, 3}^IntegerExponent[#, {2, 3}]), 4] == 3 &] (* Amiram Eldar, Nov 14 2023 *)

PARI

for(n=1,200,t=n/(2^valuation(n,2)*3^valuation(n,3));if((t%4==3),print1(n,",")))

Python

from itertools import count
from sympy import integer_log
def A225858(n):
    def f(x):
        c = n
        for i in range(integer_log(x,3)[0]+1):
            i2 = 3**i
            for j in count(0):
                k = i2<x:
                    break
                m = x//k
                c += (m-1)//12+(m-5)//12+2
        return c
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Feb 24 2025

Extensions

Name clarified by Peter Munn, Nov 11 2023