A339690 Positive integers of the form 4^i*9^j*k with gcd(k,6)=1.
1, 4, 5, 7, 9, 11, 13, 16, 17, 19, 20, 23, 25, 28, 29, 31, 35, 36, 37, 41, 43, 44, 45, 47, 49, 52, 53, 55, 59, 61, 63, 64, 65, 67, 68, 71, 73, 76, 77, 79, 80, 81, 83, 85, 89, 91, 92, 95, 97, 99, 100, 101, 103, 107, 109, 112, 113, 115, 116, 117, 119, 121
Offset: 1
Keywords
Examples
Numbers are removed by the sieve only due to the presence of a smaller number, so 1 is in the sequence as the smallest positive integer. The sieve removes 2, as it is twice 1, which is in the sequence; so 2 is not in the sequence. The sieve removes 3, as it is three times 1, which is in the sequence, so 3 is not in the sequence. There are no integers m for which 3m = 4 or 6m = 4; 2m = 4 for m = 2, but 2 is not in the sequence; so the sieve does not remove 4, so 4 is in the sequence.
Links
- Jan Snellman, Greedy Regular Convolutions, arXiv:2504.02795 [math.NT], 2025.
- Eric Weisstein's World of Mathematics, Group.
- Eric Weisstein's World of Mathematics, Squarefree Part.
- Index entries for sequences generated by sieves
Crossrefs
Programs
-
Mathematica
Select[Range[117], EvenQ[IntegerExponent[#, 2]] && EvenQ[IntegerExponent[#, 3]] &] f[p_, e_] := p^Mod[e, 2]; core[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[121], CoprimeQ[core[#], 6] &] (* Amiram Eldar, Feb 06 2021 *)
-
PARI
isok(m) = core(m) % 6 == 1 || core(m) % 6 == 5;
-
Python
from itertools import count from sympy import integer_log def A339690(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 = n+x for i in range(integer_log(x,9)[0]+1): i2 = 9**i for j in count(0,2): k = i2<
x: break m = x//k c -= (m-1)//6+(m-5)//6+2 return c return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025
Formula
{a(n) : n >= 1} = {m : A307150(m) = 6m, m >= 0}.
{a(n) : n >= 1} = {k : k = A052330(4m), m >= 0}.
A329575(n) = a(n) * 3.
{A036668(n) : n >= 0} = {a(n) : n >= 1} U {6 * a(n) : n >= 1}.
{A003159(n) : n >= 1} = {a(n) : n >= 1} U {3 * a(n) : n >= 1}.
{A007417(n) : n >= 1} = {a(n) : n >= 1} U {2 * a(n) : n >= 1}.
a(n) ~ 2n.
Comments