A108347 Numbers of the form (3^i)*(5^j)*(7^k), with i, j, k >= 0.

1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 49, 63, 75, 81, 105, 125, 135, 147, 175, 189, 225, 243, 245, 315, 343, 375, 405, 441, 525, 567, 625, 675, 729, 735, 875, 945, 1029, 1125, 1215, 1225, 1323, 1575, 1701, 1715, 1875, 2025, 2187, 2205, 2401, 2625, 2835, 3087

Offset: 1

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jul 01 2005

Keywords

• nonn

Comments

The Heinz numbers of the partitions into parts 2,3, and 4 (including the number 1, the Heinz number of the empty partition). We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [2,3,3,4] the Heinz number is 3*5*5*7 = 525; it is in the sequence. - Emeric Deutsch , May 21 2015

Numbers m | 105^e with integer e >= 0. - Michael De Vlieger, Aug 22 2019

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)

Crossrefs

Cf. A003586, A003591-A003595, A051037, A108319, A108513, A215366.

The odd terms of A002473.

Programs

Magma

[n: n in [1..4000] | PrimeDivisors(n) subset [3,5,7]]; // Bruno Berselli, Sep 24 2012

Maple

with(numtheory): S := {}: for j to 3100 do if `subset`(factorset(j), {3, 5, 7}) then S := `union`(S, {j}) else end if end do: S; # Emeric Deutsch, May 21 2015
# alternative
isA108347 := proc(n)
      if n = 1 then
        true;
    else
        return (numtheory[factorset](n) minus {3, 5, 7} = {} );
    end if;
end proc:
A108347 := proc(n)
     option remember;
     if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA108347(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A108347(n),n=1..80); # R. J. Mathar, Jun 06 2024

Mathematica

With[{n = 3087}, Sort@ Flatten@ Table[3^i * 5^j * 7^k, {i, 0, Log[3, n]}, {j, 0, Log[5, n/2^i]}, {k, 0, Log[7, n/(3^i*5^j)]}]] (* Michael De Vlieger, Aug 22 2019 *)

Python

from sympy import integer_log
def A108347(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: 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,7)[0]+1):
            for j in range(integer_log(m:=x//7**i,5)[0]+1):
                c -= integer_log(m//5**j,3)[0]+1
        return c
    return bisection(f,n,n) # Chai Wah Wu, Sep 16 2024

Formula

Sum_{n>=1} 1/a(n) = (3*5*7)/((3-1)*(5-1)*(7-1)) = 35/16. - Amiram Eldar, Sep 22 2020

a(n) ~ exp((6*log(3)*log(5)*log(7)*n)^(1/3)) / sqrt(105). - Vaclav Kotesovec, Sep 23 2020