A108238 Numbers of the form (7^i)*(12^j), with i, j >= 0.

1, 7, 12, 49, 84, 144, 343, 588, 1008, 1728, 2401, 4116, 7056, 12096, 16807, 20736, 28812, 49392, 84672, 117649, 145152, 201684, 248832, 345744, 592704, 823543, 1016064, 1411788, 1741824, 2420208, 2985984, 4148928, 5764801, 7112448

Offset: 1

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 17 2005

Keywords

• nonn
• easy

Links

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

Crossrefs

Cf. A036566, A025631, A025632, A003599, A108056, A108201, A064476.

Programs

Mathematica

Take[7^#[[1]]*12^#[[2]]&/@Tuples[Range[0,10],2]//Union,40] (* Harvey P. Dale, Mar 05 2017 *)
n = 10^6; Flatten[Table[7^i*12^j, {i, 0, Log[7, n]}, {j, 0, Log[12, n/7^i]}]] // Sort (* Amiram Eldar, Sep 26 2020 *)

Python

from sympy import integer_log
def A108238(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(integer_log(x//12**i,7)[0]+1 for i in range(integer_log(x,12)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 26 2025

Formula

Sum_{n>=1} 1/a(n) = (7*12)/((7-1)*(12-1)) = 14/11. - Amiram Eldar, Sep 26 2020

a(n) ~ exp(sqrt(2*log(7)*log(12)*n)) / sqrt(84). - Vaclav Kotesovec, Sep 26 2020