A107466 Numbers of the form (5^i)*(13^j).

1, 5, 13, 25, 65, 125, 169, 325, 625, 845, 1625, 2197, 3125, 4225, 8125, 10985, 15625, 21125, 28561, 40625, 54925, 78125, 105625, 142805, 203125, 274625, 371293, 390625, 528125, 714025, 1015625, 1373125, 1856465, 1953125, 2640625

Offset: 1

Author

Douglas Winston (douglas.winston(AT)srupc.com), May 27 2005

Keywords

• nonn

Links

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

Crossrefs

Cf. A003586, A003592, A003593, A003591, A003594, A003595, A003596, A003597, A003598, A003599, A107326, A107364.

Programs

Mathematica

mx = 2700000; Sort@ Flatten@ Table[5^i*13^j, {i, 0, Log[5, mx]}, {j, 0, Log[13, mx/5^i]}] (* Robert G. Wilson v, Aug 17 2012 *)

PARI

list(lim)=my(v=List(),N);for(n=0,log(lim)\log(13),N=13^n;while(N<=lim,listput(v,N);N*=5));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

Python

from sympy import integer_log
def A107466(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//13**i,5)[0]+1 for i in range(integer_log(x,13)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Formula

Sum_{n>=1} 1/a(n) = (5*13)/((5-1)*(13-1)) = 65/48. - Amiram Eldar, Sep 23 2020

a(n) ~ exp(sqrt(2*log(5)*log(13)*n)) / sqrt(65). - Vaclav Kotesovec, Sep 23 2020