A135581 The 5th divisor of numbers with 25 divisors.

6, 8, 8, 15, 21, 11, 13, 27, 16, 35, 16, 27, 16, 27, 55, 27, 16, 16, 16, 65, 27, 16, 77, 16, 85, 16, 29, 91, 31, 16, 95, 16, 37, 115, 16, 119, 16, 41, 43, 133, 16, 47, 16, 143, 125, 16, 125, 16, 53, 161, 16, 59, 16, 61, 125, 187, 16, 67, 16, 203, 125, 16, 209, 71, 16, 125

Offset: 1

Author

G. H. Ens (GerardEns(AT)gmail.com), Feb 24 2008

Keywords

• nonn
• look
• easy

Comments

n=1 means the first number that has 25 divisors (1296), 6 is the 5th divisor of 1296. The second number with 25 divisors is 10000 and its 5th divisor is 8

This is one example of such a sequence where the divisor index is the square root of the total number of divisors (self included).

Other examples would be the 6th divisor of numbers with 36 divisors, 7th divisor of numbers with 49 divisors, etc.

Choice of the square root is arbitrary.

All but 16 primes {2, 3, 5, 7, 17, 19, 23, 83, 89, 97, 101, 103, 107, 109, 113} are in this sequence; p^3 and p^4 are in this sequence for all prime p; pq is in this sequence for all prime p and q with p < q < p^2. No other terms are members. - Charles R Greathouse IV, Nov 28 2011

Examples

			a(1) = 6 because 6 is the 5th divisor of 1296 and 1296 is the first number with 25 divisors.
a(2) = 8 because 8 is the 5th divisor of 10000 and 10000 is the second number with 25 divisors.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Laurens Lapré, Natural division.
• Wikipedia, Divisor function

Crossrefs

Cf. A005179, A137488, A200722, A201266.

Programs

Haskell

a135581 n = [d | d <- [1..], a137488 n `mod` d == 0] !! 4
-- Reinhard Zumkeller, Nov 29 2011

Mathematica

upto=10^10;With[{max1=Ceiling[Power[upto, (4)^-1]],max2=Ceiling[ Power[ upto, (24)^-1]]},Take[Divisors[#][[5]]&/@Select[Union[Join[ Range[ max2]^24, Times@@@(Subsets[Range[max1],{2}]^4)]],DivisorSigma[0,#] == 25&], Ceiling[max1/4]]] (* Harvey P. Dale, Nov 25 2011 *)

Python

from math import isqrt
from sympy import primepi, integer_nthroot, primerange, divisors
def A135581(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 int(n+x+(t:=primepi(s:=isqrt(y:=integer_nthroot(x,4)[0])))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)))-primepi(integer_nthroot(x,24)[0])
    return divisors(bisection(f,n,n))[4] # Chai Wah Wu, Feb 22 2025

Extensions

Corrected and extended by R. J. Mathar, Apr 21 2008. The original entries were wrong from the 16th term onwards.