A380574 For an integer k with prime factorization p_1*p_2*p_3* ... *p_m let k* = (p_1+1)*(p_2+1)*(p_3+1)* ... *(p_m+1); sequence gives k* such that k* is divisible by k.

1, 12, 36, 144, 432, 1296, 1728, 5184, 15552, 20736, 46656, 62208, 186624, 248832, 559872, 746496, 1679616, 2239488, 2985984, 6718464, 8957952, 20155392, 26873856, 35831808, 60466176, 80621568, 107495424, 241864704, 322486272, 429981696, 725594112, 967458816

Offset: 1

Author

Chai Wah Wu, Mar 26 2025

Keywords

• nonn
• easy

Comments

Terms of A064518 in increasing order.

Numbers of the form 3^i*4^j with j <= i <= 2j.

Subsequence of A064476.

Links

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

Crossrefs

Cf. A064518, A064514, A064515, A064476.

Programs

Mathematica

With[{max = 10^9}, Select[Flatten[Table[3^i*4^j, {j, 0, Log[12, max]}, {i, j, 2*j}]] // Sort, # <= max &]] (* Amiram Eldar, Mar 29 2025 *)

Python

from sympy import integer_log
def A380574(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(max(0,min((j:=i<<1),integer_log(x>>j,3)[0])-i+1) for i in range(x.bit_length()+1>>1))
    return bisection(f,n,n)

Formula

Sum_{n>=1} 1/a(n) = 432/385. - Amiram Eldar, Mar 29 2025