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

1,2

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

Formula

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

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 *)

Prog

(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)

Crossrefs

Cf. A064518, A064514, A064515, A064476.

Sequence in context: A073403 A191817 A270840 * A064518 A238923 A380033

Adjacent sequences: A380571 A380572 A380573 * A380575 A380576 A380577

Keyword

nonn,easy

Author

Chai Wah Wu, Mar 26 2025

Status

approved