A393012 List giving least 5-rough number (A007310) of each prime signature.

1, 5, 25, 35, 125, 175, 385, 625, 875, 1225, 1925, 3125, 4375, 5005, 6125, 9625, 13475, 15625, 21875, 25025, 30625, 42875, 48125, 67375, 78125, 85085, 109375, 125125, 148225, 153125, 175175, 214375, 240625, 336875, 390625, 425425, 471625, 546875, 625625, 741125

Offset

1,2

Comments

Sequences of the form "List giving least p-rough number of each prime signature." for prime p can help in a search for numbers k <= U such that sigma(k) >= b*k where U is some chosen upper bound and b is some chosen constant. If such sequence is S(p, U) then we can find the largest ratio sigma(k)/k for k in S(p, U) and use that to decide to continue searching.

b = 2 describes nonabundant numbers (A023196). p = 2 describes A025487. p = 3 describes A147516. This sequence has p = 5.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Michael De Vlieger, Mathematica algorithm that generates least prime(k)-rough numbers less than a limit.

Formula

Sum_{n>=1} 1/a(n) = Product_{n>=3} 1/(1 - 3/A070826(n)) = 1.290389472537278155572... . - Amiram Eldar, Feb 13 2026

Example

175 is in the sequence as it is a product of consecutive primes starting at 5 and for any two distinct prime powers p^ep and q^eq where p < q we have ep >= eq.

Prog

(PARI) is(n) = {if(n == 1, return(1)); e = valuation(n, 5); if(e == 0, return(0)); n\=5^e; if(n==1, return(1)); forprime(p = 7, oo, v = valuation(n, p); if(v > e, return(0)); if(v == 0, return(0)); n\=p^v; if(n==1, return(1)); e = v)}
(Python)
from itertools import count
from functools import lru_cache
from sympy import prime, integer_log
from oeis_sequences.OEISsequences import bisection
def A393012(n):
    @lru_cache(maxsize=None)
    def g(x, m, j): return sum(g(x//(prime(m)**i), m-1, i) for i in range(j, integer_log(x, prime(m))[0]+1)) if m>3 else max(0, integer_log(x, 5)[0]+1-j)
    def f(x):
        c, p = n-1+x, 1
        for k in count(3):
            p *= prime(k)
            if p>x:
                break
            c -= g(x, k, 1)
        return c
    return bisection(f, n, n) # Chai Wah Wu, Mar 31 2026
(Python)
from itertools import islice
from heapq import heappop, heappush
from sympy import factorint, prevprime, nextprime
def A393012_gen(): # generator of terms if the first n terms are desired.
    h, hset = [1], {1}
    while True:
        yield (m:=heappop(h))
        ps = factorint(m)
        for p in ps:
            if p == 5 or ps[prevprime(p)]>ps[p]:
                mp = m*p
                if mp not in hset:
                    heappush(h, mp)
                    hset.add(mp)
        mp = m*nextprime(max(ps.keys(), default=3))
        if mp not in hset:
            heappush(h, mp)
            hset.add(mp)
A393012_list = list(islice(A393012_gen(), 40)) # Chai Wah Wu, Mar 31 2026

Crossrefs

Cf. A007310, A023196, A025487, A070826, A147516.

Sequence in context: A396206 A084967 A249827 * A343734 A279541 A361615

Adjacent sequences: A393009 A393010 A393011 * A393013 A393014 A393015

Keyword

nonn

Author

David A. Corneth, Feb 07 2026

Status

approved