A251728 Semiprimes p*q for which p <= q < p^2.

4, 6, 9, 15, 21, 25, 35, 49, 55, 65, 77, 85, 91, 95, 115, 119, 121, 133, 143, 161, 169, 187, 203, 209, 217, 221, 247, 253, 259, 287, 289, 299, 301, 319, 323, 329, 341, 361, 377, 391, 403, 407, 437, 451, 473, 481, 493, 517, 527, 529, 533, 551, 559, 583, 589, 611, 629, 649, 667, 671, 689, 697, 703

Offset: 1

Author

Antti Karttunen, Dec 16 2014

Keywords

• nonn

Comments

Semiprimes p*q for which there exists r <= q such that r^k <= p <= q < r^(k+1), for some k >= 1, i.e., semiprimes whose both prime factors fit inside a semiopen range of two consecutive powers of some natural number r which itself is not greater than the larger prime factor. If such r exists, then it must be <= p (the smaller prime factor of n), which forces q to be less than p^2. On the other hand, when p <= q < p^2, then setting r = p and k = 1 satisfies the equation r^k <= p <= q < r^(k+1).

Assuming that A054272(n), the number of primes in interval [p(n), p(n)^2], is nondecreasing (implied for example if Legendre's or Brocard's conjecture is true), it follows that for any a(n), A003961(a(n)) is also in sequence. In other words, whenever prime(i)*prime(j) is in the sequence, then so is also prime(i+1)*prime(j+1).

From above would follow also that these are all the "settled semiprimes" that occur in a square array A083221 constructed from the sieve of Eratosthenes, from the level A251719 downward. Furthermore, this sequence would then be an infinite disjoint union of sequences of A003961-iterates starting from the initial values given in A251724.

See also the comments in the complementary sequence of semiprimes, A138511.

Composite numbers n with all prime factors greater than the cube root of n. - Doug Bell, Oct 27 2015

If "p <= q" in the definition were changed to "p < q" then the squares of primes (A001248) would be removed, yielding A138109. - Jon E. Schoenfield, Dec 27 2022

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Carlos Rivera, Conjecture 90. Semiprimes/Primes ->log(2), The Prime Puzzles & Problems Connection.
• Wikipedia, Brocard's conjecture

Crossrefs

An intersection of A251726 and A001358 (semiprimes).

Complement of A138511 in A001358.

A251724 after the initial 2 is a subsequence.

Cf. A003961, A006530, A054272, A078898, A083221, A138109, A138510, A243055, A250474, A251719.

Programs

Haskell

a251728 n = a251728_list !! (n-1)
a251728_list = filter f [1..] where
                      f x = q < p ^ 2 && a010051' q == 1
                            where q = div x p; p = a020639 x
-- Reinhard Zumkeller, Jan 06 2015

Mathematica

fQ[n_] := Block[{pf = FactorInteger@ n, p, q}, p = pf[[1, 1]]; q = pf[[-1, 1]]; And[p <= q < p^2, PrimeOmega@ n == 2]]; Select[Range@ 720, fQ] (* Michael De Vlieger, Oct 27 2015 *)

PARI

lista(nn) = forcomposite(n=1, nn, my(f = factor(n));if (#select(x->(x > n^(1/3)), f[,1]) == #f~, print1(n, ", "))); \\ Michel Marcus, Oct 27 2015

PARI

list(lim)=my(v=List()); forprime(q=2,sqrtnint((lim\1)^2,3), forprime(p=sqrtint(q)+1,min(q,lim\q), listput(v,p*q))); Set(v) \\ Charles R Greathouse IV, Oct 27 2015

Python

from math import isqrt
from sympy import primepi, primerange
def A251728(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(x)))*(t-1)>>1)-sum(primepi(min(x//p,p**2)) for p in primerange(s+1)))
    return bisection(f,n,n) # Chai Wah Wu, Mar 05 2025

Formula

For all n >= 1, A078898(a(n)) = A243055(a(n)) + 2.

Limit_{n->oo} n*log(a(n))/a(n) = log(2). - Alain Rocchelli, Nov 10 2022