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A349241
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Numbers N = pqrs such that |pqr - s| > |ps - qr|, where p <= q <= r <= s are the 4 prime factors of N.
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1
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16, 24, 36, 54, 60, 81, 90, 100, 126, 135, 140, 150, 189, 196, 210, 225, 250, 294, 308, 315, 330, 350, 364, 375, 390, 441, 462, 484, 490, 495, 525, 546, 550, 572, 585, 625, 650, 676, 686, 693, 714, 726, 735, 748, 770, 798, 819, 825, 836, 850, 858, 875, 884, 910, 950, 975, 988
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OFFSET
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1,1
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COMMENTS
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The set A014613 of numbers n with bigomega(n) = A001222(n) = 4, can be partitioned in these here and their complement A349242. It was suggested (cf. math-fun post in LINKS) to call these here the "trans"- and the others the "cis"-type.
These here include squares of semiprimes (A074985), and in particular 4th powers of primes (A030514), for which |ps - qr| = 0.
Within the 4-almost primes below 10^k, k = 2, 3, ...,8, we have (8, 57, 497, 4960, 49228, 491397, 4869917, ...) of trans type, and more than twice (or even three times) as many of cis type.
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LINKS
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Marc LeBrun, four factor fun, math-fun mailing list (available for subscribers only), Nov 10 2021
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FORMULA
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{ N in A014613 | |g - N/g| > |sg - N/sg| }, where g = gpf(N) = A006530(N) is the greatest, and s = spf(N) = A020639(N) is the smallest prime factor.
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EXAMPLE
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16 = 2^4 = u*v with u = v = 2*2 closer (equal) than u = 2*2*2, v = 2 (difference 8 - 2 = 6).
24 = 2^3*3 = u*v with u = 2*2, v = 2*3 closer (distance 6 - 4 = 2) than u = 2*2*2, v = 3 (distance 8 - 3 = 5).
36 = 2^2*3^2 = u*v with u = v = 2*3 closer (equal) than u = 2^2*3, v = 3 (difference 12 - 3 = 9).
The 4-almost prime 40 = 2^3*5 is not in this sequence because the factorization 40 = u*v with u = 2^3, v = 5 has closer factors (distance 8 - 5 = 3) than u = 2*2, v = 2*5 (distance 10 - 4 = 6).
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PROG
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(PARI) select( {is_A349241(n, a(u)=abs(u-n\u))=bigomega(n)==4 && a((s=factor(n)[, 1])[#s])>a(s[1]*s[#s])}, [1..1000])
(Python)
from itertools import chain
from sympy import factorint
def expand(n):
return list(chain.from_iterable([[i[0] for j in range(i[1])] for i in factorint(n).items()]))
def is_ok(p, q, r, s):
return abs(p*q*r-s) > abs(p*s-q*r)
print([i for i in range(2, 1000) if len(expand(i)) == 4 and is_ok(*expand(i))]) # Gleb Ivanov, Nov 12 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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