A349241 Numbers N = pqrs such that |pqr - s| > |ps - qr|, where p <= q <= r <= s are the 4 prime factors of N.
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
Offset: 1
Keywords
Examples
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).
Links
- Marc LeBrun, four factor fun, math-fun mailing list (available for subscribers only), Nov 10 2021
Programs
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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
Comments