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%I #19 Aug 01 2020 01:17:35
%S 1,60,280,420,840,1260,2520,6930,9240,13860,27720,55440,60060,120120,
%T 180180,240240,360360,720720,1021020,1801800,2042040,2282280,2762760,
%U 3063060,4084080,4564560,6126120,12252240,19399380,24504480,30630600,36756720,38798760,58198140,77597520
%N Positive integers where the number of triples of divisors (d1, d2, d3) such that d1 < d2 < d3 < 2*d1 and each pair of these divisors is pairwise coprime, sets a new record.
%C Records are 0, 1, 2, 3, 4, 5, 8, 9, 11, 13, 19, ...
%C Are terms > 4564560 products of primorials (cf. A025487)? Terms 4564560 < k <= 54765047434897800 are.
%C In a triple (d1, d2, d3) such that lcm(d1, d2, d3) = d1*d2*d2 <= k we must have d1^3 < k. Proof: Suppose d1^3 >= n. Then d1 * d2 * d3 > n since d2 > d1 and d3 > d1. Since any pair is coprime d1 * d2 * d3 = LCM(d1,d2,d3) is a divisor of n. A contradiction. - _David A. Corneth_ and _Amiram Eldar_, Jul 28 2020
%H David A. Corneth, <a href="/A333966/a333966_1.gp.txt">records; number of such triples in divisors of a(n)</a>
%e 280 has two such divisor triples; (4, 5, 7) and (5, 7, 8) and no number less than 280 has at least two such triples so 280 is in the sequence.
%o (PARI) upto(n) = { v = vectorsmall(n); for(i = 2, sqrtnint(n, 3), for(j = i + 1, min(sqrtint(n \ i), 2*i-2), g = gcd(i, j); if(g == 1, l = i * j / g; for(k = j + 1, min(2*i-1, n \ (i*j)), if(gcd(l, k) == 1, p = l*k; forstep(m = p, n, p, v[m]++ ); t++ ))))); my(res=List(1), r=v[1]); for(i=2, #v, if(v[i]>r, r=v[i]; listput(res,i))); res }
%Y Cf. A336442, A336443, A336628, A336629.
%K nonn
%O 1,2
%A _David A. Corneth_, Jul 22 2020
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