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A333966
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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.
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3
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1, 60, 280, 420, 840, 1260, 2520, 6930, 9240, 13860, 27720, 55440, 60060, 120120, 180180, 240240, 360360, 720720, 1021020, 1801800, 2042040, 2282280, 2762760, 3063060, 4084080, 4564560, 6126120, 12252240, 19399380, 24504480, 30630600, 36756720, 38798760, 58198140, 77597520
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OFFSET
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1,2
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COMMENTS
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Records are 0, 1, 2, 3, 4, 5, 8, 9, 11, 13, 19, ...
Are terms > 4564560 products of primorials (cf. A025487)? Terms 4564560 < k <= 54765047434897800 are.
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
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LINKS
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EXAMPLE
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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.
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PROG
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(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 }
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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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