cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A333966 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.

Original entry on oeis.org

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
Offset: 1

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Author

David A. Corneth, Jul 22 2020

Keywords

Comments

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

Examples

			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.
		

Crossrefs

Programs

  • 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 }

A336628 Numbers k that have 3 divisors d1, d2, d3 such that d1 < d2 < d3 < 2*d1 and are pairwise coprime and d1*d2*d3 = k.

Original entry on oeis.org

60, 140, 210, 280, 315, 360, 462, 504, 616, 630, 693, 728, 770, 792, 819, 910, 924, 936, 990, 1001, 1092, 1144, 1170, 1287, 1320, 1386, 1430, 1530, 1560, 1584, 1638, 1683, 1716, 1870, 1872, 1989, 2002, 2090, 2142, 2145, 2210, 2244, 2288, 2310, 2431, 2448, 2470, 2508
Offset: 1

Views

Author

David A. Corneth and Amiram Eldar, Jul 28 2020

Keywords

Comments

(k/4)^(1/3) < d1 < k^(1/3). Proof: as k = d1 * d2 * d3 < d1 * (2*d1) * (2*d1) = 4*d1^3 we have (k/4)^(1/3) < d1 and as k = d1 * d2 * d3 > d1 * d1 * d1 = d1^3 we have k^(1/3) > d1. Q.e.d.

Examples

			210 is in the sequence because 5*6*7 = 210 and each of these factors are pairwise coprime and 5 < 6 < 7 < 2*5 = 10.
		

Crossrefs

Showing 1-2 of 2 results.