A287935 - OEIS

%I #17 Jun 27 2021 03:40:19

%S 2,6,105,23023,16028909,40502880317

%N a(n) is the smallest squarefree number k with n prime factors such that gcd(k, d2-d1) = 1 for all coprime pairs of divisors of k, 1 < d1 < d2 < k.

%C a(2) - a(5) were calculated by Erdős and Evans. They formulated the sequence in terms of sets of primes. They proved that a(n) exists for all n >= 2, and that 1-2/n < log_2(a(n))/n^2 < log_2(3 + e), where the upper bound holds for any e > 0 for sufficiently large n.

%C The prime factors of the first terms are:

%C a(1) = 2

%C a(2) = 2 * 3

%C a(3) = 3 * 5 * 7

%C a(4) = 7 * 11 * 13 * 23

%C a(5) = 13 * 17 * 29 * 41 * 61

%C a(6) = 19 * 37 * 47 * 53 * 101 * 229

%C a(7) <= 1378987557700217. - _Giovanni Resta_, Jun 03 2017

%H Paul Erdős and Anthony B. Evans, <a href="http://dx.doi.org/10.1006/jnth.1996.0135">Sets of Prime Numbers Satisfying a Divisibility Condition</a>, Journal of Number Theory, Vol. 61, No. 1, (1996), pp. 39-43.

%e 105 = 3 * 5 * 7, gcd(5-3, 105) = gcd(7-3, 105) = gcd(7-5, 105) = gcd(3*5-7, 105) = gcd(3*7-5, 105) = gcd(5*7-3, 105) = 1. 105 is the smallest product of 3 different primes with this property, thus a(3) = 105.

%t aQ[n_] := Module[{g = True}, d = Drop[Drop[Divisors[n], 1], -1]; nd = Length[d]; For[k1 = 0, k1 < nd - 1, k1++; g1 = 1; d1 = d[[k1]]; For[k2 = k1, k2 < nd, k2++; d2 = d[[k2]]; If[GCD[d1, d2] > 1, Continue[]];   If[GCD[n, d2 - d1] > 1, g1 = 0; Break[]]]; If[g1 == 0, g = False; Break[]]]; g]; m = 2; k = 1; a = {}; While[Length[a] < 5, If[SquareFreeQ[k] && PrimeNu[k] == m && aQ[k], a = AppendTo[a, k]; m++, k++]];a

%K nonn,more

%O 1,1

%A _Amiram Eldar_, Jun 03 2017