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A287935
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.
0
2, 6, 105, 23023, 16028909, 40502880317
OFFSET
1,1
COMMENTS
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.
The prime factors of the first terms are:
a(1) = 2
a(2) = 2 * 3
a(3) = 3 * 5 * 7
a(4) = 7 * 11 * 13 * 23
a(5) = 13 * 17 * 29 * 41 * 61
a(6) = 19 * 37 * 47 * 53 * 101 * 229
a(7) <= 1378987557700217. - Giovanni Resta, Jun 03 2017
LINKS
Paul Erdős and Anthony B. Evans, Sets of Prime Numbers Satisfying a Divisibility Condition, Journal of Number Theory, Vol. 61, No. 1, (1996), pp. 39-43.
EXAMPLE
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.
MATHEMATICA
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
CROSSREFS
Sequence in context: A294906 A284262 A374006 * A357090 A181036 A222854
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Jun 03 2017
STATUS
approved