%I #13 May 17 2016 09:22:51
%S 4,6,6,12,12,24,24,36,36,48,60,60,60,60,60,60,60,120,120,120,120,120,
%T 120,120,180,180,180,180,180,180,240,240,240,240,240,240,240,240,240,
%U 240,240,360,360,360,360,360,360
%N Smallest k in the interval [prime(n), 2*prime(n)], such that k has the maximal number of divisors in this interval.
%C Conjecturally the different values of the sequence are highly composite numbers (A002182, n>=3).
%H Peter J. C. Moses, <a href="/A272771/b272771.txt">Table of n, a(n) for n = 1..1000</a>
%e Let n=5, prime(n)=11. In interval [11,22] we have 3 numbers 12,18 and 20 with the maximal number of divisors in this interval(6). Since 12 is the smallest of them, then a(5)=12.
%t Table[Function[p, First@ FirstPosition[#, Max@ #] + p - 1 &@ Map[DivisorSigma[0, #] &, Range[p, 2 p]]]@ Prime@ n, {n, 80}] (* _Michael De Vlieger_, May 07 2016, Version 10 *)
%Y Cf. A000005, A000040.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, May 06 2016