%I #42 Mar 17 2021 22:56:35
%S 3,3,9,15,1287,63,729,195,68475,18224,59049,58695,81344,714879,58564,
%T 65535,407835,3569264,5702444,18224,1894815,994175,4741659,456975,
%U 26790015,17589375,9617750,8959544,7928375,17343104,34574175,1476224,89513864,74413899,95242959
%N Let d(n,k) be the nth divisor of a number k. a(n) is the smallest k such that d(n+1,k+1) = d(n,k) + 1.
%C Or a(n) is the smallest number k such that, if d is the nth divisor of k, then d+1 is the (n+1)th divisor of k+1.
%C We observe two classes of numbers:
%C 1) Numbers k such that d(n,k) < k and d(n+1,k+1) < k+1. The numbers of the sequence having this property are 1287, 68475, 18224, 81344, ...
%C 2) Numbers k where d(n,k) = k and d(n+1,k+1) = k+1 are trivial divisors. The numbers of the sequence having this property are 3, 9, 15, 63, 729, 195, 59049, 58564, 65535, 18224, 456975, ...
%C We observe that a(10) = a(20) = 18224 belongs to both classes.
%C From this remark, we introduce the notion of order O(a(n)) = number of occurrences of a(n). For instance, O(a(3))= 1, O(a(10)) = O(a(20)) = 2. Is there n such that O(a(n)) > 2?
%e a(5) = 1287 because the 5th divisor of 1287 is 13 and the 6th divisor of 1288 is 14. Hence, d(6,1288) = d(5,1287)+ 1 = 14.
%t Do[k=3;While[!(Length[Divisors[k]]>=n&&Length[Divisors[k+1]]>=n+1&&Part[Divisors[k],n]+1==Part[Divisors[k+1],n+1]),k++];Print[n," ",k],{n,1,50}]
%o (PARI) d(n,m) = {vd = divisors(m); if (n > #vd, return (0)); vd[n];}
%o a(n) = {m = 2; while(!((db = d(n+1, m+1)) && (da = d(n,m)) && (db == da+1)), m++); m;} \\ _Michel Marcus_, May 17 2017
%Y Cf. A027750.
%K nonn
%O 1,1
%A _Michel Lagneau_, May 16 2017
