

A285883


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.


0



3, 3, 9, 15, 1287, 63, 729, 195, 68475, 18224, 59049, 58695, 81344, 714879, 58564, 65535, 407835, 3569264, 5702444, 18224, 1894815, 994175, 4741659, 456975, 26790015, 17589375, 9617750, 8959544, 7928375, 17343104, 34574175, 1476224, 89513864, 74413899, 95242959
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OFFSET

1,1


COMMENTS

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.
We observe two classes of numbers:
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, ...
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, ...
We observe that a(10) = a(20) = 18224 belongs to both classes.
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?


LINKS



EXAMPLE

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.


MATHEMATICA

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}]


PROG

(PARI) d(n, m) = {vd = divisors(m); if (n > #vd, return (0)); vd[n]; }
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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



