%I #21 Aug 30 2019 22:09:16
%S 1,1,61,61,11371,11371,7392171,168776043,1584614377,38045133481,
%T 30386250649371,1848289766450821
%N a(1) = 1, then a(n) = the smallest number k such that the number of divisors of the n numbers from k through k+n-1 are in strictly ascending order.
%C tau(k) < tau(k+1) < ... < tau(k+n-1).
%C a(11) > 10^12. - _Donovan Johnson_, Oct 13 2009
%C a(11) > 10^13. - _Giovanni Resta_, Jul 25 2013
%C a(13) > 2.64*10^15. - _Jud McCranie_, Mar 27 2019
%e a(3) = 61 = a(4) as tau(61) = 2 < tau(62) = 4 < tau(63) = 6 < tau(64) = 7.
%Y Cf. A075029, A075031.
%K nonn,more
%O 1,3
%A _Amarnath Murthy_, Sep 02 2002
%E More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 19 2003
%E a(7)-a(10) from _Donovan Johnson_, Oct 13 2009
%E a(11)-a(12) from _Jud McCranie_, Mar 27 2019