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%I #12 Jul 13 2019 05:23:35
%S 1,2,3,5,8,14,15,20,35,63,80,99,104,195,224,384,440,560,935,1224,1539,
%T 2015,2079,5264,5984,12375,21735,41040,78624,123200,126224,156519,
%U 176175,201824,313599,338624,395199,453375,638000,1154439,1890944
%N a(n) is the smallest positive integer such that d(a(n))*d(a(n)+1) > d(a(n-1))*d(a(n-1)+1), where d(m) is the number of divisors of m and n > 1; a(1) = 1.
%C a(n) also equals the smallest positive integer such that d(a(n)(a(n)+1)) > d(a(n-1)(a(n-1)+1)).
%C That is, this is the sequence of indices of oblong numbers that have more divisors than the preceding oblong numbers. - _Michel Marcus_, Jul 13 2019
%e Since a(7) = 15, we want for a(8) the smallest positive integer m such that d(m)*d(m+1) > d(15)d(16) = 4*5=20. Checking: d(16)*d(17)=10, d(17)*d(18)=12, d(18)*d(19)=12, d(19)*d(20)=12. All of these are <= 20. But d(20)*d(21) = 6*4=24, which is > 20. So a(8) = 20.
%o (PARI) lista(nn) = {my(m = 0, nm); for (n=1, nn, if ((nm = numdiv(n*(n+1))) > m, m = nm; print1(n, ", ")););} \\ _Michel Marcus_, Jul 13 2019
%Y Cf. A002378 (oblong numbers), A092517.
%K nonn
%O 1,2
%A _Leroy Quet_, Jul 06 2008
%E More terms from _Max Alekseyev_, Apr 26 2010
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