A026467|A113626

[Richard Mathar]

Maximilian Hasler wrote:
>>  a(n) = smallest positive integer such that
>>  d(a(n)) = d(a(n)+2n) = 2n,
>>  where d(m) is the number of positive divisors of
>>  m.
>>
>>  The sequence begins: 3,6,12,70,...
> 
> lq080419(n,a)={n*=2;until(numdiv(a)==n && numdiv(a+n)==n,a++);a}
> 
> for(i=1,99,print1(lq080419(i)", "))
> 3, 6, 12, 70, 600281, 60, 1458, 264, 450, 266875,
>   *** until: user interrupt after 1mn, 18,250 ms.
> (next term seems quite large, maybe the function should be made a bit
> more intelligent....)
> Maximilian
> 
> 


11th term is <= 12670498046853.

Proving whether that is the minimal one should be doable
without too much trouble...