[seqfan] A075040

[Don Reble]

Seqfans:

> %I A075040
> %S 0,33,242,230,7939375,1274,76571890623,8294,959075,248750
> %N a(1) = 0;
>    a(n) = smallest of three consecutive numbers all with 2n divisors.
> %C a(11) <= 104228508212890623,
>    a(13) <= 1489106237081787109375. - _Giovanni Resta_, Aug 14 2013

    a(11): Numbers with 22 divisors are either p^21 or q^10 * r,
    with p,q,r prime. From that bound on a(11), we have
    p<7 and q<47 (since r>=2).

    So a(11) is near one of three p^21 values, or (since 43 is the 14th
    prime), there are only 14*13*12 possible combinations of q^10
    factors for a(11),a(11)+1,a(11)+2. One can use CRT, and search
    through very sparse arithmetic sequences. I confirm that
    a(11) = 104228508212890623

    More generally, a(N) computes well when N is prime. Assume that the
    big factors of X,X+1,X+2 are powers of 2,3,5, in some order (as Dr.
    Resta probably did), and seek that kind of solution. Then use it to
    compute limits on p and q, and finish the search.

    a(13) = 1489106237081787109375
    a(17) = 273062471666259918212890623
    a(19) = 804505911103256259918212890623
    a(23) = 490685203356467392256259918212890623
    a(29) = 6794675247932944436619977392256259918212890623
    a(31) = 329757106427071213106619977392256259918212890623
    a(37) = 4459248710164424946384890995893380022607743740081787109375
    a(41) = 3685099958690838758895720896109004106619977392256259918212890623
    a(43) = 1038001791494840815734697769103890995893380022607743740081787109375
    a(47) = 12229485870130123102579152313423230896109004106619977392256259918212890623

    Unsurprisingly, each solution so far has 2,3,5^(n-1) factors.

    For small values a brute-force search does it.
    a(18) = 46681074
    a(20) = 41069104

    And then
    a(21) <= 384153084109375 (sigh)

-- 
Don Reble  djr at nk.ca