|
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 the Chinese Remainder Theorem, and search through very sparse arithmetic sequences. I confirm that (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(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(21) <= 384153084109375.
(End)
a(26) = 13343831081787109374
a(34) = 6445231882519836425781248
a(38) = 4985683002487480163574218750
a(46) = 14840091517264784512519836425781248
a(58) = 43726550089078883239954784512519836425781248
a(62) = 37552673229782602893380022607743740081787109375
(End)
| 