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%I #42 Dec 24 2021 21:55:48
%S 0,33,242,230,7939375,1274,76571890623,8294,959075,248750,
%T 104228508212890623,72224,1489106237081787109375,23513890624,
%U 145705879375,318680,273062471666259918212890623,46681074,804505911103256259918212890623,41069104,384153084109375
%N a(1) = 0; a(n) = smallest of three consecutive numbers all with 2n divisors.
%C From _Don Reble_, Jan 22 2015: (Start)
%C 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.
%C 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.
%C a(23) = 490685203356467392256259918212890623
%C a(29) = 6794675247932944436619977392256259918212890623
%C a(31) = 329757106427071213106619977392256259918212890623
%C a(37) = 4459248710164424946384890995893380022607743740081787109375
%C a(41) = 3685099958690838758895720896109004106619977392256259918212890623
%C a(43) = 1038001791494840815734697769103890995893380022607743740081787109375
%C a(47) = 12229485870130123102579152313423230896109004106619977392256259918212890623
%C Unsurprisingly, each solution so far has 2,3,5^(n-1) factors. For small values a brute-force search does it.
%C a(21) <= 384153084109375.
%C (End)
%C From _Chai Wah Wu_, Mar 14 2019: (Start)
%C a(26) = 13343831081787109374
%C a(34) = 6445231882519836425781248
%C a(38) = 4985683002487480163574218750
%C a(46) = 14840091517264784512519836425781248
%C a(58) = 43726550089078883239954784512519836425781248
%C a(62) = 37552673229782602893380022607743740081787109375
%C (End)
%D _Don Reble_, Posting to Sequence Fans Mailing List, Jan 22, 2015
%H Chai Wah Wu, <a href="/A075040/b075040.txt">Table of n, a(n) for n = 1..24</a>
%H Vladimir A. Letsko, <a href="http://arxiv.org/abs/1510.07081">Some new results on consecutive equidivisible integers</a>, arXiv:1510.07081 [math.NT], 2015.
%e a(3) = 242 as tau(242) = tau(243) = tau(244) = 6.
%Y A306879 is a subsequence.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Sep 03 2002
%E More terms from _Jason Earls_, Sep 05 2002
%E a(7), a(9), a(12), a(14)-a(16) from _Donovan Johnson_, Oct 13 2009
%E a(11), a(13) conjectured by _Giovanni Resta_, Aug 14 2013, confirmed by _Don Reble_, Jan 22 2015
%E a(17)-a(20) from _Don Reble_, Jan 22 2015
%E Edited by _Max Alekseyev_, Jan 23 2015
%E a(21) confirmed and a(22), a(24) added by _Chai Wah Wu_, Mar 14 2019
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