A075040 a(1) = 0; a(n) = smallest of three consecutive numbers all with 2n divisors.

0, 33, 242, 230, 7939375, 1274, 76571890623, 8294, 959075, 248750, 104228508212890623, 72224, 1489106237081787109375, 23513890624, 145705879375, 318680, 273062471666259918212890623, 46681074, 804505911103256259918212890623, 41069104, 384153084109375

Offset: 1

Author

Amarnath Murthy, Sep 03 2002

Keywords

• nonn

Comments

From Don Reble, Jan 22 2015: (Start)

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)

From Chai Wah Wu, Mar 14 2019: (Start)

a(26) = 13343831081787109374

a(34) = 6445231882519836425781248

a(38) = 4985683002487480163574218750

a(46) = 14840091517264784512519836425781248

a(58) = 43726550089078883239954784512519836425781248

a(62) = 37552673229782602893380022607743740081787109375

(End)

Examples

			a(3) = 242 as tau(242) = tau(243) = tau(244) = 6.
		

References

• Don Reble, Posting to Sequence Fans Mailing List, Jan 22, 2015

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..24
• Vladimir A. Letsko, Some new results on consecutive equidivisible integers, arXiv:1510.07081 [math.NT], 2015.

Crossrefs

A306879 is a subsequence.

Extensions

More terms from Jason Earls, Sep 05 2002

a(7), a(9), a(12), a(14)-a(16) from Donovan Johnson, Oct 13 2009

a(11), a(13) conjectured by Giovanni Resta, Aug 14 2013, confirmed by Don Reble, Jan 22 2015

a(17)-a(20) from Don Reble, Jan 22 2015

Edited by Max Alekseyev, Jan 23 2015

a(21) confirmed and a(22), a(24) added by Chai Wah Wu, Mar 14 2019