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A075040 a(1) = 0; a(n) = smallest of three consecutive numbers all with 2n divisors. 4
0, 33, 242, 230, 7939375, 1274, 76571890623, 8294, 959075, 248750, 104228508212890623, 72224, 1489106237081787109375, 23513890624, 145705879375, 318680, 273062471666259918212890623, 46681074, 804505911103256259918212890623, 41069104, 384153084109375 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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 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(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)

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.

EXAMPLE

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

CROSSREFS

A306879 is a subsequence.

Sequence in context: A127870 A142993 A230186 * A274639 A306879 A178448

Adjacent sequences:  A075037 A075038 A075039 * A075041 A075042 A075043

KEYWORD

nonn

AUTHOR

Amarnath Murthy, Sep 03 2002

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

STATUS

approved

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Last modified April 15 03:00 EDT 2021. Contains 342974 sequences. (Running on oeis4.)