Problems & Puzzles: Puzzles

Puzzle 878. Consecutive abundant integers

Emmanuel Vantieghem pointed out the following issue during the last week: Consecutive abundant integers.

An abundant integer N is such that sigma(N)>2N

This is -apparently- the status of this issue:

  • There are at least 10000 pairs of known consecutive abundant integers. See A096399 and this file by T. D. Noe.
  • The triple 171078830, 171078831, 171078832 was apparently found by Laurent Hodges and Michael Reid in 1995.

  • There are at least 1000 triples of consecutive abundante numbers. See A096536 and this file by Donovan Johnson.

  • The starting term of the smallest consecutive 4-tuple of abundant numbers is at most 141363708067871564084949719820472453374 (39 digits) - by Bruno Mishutka, Nov 01 2007. See A094628.

  • According to CYF36, Luke Pebody a little bit previously (21-08-2007) found a larger 4-tuple starting at 36689206836378866312597986802568592633785433489374 (50 digits).

  • This last reference could be interesting because Pebody describes the method he employed to get his result.

Q1. Can you produce a smaller 4-tuple of consecutive abundant integers, than the Bruno's one?
Q2. Can you produce a 5-tuple, as small as you may?
Q3. I noticed that none of the
 starting integer for the 1000 triplets reported by D.J. is odd, or is even and has an ending with the digits 2 or 6 and none Why is this?


Contribution came from Jens K. Andersen


Andersen wrote:

I worked on CYF36 in 2014 but never submitted my results.
The sum of reciprocals of the primes diverges.
It can be shown that this means the chinese remainder theorem can be used
to generate arbitrarily many consecutive abundant integers by placing
selected prime factors in them, but the size grows quickly.

Q1. The smallest I found is 4 36-digit integers starting at

Q2. The smallest I found is 5 171-digit integers starting at

I also found 6 3483-digit integers, and 7 69267-digit integers.

All results are from searches trying many applications of the chinese
remainder theorem to look for solutions a little smaller than a single
application could be expected to give.
I only know partial factorizations for 5 to 7 integers but enough to prove
the integers must be abundant.

Q3. It's a big advantage to have many small prime factors when integers
have to be abundant.
If a triplet starts with an odd integer then 2 only divides 1 of the 3 integers.
If a triplet starts with an even integer ending in 2 or 6 then 5 does not
divide any of the 3 integers.

Complete results here.


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