Problems & Puzzles:
Consecutive abundant integers
Vantieghem pointed out the following issue during the last week:
Consecutive abundant integers.
abundant integer N is such that
-apparently- the status of this issue:
There are at least 10000 pairs of
consecutive abundant integers. See
and this file
by T. D. Noe.
171078830, 171078831, 171078832 was apparently found by
Laurent Hodges and Michael Reid in 1995.
There are at least 1000 triples of consecutive abundante
A096536 and this
starting term of the smallest consecutive 4-tuple of
abundant numbers is at most
141363708067871564084949719820472453374 (39 digits) - by
Bruno Mishutka, Nov 01 2007. See
CYF36, Luke Pebody a little bit previously (21-08-2007)
found a larger 4-tuple starting at
This last reference could be interesting
because Pebody describes the method he employed to get his
Q1. Can you produce a
smaller 4-tuple of consecutive abundant integers, than the
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
ending with the digits 2 or 6 and none Why is this?
Contribution came from Jens K. Andersen
I worked on CYF36 in 2014 but never submitted my
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
Q1. The smallest I found is 4 36-digit integers
Q2. The smallest I found is 5 171-digit integers
I also found 6 3483-digit integers, and 7 69267-digit
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.