Problems & Puzzles:
Consecutive odd abundant integers.
we propose to find the minimal pair of odd consecutive
do we know about these kind of pairs?
In our Puzzle 878, J. K.
Andersen reported the minimal known [May 2017]
quadruplet of consecutive abundant numbers,
so two of the members of these quadruplet are a pair of
consecutive odd abundant numbers. As a matter of fact,
the start of the quadruplet reported by JKA is
n+1 and n+3 are the smallest known (May 12017)
pair of odd
consecutive abundant integers. But JKA found this
quadruplet seeking precisely quadruplet of consecutive
integers (using the Chinese Remaninder Theorem). So,
it's reasonable to expect smaller pairs of odd
consecutive abundant integers if the search is focused
just in pairs and not in quadruplets.
2. How large is this JKA couple from the minimal
possible? According to
John L. Drost, Aug
13 2004, author of the sequence
related to triplets of consecutive abundant integers, n,
n+1 & n+2,
"the first odd n would have to have sigma(n*(n+2)) > 4n
so n > 10^19".
3. Thus, if the statement by Drost is correct, our target is to find a pair of consecutive odd
abundant integers in the range [10^19-10^35]
Q1. Please send your
minimal pair of consecutive odd abundant integers.
Q2. Can you
explain the Drost's statement?
Contributions came from J. K. Andersen and Emmanuel
Vantieghem and Antoine Verroken
Q1. The smallest I found is
(30 digits) is a near miss.
n+2 is abundant but sigma(n) is only around 1.99n.
Q2. The statement should have said sigma(n*(n+2)) >
n is odd so n and n+2 are coprime, and sigma(n*(n+2))
n and n+2 are abundant so we get sigma(n*(n+2)) > 2n
* 2(n+2) = 4*(n*(n+2)).
the least odd k such that sigma(k) >= 4k is
k = 1853070540093840001956842537745897243375.
This means n*(n+2) >= k, and n+2 > sqrt(k), which is
The smallest pair (n,n+2) of odd abundant number I could find
is the one with
n = 76728582876430878992529528245373
(32 digits) n+1 is deficient. I think there can be smaller
The part of a sentence
"sigma(n*(n+2)) > 4n" is an error (no
doubt) and should be replaced by :
"sigma(n*(n+2)) > 4*n(n+2)".
The statement that this would imply n > 10^19 can be
clarified as follows :
The smallest odd integer m such that sigma(m)
> 4*m is 1853070540093840001956842537745897243375.
(I found that number myself, but since it
is in the OEIS it relieves me from the duty to explain
how I got it).
So if, in addition, m must be of the
form n(n+2), this implies n > Sqrt(m), which is about
1. A multiple of an abundant integer is an abundant integer
Problem : A * a - B * b = 2 (1)
A,B abundant odd integers coprime
A*a , B*b consecutive odd abundant integers
A : 11 * 13 * ( p33)# / p(4)# A : not a multiple of 3,5,7
(px)# means product of the first ‘x’ primes
B : 945 = 3² * 5 * 7
2. Solving (1)
A * p – B * q = 1 by continied fractions :
p : 212
q : 11006307076227998864821427633189578324791153067438995
A*a : 2 * A * p + A * B B * b : 2 * B * q + A * B
A*a : 67164691019111053167046250170733601069216691882599835037
B*b : 67164691019111053167046250170733601069216691882599835035
sigma(A*a) / 2 / A*a = 1.002835119 sigma : sum of divisors
sigma(B*b) / 2 / B*b = 1.015923383