Re: sums of divisors
- From: Mark Griffith <contact@xxxxxxxxxxxxxxxxxx>
- Date: Mon, 15 Aug 2005 15:00:53 EDT
..
No odd squares are both 'neat' and 'slim'
1) Definition: Call a number n 'slim' if only one
prime divides into sigma(n) and not into n itself.
3.3.13.13.61.61 is an odd slim square, for example.
Sigma(n) is the sum of the divisors of n, including n
itself.
2) Definition: Call a number m 'neat' if 2.m -
sigma(m) = f, a factor of m. 3.3.7.7.11.11.13.13 is an
odd neat square, for example.
3) To find if any odd squares are both slim and neat,
start by considering numbers of the form of n_1, an
arbitrary odd, deficient square = a^2.b^2.c^2 where
1 + a + a^2 = b, 1 + b + b^2 = c, 1 + c + c^2 = p.a^2.b. All of a, b, c, p are assumed to be odd primes.
Definition: Note that for n_1, since c does not divide
into 1 + c + c^2, and assuming by neatness all primes
dividing n_1 divide sigma(n_1) or else almost-divide p
[say that c 'almost-divides' 2.c - 1], then 2.c - 1 is
the smallest value p can have, and (2.c - 1).a^2.b is
the smallest value p-times-cofactors can have.
4) Definitions: Call the primes that divide into any
but the last term of sigma(n1) the 'outside primes'.
Call primes that divide into the last term =
p-times-cofactors (or are almost-factors of p) the
'inside primes'. Some primes may occur in both lists.
5) With n_1, dividing both sides of
(2.c - 1).a^2.b = 1 + c + c^2
by the inside primes (a^2.b.c) gives the
result that (2.c - 1)/c > right-hand sum of fractions
if the right side is divided and then substituted by
each prime in turn, so dividing by c, then replacing c
with 1 + b + b^2, etc.
This is because values (3, 5, 7) smaller than the
smallest possible values for a, b, c, give
1 + 7/8 > 1 + 1/3 + 2/9 + 2/45 + 2/315 + ... where
each right-hand fraction after 1 is less than the
corresponding fraction in 1 + 1/2 + 1/4 + 1/8 + ...
Larger values for a, b, c produce smaller sums on
the right-hand side. Larger sets of primes, all raised
to the power of 2, or to powers higher than 2, also
preserve the inequality, if any new odd square n_2
keeps to the form of
1 + a + a^2 + ... = b, 1 + b + b^2 + ... = c...
6) Definition: Call this last form in (5) the 'steepest' form of n_1, and abandon this condition on n.
Suppose n_2 is as follows:
1 + a + a^2 = b, 1 + b + b^2 = c.a,
giving that (2.c - 1).a.b = 1 + c + c^2 if n2 is to be
both neat and slim.
Note that c = (1 + b + b^2)/a, and so the largest,
leftmost, terms of the right-hand side (those to the
left of and including the /a term) will still have
a^2.b.c as a divisor after both sides are divided by
a.b.c.
At the first step (1 + c + c^2 + ...) every member of
the right-hand sequence is different. Substituting in
each smaller prime and then dividing in turn (dividing
by c, then replacing c with (1 + b + b^2)/a, etc)
introduces at most one new 1 at any stage, so there
are never more than two instances of 1 at any stage.
Therefore no fraction has a numerator larger than 2.
Therefore the sequence 1 + 1/3 + 2/9 + 2/45 + ... is
an upper bound for any right-hand sequence after all
inside primes are divided into both sides in turn and
the right-side outside prime divisors are revealed
during substitution by progressively smaller primes.
After 1 = 1, there is 1/3 < 1/2, 2/9 < 1/4, 2/45 <
1/8... showing that, with denominators multiplied by
progressively larger odd primes, and numerators never
larger than 2, all right-hand sums are < 1 + 1/2 + 1/4
+ 1/8 + ..., which = (2.c - 1)/c for c = largest power
of 2 used but not for c = an odd prime > 8.
This gives that the left-hand side of
p-times-cofactors=expansion-of-largest-prime is always larger than the right-hand side (1 + c + c^2 + ...). So no such squares exist.
7) Therefore no odd squares are both neat and
slim. Hence there are no odd perfects, since an odd
perfect number n must divide by an odd, neat, slim
square.
(odd perfect n = s.p -> s is deficient, s =
factors-of-s + f, p.factors-of-s = (p-2).s + f, p.f =
s + (s-f) = sigma(s) -> p divides into sigma(s), f not
= 1 (since more than one prime divides s so sigma(s)
is many-term), f divides into s so s neat, p is prime
so s is slim.)
Mark Griffith
2005 July
.
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