Paul,
>> But am I assuming anything about t that should be anathema to the
>> methodologist, if I just assume that the sum of all t(n) and of all the
>> binonmial coefficents for k=2, for n in [x, y], are equal to their
sums
>> in
>> [x', y']? The value of t(n) is still arbitrary.
>
> As near as I can tell, you want to proceed like this:
>
> Prop 1. Let t be a function from |N to |N with positive values. Let
> [x, y] and [x', y'] be integer intervals with y - x = y' - x'. Further
> assume sum(t(n), n in [x, y]) = sum(t(k), k in [x', y']). Then blah,
> blah, blah ... there is a bijection h ... blah, blah, blah.
>
> Prop 2. Let J be a set of primes. Let [x, y] and [x', y'] be integer
> intervals with y - x = y' - x'. Then, by Prop 1 with t = t_j, let h be
> a bijection blah, blah, blah.
>
> The careful reader will take about a nanosecond to say "Hey, I know a J
> and some intervals [x, y] and [x', y'] for which the sums are _not_
> equal - Prop 1 does not apply".
>
> You're sunk.
>
> All you can do is handle the triples J, [x, y], [x', y'] on a case by
> case basis showing in each case that the sums are equal so that you can
> apply Prop 1.
>
My Prop 2 has the relevant conditions built into it.
>> BTW I wonder whether you had any thought about this:
>>
>> > But if my function takes the form f : G -->G' (as it does), how can
one
>> > sneak into that expression the principle that for [x,y] and [x',y'],
>> > sum((g,n) in G) is equal to sum (h(g,n) in G'))?
>>
>> Perhaps I'm barking up the wrong tree?
>
> I would think so since you can't add the (g, n)'s and you don't know
> there _is_ an h.
I was saying on the Prime Number theorem thread that if for any p in J, (y
-
x + 1)/p = i + v, where i is a positive integer and 0 < v < 1, then if
there
are i+1 n in [x,y] for which p|n, then for the two adjacent intervals, of
length y-x+1, either side, there will be i integers for which p|n. So if
for
all p in J, [x,y] there exists some integer i and some value v for which
(y - x + 1)/p = i + v and there are i + 1 integers that divide pq in
[x,y],
then we have the two intervals either side of [x,y] for which the numbers
of
n for which p|n are equal. The same applies if p, above, is replaced with
pq.
All one needs to obtain the cluster is to have members of J that are
pro****tionally very close together.
Would my h : G --> G' now be OK?
With thanks.
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