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Topic: Early (1930's) work on arithmetic progressions
Replies: 10   Last Post: Oct 1, 2017 11:40 AM

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David Bernier

Posts: 3,892
Registered: 12/13/04
Re: Early (1930's) work on arithmetic progressions
Posted: Oct 1, 2017 11:40 AM
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On 09/28/2017 01:07 PM, Quadibloc wrote:
> On Thursday, September 28, 2017 at 11:02:16 AM UTC-6, Quadibloc wrote:
>> I'm confused myself.
>> A sequence of increasing integers containing "no k consecutive elements in an arithmetic progression" would, it seems to me, have maximum length 1 if k=2; and if k=3, it could have unbounded length: for example, the sequence
>> 1, 2, 4, 5, 7, 8, 10, 11, 13...
>> where a(n+1) = a(n) + 1 for n odd, and a(n+1) = a(n) + 2 for n even.
>> Thus, I must have misunderstood the kind of sequence he is looking for.

> ...ah, if what it doesn't contain is items that are consecutive *within the
> arithmetic progression*, as opposed to being consecutive *in the sequence*, then
> that is a different matter.
> But in _that_ case, I think of Liouville's number, and suggest that a series
> starting with 1 and with the distance between terms going as, say (n!)! + 1 or
> something similar would be unbounded in size and work for any k greater than or equal to 3.
> John Savard

Yes, that's certainly true.

Already, the case of no APs of length 3 terms is instructive.

If A is a set of positive integers, and all elements of A are between
1 and 'n', and the set A has no A.P.s of length 3,
how large can A be in cardinality?

If Max_3(n) is max |A|, A subset of ( [1, n] /\ Naturals) s.t. A has
zero A.P.s of length 3 or more, as I recall, it was eventually shown that
Max_3(n) is o(n) as n --> oo .

Then there are subsets of ( [1, n] /\ Naturals) that "only" avoid
APs of length 4, or 42, or 1 million. Those can be larger and larger,
going from 3, to 4, to 42, to 1 million.

David Bernier

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