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.