Converging sum of reciprocals
JoeSh...@aol.com
integers that are the sum of two earlier ones in the set:
1,2,4,5,10,11,13,14,28,29,31,32,37,38,40,41,82,... X_3 contains no
arithmetic progressions of length 3.
The sum of the reciprocals of the first 2^16 elements of X_3 is
3.004209955... The Erdos-Turan conjecture states that any set of
positive integers whose sum of reciprocals converges contains
arbitrarily long arithmetic progressions, and I know this is open even
for 3-term arithmetic progressions, so it can't have been proven that
the sum of the reciprocals of X_3 diverges.
My question is, assuming the Erdos-Turan conjecture is true, does X_3
have the maximal reciprocal-sum among all sets of positive integers
containing no 3-term arithmetic progression? More generally, does the
set constructed "greedily" to be the lexicographically first one with
no k-term arithmetic progressions have the maximal reciprocal-sum
among such progressions?
The first test case is the sequence beginning 1,3,4,6,10,12... which
skips 2 but then proceeds "greedily". Can one prove that the partial
sums of the reciprocals of this sequence are always less than the
partial sums of the reciprocals of X_3?
-- Joe Shipman
Gerry Myerson
"joesh...@aol.com" <JoeSh...@aol.com> wrote:
> Let X_3 be the set of positive integers constructed by sieving out all
> integers that are the sum of two earlier ones in the set:
> 1,2,4,5,10,11,13,14,28,29,31,32,37,38,40,41,82,...
Then 5 shouldn't be in there, should it?
> X_3 contains no arithmetic progressions of length 3.
Is there some connection between no number a sum of two others
and no 3-term arithmetic progressions?
> The sum of the reciprocals of the first 2^16 elements of X_3 is
> 3.004209955... The Erdos-Turan conjecture states that any set of
> positive integers whose sum of reciprocals converges contains
> arbitrarily long arithmetic progressions,
I think you mean "diverges."
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Christopher Henrich
> Let X_3 be the set of positive integers constructed by sieving out all
> integers that are the sum of two earlier ones in the set:
> 1,2,4,5,10,11,13,14,28,29,31,32,37,38,40,41,82,... X_3 contains no
> arithmetic progressions of length 3.
The part of X_3 that you quote does not seem to me to agree with your
definition. For instance, 5 = 1 + 4.
Here, in detail (which may reveal how I am not understanding you) is my
construction of the first few members of the set.
1 is a member.
2 is a member.
xxxxxxx 3 is not a member (1 + 2).
4 is a member.
xxxxxxx 5 is not a member (1 + 4).
xxxxxxx 6 is not a member (2 + 4).
7 is a member.
xxxxxxx 8 is not a member (1 + 7).
xxxxxxx 9 is not a member (2 + 7).
10 is a member.
But now I note that (4, 7, 10) is an arithmetic progression of length 3.
What's going on here?
>
> The sum of the reciprocals of the first 2^16 elements of X_3 is
> 3.004209955... The Erdos-Turan conjecture states that any set of
> positive integers whose sum of reciprocals converges contains
> arbitrarily long arithmetic progressions, and I know this is open even
> for 3-term arithmetic progressions, so it can't have been proven that
> the sum of the reciprocals of X_3 diverges.
>
> My question is, assuming the Erdos-Turan conjecture is true, does X_3
> have the maximal reciprocal-sum among all sets of positive integers
> containing no 3-term arithmetic progression? More generally, does the
> set constructed "greedily" to be the lexicographically first one with
> no k-term arithmetic progressions have the maximal reciprocal-sum
> among such progressions?
>
> The first test case is the sequence beginning 1,3,4,6,10,12... which
> skips 2 but then proceeds "greedily". Can one prove that the partial
> sums of the reciprocals of this sequence are always less than the
> partial sums of the reciprocals of X_3?
>
> -- Joe Shipman
--
Christopher J. Henrich
chen...@monmouth.com
htp://www.mathinteract.com
Robert Israel
> In article <fvaq6r$89i$1...@news.ks.uiuc.edu>,
> "joesh...@aol.com" <JoeSh...@aol.com> wrote:
>
> > Let X_3 be the set of positive integers constructed by sieving out all
> > integers that are the sum of two earlier ones in the set:
> > 1,2,4,5,10,11,13,14,28,29,31,32,37,38,40,41,82,...
>
> Then 5 shouldn't be in there, should it?
>
> > X_3 contains no arithmetic progressions of length 3.
>
> Is there some connection between no number a sum of two others
> and no 3-term arithmetic progressions?
I thing Joe is referring to A003278 in The On-Line Encyclopedia of Integer
Sequences
<http://www.research.att.com/~njas/sequences/A003278>
> > The sum of the reciprocals of the first 2^16 elements of X_3 is
> > 3.004209955... The Erdos-Turan conjecture states that any set of
> > positive integers whose sum of reciprocals converges contains
> > arbitrarily long arithmetic progressions,
>
> I think you mean "diverges."
The other description of this sequence is
a(n)-1 in ternary = n-1 in binary
Thus the members x_n of this sequence all have the property that x_n-1
contains no 2's in base 3. There are 2^m of them between 3^m + 2 and
3^(m+1) + 1. From that it is easy to show that the sum of reciprocals converges.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
JoeSh...@aol.com
> In article <fvaq6r$89...@news.ks.uiuc.edu>,
> htp://www.mathinteract.com- Hide quoted text -
>
> - Show quoted text -
I apologize for my imprecision. I meant to only consider sets of
positive integers with no subsets that form 3-term arithmetic
progressions -- in other words, there are no triples a, b, and c in
the set such that a+c=b+b (but a+b=c is OK!).
And, of course, I should have stated the Erdos-Turan conjecture as
"Any set of inegers whose reciprocal-sum DIverges contains arbitrarily
long arithmetic progressions".
My main question remains the same. Is it known whether the Erdos-Turan
conjecture implies that the sum of the reciprocals of X_3 [which
consists of the numbers 1, 2, 4, 5, 10, 11, 13, 14, 28, ... in other
words all integers whose predecessors have no digit "2" when expressed
in base 3] is maximal among all reciprocal-sums of sets of positive
integers with no 3-term arithmetic progression?
JoeSh...@aol.com
<isr...@math.MyUniversitysInitials.ca> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
I expected that this would converge, but didn't realize the proof was
so easy. Is it really true that no one can prove that every set with
no 3-term arithmetic progressions has a convergent reciprocal-sum? It
seems that something stronger is true, namely that the set of such
reciprocal-sums not only contains only finite real numbers, but that
it is bounded from above. A still stronger conjecture is that the
reciprocal sequence 1/1, 1/2, 1/4, 1/5, 1/10, ... not only has maximal
sum among all such sequences, but that it dominates every other such
sequence for every partial sum. Strongest of all is the conjecture
that it dominates term by term (this is the only one of these
conjectures I'm not sure is true!).
If this (Erdos-Turan for 3-term progressions) is really a very hard
problem, one feels that it is almost useless to try to prove ANY
theorem! My final conjecture is that 3-term Erdos-Turan is not only
provable, but will eventually be found to have a very short proof.
-- Joe Shipman
David C. Ullrich
>>[...]
>
>I apologize for my imprecision. I meant to only consider sets of
>positive integers with no subsets that form 3-term arithmetic
>progressions -- in other words, there are no triples a, b, and c in
>the set such that a+c=b+b (but a+b=c is OK!).
Not to beat a dead horse, but doesn't 10 + 4 equal 7 + 7 ?
>And, of course, I should have stated the Erdos-Turan conjecture as
>"Any set of inegers whose reciprocal-sum DIverges contains arbitrarily
>long arithmetic progressions".
>
>My main question remains the same. Is it known whether the Erdos-Turan
>conjecture implies that the sum of the reciprocals of X_3 [which
>consists of the numbers 1, 2, 4, 5, 10, 11, 13, 14, 28, ... in other
>words all integers whose predecessors have no digit "2" when expressed
>in base 3] is maximal among all reciprocal-sums of sets of positive
>integers with no 3-term arithmetic progression?
David C. Ullrich
tc...@lsa.umich.edu
joesh...@aol.com <JoeSh...@aol.com> wrote:
>Is it really true that no one can prove that every set with
>no 3-term arithmetic progressions has a convergent reciprocal-sum?
It seems so. In particular, your original question was raised by MacKinnon
and Eastmond, "An attack on the Erdos conjecture," Mathematical Gazette
71 (1987), 14-19.
>If this (Erdos-Turan for 3-term progressions) is really a very hard
>problem, one feels that it is almost useless to try to prove ANY
>theorem! My final conjecture is that 3-term Erdos-Turan is not only
>provable, but will eventually be found to have a very short proof.
My favorite example of a conjecture that inspires the same reaction
(if we can prove anything, surely we can prove *this*!) is Frankl's
conjecture: Given any nonempty family F of subsets of a finite set S,
there exists an element of S that belongs to at least half the members
of F.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
tc...@lsa.umich.edu
>My favorite example of a conjecture that inspires the same reaction
>(if we can prove anything, surely we can prove *this*!) is Frankl's
>conjecture: Given any nonempty family F of subsets of a finite set S,
>there exists an element of S that belongs to at least half the members
>of F.
It has been pointed out to me that I garbled the statement of this
conjecture. F should be closed under taking unions. That is, if
X and Y are in F then X union Y must be in F. Also, of course, S
must be nonempty.
isr...@math.ubc.ca
> My favorite example of a conjecture that inspires the same reaction
> (if we can prove anything, surely we can prove *this*!) is Frankl's
> conjecture: Given any nonempty family F of subsets of a finite set S,
> there exists an element of S that belongs to at least half the members
> of F.
That's not Frankl's conjecture (for one thing, it's trivially false,
take F = {{1},{2},{3}}). The actual
conjecture requires F to be closed under unions. See <http://
en.wikipedia.org/wiki/Frankl%27s_conjecture>.