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The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions. Bloom and Sisa

by y7 6y ago

The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions. Bloom and Sisask now proved that A contains infinitely many length-3 arithmetic progressions, following from their main result which is an improved upper bound for Roth's theorem. https://arxiv.org/abs/2007.03528 https://arxiv.org/abs/2007.03528

ColinWright 6y ago
And just to expand a smidgen on that, the maths expression: A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges is one way of saying that the set A of integers is not too sparse. The set of powers of 2 does not satisfy this condition ... it's too sparse. The set of primes does satisfy this condition ... it's not too sparse, primes turn up "reasonably often".
- Ar-Curunir 6y ago
  Hmm is there an asymptotic statement underlying this? The powers of 2 are exponentially sparse (N integers contain at most 1/logN powers of 2), whereas primes are polynomially sparse (N integers contains N/LogN primes)
  - ColinWright 6y ago
    I was trying to give people a sense of the statement, giving two examples of "not very dense" and "dense enough", but I don't understand what you're trying to say here. I can't work out whether you are asking a question, or making a conjecture, or ... what. Would you care to clarify? If it's a question then I'll try to answer, but if it's a conjecture, can you make it more precise?
    - notafraudster 6y ago
      He is asking "what is the line between too sparse and not too sparse" and attempting to characterize the example you gave of each in terms of approximate relative sparsity.
      - ColinWright 6y ago
        Right ... understood. I don't have an answer, it's not my area, but it does feel a lot like the condition "sum(1/n) diverges" is likely to be close, but that's just a gut feeling.
      - RaptorJ 6y ago
        Ah well there isn't one, from any divergent series you can construct a more slowly growing (but still divergent) series.
  - Certhas 6y ago
    So let's look at n^m. The condition is sum_n n^-m ; and that diverges for m smaller or equal 1. If you have a sequence that grows asymptotically like n^m it means that f(n)/n^m goes to a constant asymptotically. So the question is if this implies that sum 1/f(n) converges. I feel intuitively that it should be possible to prove that. A sketch: For every epsilon there is an N such that |1/f(n) - 1/C n^-m| < epsilon n^-m for n > N. So if we take the difference between the reciprocal partial sums, then that is bounded by eps * the partial sum. That is finite exactly if the reciprocal sums converge. Thus the convergence behaviour is the same for this case...
  - hansvm 6y ago
    There can be. Let f(n) be the count of members of the set in question which are at most n. If ε>0 and f(n)=O(n^(1-ε)) then the sum of reciprocals converges. For denser sets I'd have to think a bit longer about it. It's probably also worth noting that weird density distributions are possible. E.g. one could imagine a kind of oscillation where you include enough values to get the total density up to that point up to some decreasing threshold (e.g. 1-2^-k if we've flip-flopped k times) then omit values to get below a different threshold (e.g. 2^-k if we've flip-flopped k times), and repeat the process indefinitely. The count up to a point n for this construction has the interesting property that it isn't greater in a big-O sense than the count for exponentially sparse sets while also not being less than the count for any sublinear density function in a big-omega sense (using the Knuth interpretation).
    - hansvm 6y ago
      Edit: not greater than exponentially sparse sets, something a bit denser -- exponential sparsity is manageable as well, but not with the 2^-k construction I gave.
- gigatexal 6y ago
  Can you dumb it down a bit more using more English than symbols? My math skills are not the greatest but I am keen to hear more.
  - ColinWright 6y ago
    OK, let's have a go. We are looking for set of three numbers that are "equally spaced". So {4, 7, 10} are equally spaced, differing by 3 each time. Another set might be {20, 30, 40}, this time differing by 10. We'll call such a set "Equally Spaced Triples", or "EST" for short. If you have the positive even integers - 2, 4, 6, 8, ... - then clearly you can find infinitely many ESTs. You have {2,4,6}, {6,10,14}, and so on. However, we can show that if you take the powers of 2 - 1, 2, 4, 8, 16, 32, 64, ... - then we cannot find an EST. So, when can we do this? When can we be guaranteed always to find infinitely many ESTs? Suppose you have a set of numbers - n0, n1, n2, n3, n4, ... - is there a test to see if we are guaranteed to have infinitely many ESTs? The answer is yes, and that's the result that has been proved. The result says this: Take any set of positive integers, take their inverses, and add them all together. If the result has an upper bound, then the set might not have infintely many ESTs. However, if the sum grows without bound, then you are guaranteed to have infinitely many ESTs. Unpacking that with our examples, taking the inverses of the powers of two and adding them up we get 1/1 + 1/2 + 1/4 + 1/8 + 1/16 + ... and we can show that the total never exceeds 2. In fact, the total never reaches 2. So since the total is bounded, we do not have infinitely many ESTs. Now look at the primes. There is a standard result that says that the sum 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + ... is unbounded above. You give me a desired total, and I can tell you how many terms you need to take to exceed that number. So the sum of the inverses is unbounded, and hence the primes will have infinitely many ESTs. So, in summary, if a set of positive integers is dense enough - if there are enough of them in some technical sense - then there are infinitely many ESTs. The test for density is to ask that the sum of the reciprocals (inverses) is unbounded. Does that help? Edit: I wanted to contact you out-of-band, but you only have a LinkedIn link in your profile, and I don't use LinkedIn. If you're interested in discussing this further then I'd be happy to help, but better by email. My contact details are in my profile. Edit 2: Thank you everyone for your kind comments. You've made me think about my write-ups. I already do a lot of writing ... I might re-visit what and how. I appreciate the kind words.
jstanley 6y ago
> The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions This is quite hard to understand for people who don't already know what it means (including me). I started trying to translate it but there were a few parts I didn't understand, starting with: 1.) Is ℕ integers >0 or >=0? Wikipedia says it can be either. Maybe it doesn't matter? Maybe it must be >0 otherwise 1/n makes no sense? 2.) When you ask whether the sum of 1/n for all n in A diverges, how do you know what order to sum them in? Does it diverge regardless of the order? Since A only contains positive integers doesn't sum of 1/n for all n in A always tend towards infinity? EDIT: I see now that sum of positive integers doesn't always tend towards infinity, thanks to ColinWright's comment about powers of 2 (for which the sum of 1/n tends towards 1). Also, since the numbers are all positive, it doesn't matter what order you sum them in, you're basically just asking whether the sum of 1/n for all the numbers in A is finite or not.
- maxov 6y ago
  Dense math can often be complex to decipher. Sometimes it feels like reading an esoteric codebase to me! 1) Yes in this case 0 is not included in N. I’ve seen N defined both ways depending on the context, so it can be confusing when it’s not given explicitly. 2) By definition, a series sum is based on the limit of partial prefix sums. E.g 1/a_1, 1/a_1+1/a_2, ... It is an interesting question mathematically, if the sum stays the same when you arbitrarily rearrange the terms of the series. In general the answer is no (see Riemann rearrangement theorem), but as this series is only made up of positive reals, it can be rearranged arbitrarily without change in how it converges (or doesn’t). To the second part of your question, take any geometric series, i.e. A = {1, r, r^2, ...}, then it will converge. There are other classes of series that will converge in this case, and the conjecture is basically asking to characterize sets with diverging series as needing to be “large and dense” in a certain sense.
  - y7 6y ago
    Note that a geometric series can also be a multiple, e.g. (a, a r, a r^2, ...). It converges if and only if |r| < 1.

by y7 6y ago

ColinWright 6y ago

And just to expand a smidgen on that, the maths expression: A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges is one way of saying that the set A of integers is not too sparse. The set of powers of 2 does not satisfy this condition ... it's too sparse. The set of primes does satisfy this condition ... it's not too sparse, primes turn up "reasonably often".

Ar-Curunir 6y ago
Hmm is there an asymptotic statement underlying this? The powers of 2 are exponentially sparse (N integers contain at most 1/logN powers of 2), whereas primes are polynomially sparse (N integers contains N/LogN primes)
- ColinWright 6y ago
  I was trying to give people a sense of the statement, giving two examples of "not very dense" and "dense enough", but I don't understand what you're trying to say here. I can't work out whether you are asking a question, or making a conjecture, or ... what. Would you care to clarify? If it's a question then I'll try to answer, but if it's a conjecture, can you make it more precise?
  - notafraudster 6y ago
    He is asking "what is the line between too sparse and not too sparse" and attempting to characterize the example you gave of each in terms of approximate relative sparsity.
    - ColinWright 6y ago
      Right ... understood. I don't have an answer, it's not my area, but it does feel a lot like the condition "sum(1/n) diverges" is likely to be close, but that's just a gut feeling.
    - RaptorJ 6y ago
      Ah well there isn't one, from any divergent series you can construct a more slowly growing (but still divergent) series.
- Certhas 6y ago
  So let's look at n^m. The condition is sum_n n^-m ; and that diverges for m smaller or equal 1. If you have a sequence that grows asymptotically like n^m it means that f(n)/n^m goes to a constant asymptotically. So the question is if this implies that sum 1/f(n) converges. I feel intuitively that it should be possible to prove that. A sketch: For every epsilon there is an N such that |1/f(n) - 1/C n^-m| < epsilon n^-m for n > N. So if we take the difference between the reciprocal partial sums, then that is bounded by eps * the partial sum. That is finite exactly if the reciprocal sums converge. Thus the convergence behaviour is the same for this case...
- hansvm 6y ago
  There can be. Let f(n) be the count of members of the set in question which are at most n. If ε>0 and f(n)=O(n^(1-ε)) then the sum of reciprocals converges. For denser sets I'd have to think a bit longer about it. It's probably also worth noting that weird density distributions are possible. E.g. one could imagine a kind of oscillation where you include enough values to get the total density up to that point up to some decreasing threshold (e.g. 1-2^-k if we've flip-flopped k times) then omit values to get below a different threshold (e.g. 2^-k if we've flip-flopped k times), and repeat the process indefinitely. The count up to a point n for this construction has the interesting property that it isn't greater in a big-O sense than the count for exponentially sparse sets while also not being less than the count for any sublinear density function in a big-omega sense (using the Knuth interpretation).
  - hansvm 6y ago
    Edit: not greater than exponentially sparse sets, something a bit denser -- exponential sparsity is manageable as well, but not with the 2^-k construction I gave.
gigatexal 6y ago
Can you dumb it down a bit more using more English than symbols? My math skills are not the greatest but I am keen to hear more.
- ColinWright 6y ago
  OK, let's have a go. We are looking for set of three numbers that are "equally spaced". So {4, 7, 10} are equally spaced, differing by 3 each time. Another set might be {20, 30, 40}, this time differing by 10. We'll call such a set "Equally Spaced Triples", or "EST" for short. If you have the positive even integers - 2, 4, 6, 8, ... - then clearly you can find infinitely many ESTs. You have {2,4,6}, {6,10,14}, and so on. However, we can show that if you take the powers of 2 - 1, 2, 4, 8, 16, 32, 64, ... - then we cannot find an EST. So, when can we do this? When can we be guaranteed always to find infinitely many ESTs? Suppose you have a set of numbers - n0, n1, n2, n3, n4, ... - is there a test to see if we are guaranteed to have infinitely many ESTs? The answer is yes, and that's the result that has been proved. The result says this: Take any set of positive integers, take their inverses, and add them all together. If the result has an upper bound, then the set might not have infintely many ESTs. However, if the sum grows without bound, then you are guaranteed to have infinitely many ESTs. Unpacking that with our examples, taking the inverses of the powers of two and adding them up we get 1/1 + 1/2 + 1/4 + 1/8 + 1/16 + ... and we can show that the total never exceeds 2. In fact, the total never reaches 2. So since the total is bounded, we do not have infinitely many ESTs. Now look at the primes. There is a standard result that says that the sum 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + ... is unbounded above. You give me a desired total, and I can tell you how many terms you need to take to exceed that number. So the sum of the inverses is unbounded, and hence the primes will have infinitely many ESTs. So, in summary, if a set of positive integers is dense enough - if there are enough of them in some technical sense - then there are infinitely many ESTs. The test for density is to ask that the sum of the reciprocals (inverses) is unbounded. Does that help? Edit: I wanted to contact you out-of-band, but you only have a LinkedIn link in your profile, and I don't use LinkedIn. If you're interested in discussing this further then I'd be happy to help, but better by email. My contact details are in my profile. Edit 2: Thank you everyone for your kind comments. You've made me think about my write-ups. I already do a lot of writing ... I might re-visit what and how. I appreciate the kind words.

jstanley 6y ago

> The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions This is quite hard to understand for people who don't already know what it means (including me). I started trying to translate it but there were a few parts I didn't understand, starting with: 1.) Is ℕ integers >0 or >=0? Wikipedia says it can be either. Maybe it doesn't matter? Maybe it must be >0 otherwise 1/n makes no sense? 2.) When you ask whether the sum of 1/n for all n in A diverges, how do you know what order to sum them in? Does it diverge regardless of the order? Since A only contains positive integers doesn't sum of 1/n for all n in A always tend towards infinity? EDIT: I see now that sum of positive integers doesn't always tend towards infinity, thanks to ColinWright's comment about powers of 2 (for which the sum of 1/n tends towards 1). Also, since the numbers are all positive, it doesn't matter what order you sum them in, you're basically just asking whether the sum of 1/n for all the numbers in A is finite or not.

maxov 6y ago
Dense math can often be complex to decipher. Sometimes it feels like reading an esoteric codebase to me! 1) Yes in this case 0 is not included in N. I’ve seen N defined both ways depending on the context, so it can be confusing when it’s not given explicitly. 2) By definition, a series sum is based on the limit of partial prefix sums. E.g 1/a_1, 1/a_1+1/a_2, ... It is an interesting question mathematically, if the sum stays the same when you arbitrarily rearrange the terms of the series. In general the answer is no (see Riemann rearrangement theorem), but as this series is only made up of positive reals, it can be rearranged arbitrarily without change in how it converges (or doesn’t). To the second part of your question, take any geometric series, i.e. A = {1, r, r^2, ...}, then it will converge. There are other classes of series that will converge in this case, and the conjecture is basically asking to characterize sets with diverging series as needing to be “large and dense” in a certain sense.
- y7 6y ago
  Note that a geometric series can also be a multiple, e.g. (a, a r, a r^2, ...). It converges if and only if |r| < 1.