Numbers of a different order? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:27:53Z http://mathoverflow.net/feeds/question/87217 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87217/numbers-of-a-different-order Numbers of a different order? Xaus 2012-02-01T10:58:03Z 2013-05-18T08:53:02Z <p>Let $d_r$ be a divergent series of positive terms and let $s_r = \sum_{i=1}^{r}d_r$. We are interested in the sequence of numbers $S_{d_r} = s_1, s_2, \ldots$. For example if $d_r = 1/r$ the $s_r = H_r$ and $S_{1/r}$ is the sequence of the harmonic numbers. Similarly when when $d_r=1$ for all $r$, we have $s_r = r$ and $S_1$ is the sequence of natural numbers. The sequence of natural numbers contains primes and in terms of our notation, we can state this fact as when $d_r=1$ for all $r$ the sequence $S_1$ contains primes. With this background, we can us consider primes in the sequence of natural numbers as primes of the simplest order $d_r = 1$. In general we define primes of the order $d_r$ as follows. </p> <p><strong>Definition</strong>: If there exists a $d_r$ such that the following conditions are satisfied:</p> <p>i) there is at least one $i$ such that $s_i$ is not divisible by any of the numbers $s_1, s_2, \ldots s_{i-1}$.</p> <p>ii) there is at least one triplet $(j,k,l)$ such that $s_j = s_ks_l$</p> <p>then $s_i$ is a prime of the order $d_r$ and it is present in the sequence $S_{d_r}$.</p> http://mathoverflow.net/questions/87217/numbers-of-a-different-order/88083#88083 Answer by Greg Martin for Numbers of a different order? Greg Martin 2012-02-10T09:08:21Z 2012-02-10T09:08:21Z <p>Maybe I'm not understanding the question correctly, but it seems trivial to construct as many of these as you want. Let $d_1=s_1$ be any real number greater than 1; let $d_2$ be any real number greater than 1 except for $d_1^2-d_1$, so that $s_2\ne s_1^2$ and so condition (i) is satisfied. Now let $d_3 = (d_1-1)(d_1+d_2)$, so that $s_3 = s_1s_2$ and so condition (ii) is satisfied. You can now define $d_4,d_5,\dots$ at will.</p> <p>If you don't want $s_2$ to be the "prime", then you can easily make $s_3$ "prime" by a similar construction.</p> http://mathoverflow.net/questions/87217/numbers-of-a-different-order/88094#88094 Answer by quid for Numbers of a different order? quid 2012-02-10T12:11:31Z 2012-02-10T12:11:31Z <p>In view of the OPs comment I will focus on the case that <code>$s_1&gt;1$</code>; but at the end make some comments on the other case.</p> <p>The question is to find all <code>$d_r$</code> such that there is at least one 'prime' of that 'order'.</p> <p>This problem can be linked (is equivalent to) certain aspects of the classical problem of product/sum free subsets. </p> <p>First, let us recast the problem. Of course, a sequence <code>$d_r$</code> uniquely defines an <code>$S_{d_r}$</code> but also the converse is true that is from the set <code>$S_{d_r} = \lbrace s_i, i=1,2,\dots \rbrace$</code> we can recover <code>$d_r$</code> by considering difference between successive elements, that is <code>$d_i= s_{i}-s_{i-1}$</code>, where we set <code>$s_0=0$</code>. A question that arises is which sets $S$ we can obtain in this form. These are precisely the infinite discrete (no accumulation point) subsets of the reals of size at least $1$ </p> <p>Let such an $S$ be written in the form <code>$S = \lbrace s_1 &lt; s_2 &lt; \dots \rbrace$</code> Now according to the definition <code>$s_1$</code> always fulfills the first condition, and thus is 'prime' <em>if</em> the second condition is met.<br> The question thus arises what about the second condition. </p> <p>Let us recall that a subset $A$ of an abelian (semi)group $(G,+)$ is called <a href="http://en.wikipedia.org/wiki/Sum-free_set" rel="nofollow">sum-free</a> if the equation $x+y=z$ has no solution in $A$. In the problem under consideration the group would be the positive reals with multiplication, or one might wish to take logarithms to tansfer to a 'truly' additive situation.</p> <p>Now to classifiy all discrete sets that are not sum-free seems a bit hopeless. In particular, the reals seem 'too large' to make the problem of sumfree sets interesting. Yet on can of course construct various things. For example one could select an arbitrary set of usual primes and take $S$ to be the subset of integers that are the product of these primes. Then the primes of order $d$ where $d$ corresponds to this $S$ will be precisely the set of the originally selected primes. Or take $P$ any finite set of reals grater $1$ that is 'sum-free' (with respect to product, or if this is seems odd terminolgy take postive reals sumfree with respect to addition and take the expoential), there are plenty, and let $S_P$ be the semigroup generated by $P$. Then the primes of that order will be precisely $P$.</p> <p>Now, if we however drop the condition that all elements are greater $1$ it is possible that there are no primes. For example one can take $S$ equal to <code>$2^{i-3}$</code> for $i=1,2,\dots$ to get this. </p> <p>In brief, it seems to me that your question is vaguely linked to interesting problems, however I think in one way or another you need to narrow it down, otherwise the situation is seems too flexible to yield interesting questions.</p>