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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.

Definition: If there exists a $d_r$ such that the following conditions are satisfied:

i) there is at least one $i$ such that $s_i$ is not divisible by any of the numbers $s_1, s_2, \ldots s_{i-1}$.

ii) there is at least one triplet $(j,k,l)$ such that $s_j = s_ks_l$

then $s_i$ is a prime of the order $d_r$ and it is present in the sequence $S_{d_r}$.

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What do you mean by divisible? what does it mean to say $s_i$ is, or isn't, divisible by $s_j$? –  Gerry Myerson Feb 1 '12 at 11:33
    
If we have $s_j$ and $s_k$ such that $s_i = s_j s_k$ then $s_i$ divisible by both $s_j$ and $s_k$ and we say that $s_i$ is a composite (of order $d_r$). If we don't have such $s_j$ and $s_k$ then $s_i$ cannot be expressed as the product of two (or more) numbers in the sequence $S_{d_r}$ and we say that $s_i$ is a prime of order $d_r$. –  user20174 Feb 1 '12 at 11:40
    
One more request for clarification: in i) you insist that si is divisible by an element that appears before. Yet in ii) the triplet is not ordered. This seems a bit inconsistent to me. Or do you want all si to be at least 1. Say, take dr such that Sr starts 1/8, 1/4, 1/2, ... than s1 = s2 s3 but s1 is not divisible by anything that appears before. (In any case what is to be done with the first element? Else this will always be prime.) And since with the first term this is sort of a special case. Say let it start with 1/10, 1/8 , 1/4, 1/2, 2,... Than s2=s3 s4 but s2 is not divisible by s1. –  quid Feb 1 '12 at 15:22
    
Dear Quid, For the case of finding primes of a different order I was implicitly assuming that $1 < s_i < s_{i+1}$. This was just be analogous to the sequence of primes over natural number in which the 1 < n-th prime < (n+1)th prime. However your doubt has given a new way of looking at the problem in which $s_1 < 1$ and I would be interested in exploring primes of this order as well. –  user20174 Feb 2 '12 at 5:08

2 Answers 2

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.

If you don't want $s_2$ to be the "prime", then you can easily make $s_3$ "prime" by a similar construction.

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In view of the OPs comment I will focus on the case that $s_1>1$; but at the end make some comments on the other case.

The question is to find all $d_r$ such that there is at least one 'prime' of that 'order'.

This problem can be linked (is equivalent to) certain aspects of the classical problem of product/sum free subsets.

First, let us recast the problem. Of course, a sequence $d_r$ uniquely defines an $S_{d_r}$ but also the converse is true that is from the set $S_{d_r} = \lbrace s_i, i=1,2,\dots \rbrace $ we can recover $d_r$ by considering difference between successive elements, that is $d_i= s_{i}-s_{i-1}$, where we set $s_0=0$. 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 $1$

Let such an $S$ be written in the form $ S = \lbrace s_1 < s_2 < \dots \rbrace$ Now according to the definition $s_1$ always fulfills the first condition, and thus is 'prime' if the second condition is met.
The question thus arises what about the second condition.

Let us recall that a subset $A$ of an abelian (semi)group $(G,+)$ is called sum-free 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.

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 $P$.

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 $2^{i-3}$ for $i=1,2,\dots$ to get this.

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.

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