1
$\begingroup$

In algebraic number theory we come across following formula:

$n= e_1f_1+\cdots+e_rf_r$

where all $e_i$ and $f_i$ are positive integers. I am sure writing a positive integer n as above must be studied.

For example : 1=1.1

2=1.1+1.1 = 1.2 = 2.1

3=1.1+1.1+1.1 = 1.1+1.2 = 1.1 +2.1 = 1.3 = 3.1

I am looking for a formula and generating function for this. Any help regarding this will be highly appreciated. Thank you.

$\endgroup$
3
  • $\begingroup$ You don't consider $1\times 2 + 1 \times 1$ to be different from $1 \times 1 + 1\times 2$? $\endgroup$ Mar 30, 2016 at 5:17
  • 1
    $\begingroup$ You don't mention it explicitly, but I guess you want the number of ways to express $n$ in this way? $\endgroup$ Mar 30, 2016 at 5:34
  • $\begingroup$ Yes 1.2 + 1.1 is same as 1.1+1.2 however not same as 1.1 + 2.1 $\endgroup$ Mar 30, 2016 at 6:48

2 Answers 2

4
$\begingroup$

1) If we do not care on the order of summands (it looks to be the case).

For each ordered pair $(e,f)$ take a sum $1+x^{ef}+x^{2ef}+\dots$, generating function is a product of these guys over all $(e,f)$. This may be further rewritten in different ways. At first, we may fix $e$, then we have $\prod_f (1-x^{ef})^{-1}=P(x^{e})$, where $P(t)=\prod (1-t^f)^{-1}=\sum p(n)t^n$ is the Euler generating function for partitions. So, the answer is $\prod_e P(x^e)$. It is appropriate to get asymptotics. We may instead fix $ef=n$, then we get $\prod_n (1-x^n)^{-\tau(n)}$, where $\tau(n)$ is a number of divisors function.

Of course, both functions are written on the OEIS page from the comment by Brendan McKay.

2) If we care on the order of summands.

For $r=1$ the number of representations is number of divisors of $n$, thus generating function equals $h(x)=\sum_{k\geqslant 1} x^k/(1-x^k)$. For arbitrary $r$ generating function if $h^r$, thus if $r$ varies, we finally get $$ h+h^2+\dots=\frac{h}{1-h}. $$

$\endgroup$
3
  • 3
    $\begingroup$ Which gives oeis.org/A129921 . $\endgroup$ Mar 30, 2016 at 6:56
  • $\begingroup$ does it count 1.2+2.1 is same as 2.1 + 1.2? $\endgroup$ Mar 30, 2016 at 6:59
  • 1
    $\begingroup$ @Anupam : No it doesn't, so it isn't correct. $\endgroup$ Mar 30, 2016 at 7:20
4
$\begingroup$

It is http://oeis.org/A006171 , which has some generating functions and other formulas.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.