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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.

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  • $\begingroup$ You don't consider $1\times 2 + 1 \times 1$ to be different from $1 \times 1 + 1\times 2$? $\endgroup$ Commented Mar 30, 2016 at 5:17
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    $\begingroup$ You don't mention it explicitly, but I guess you want the number of ways to express $n$ in this way? $\endgroup$ Commented 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 $\endgroup$ Commented Mar 30, 2016 at 6:48

2 Answers 2

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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 $$ h+h^2+\dots=\frac{h}{1-h}. $$

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    $\begingroup$ Which gives oeis.org/A129921 . $\endgroup$ Commented Mar 30, 2016 at 6:56
  • $\begingroup$ does it count 1.2+2.1 is same as 2.1 + 1.2? $\endgroup$ Commented Mar 30, 2016 at 6:59
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    $\begingroup$ @Anupam : No it doesn't, so it isn't correct. $\endgroup$ Commented Mar 30, 2016 at 7:20
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It is http://oeis.org/A006171 , which has some generating functions and other formulas.

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