What is the number $p(n)$ of partitions of an abundant number $n$ into distinct, proper divisors of $n$?

Question

For lack of a better symbol, $p_{\sigma\tau}(n)$ (feel free to suggest something better).

For example, $p_{\sigma\tau}(12) \geq 2$ since $12 = 1 + 2 + 3 + 6 = 2 + 4 + 6$.

Of course if $n$ is deficient then $p_{\sigma\tau}(n) = 0$ and if $n$ is perfect then $p_{\sigma\tau}(n) = 1$. Though there are a few abundant $n$ for which $p_{\sigma\tau}(n) = 0$.

EDIT: Earlier I neglected to put in: A suitable partition must have more than one element, so $12 = 12$ doesn't count.

Answer 1

This is exactly the content of http://oeis.org/A065205 (as Will Jagy kindly points out, the values of $p(n)$ for $n \leq 1000$ are listed in http://oeis.org/A065205/b065205.txt); for the first abundant numbers $n$, $p(n)$ (I'll suppress the suggested subscript) is given by:

$$\begin{array}{cc} n & p(n)\\ \hline 12 & 2 \\ 18 & 2 \\ 20 & 1 \\ 24 & 5 \\ 30 & 3 \\ 36 & 7 \\ \vdots & \vdots \end{array}$$

The function $p(n)$ can be implemented by the following Mathematica code due to Jean-François Alcover (2012 February 23):

a[n_] := (dd = Most[ Divisors[n] ];
    cc = Array[c, Length[dd]];
    Length[ {ToRules[Reduce[
        And @@ (0 <= # <= 1 &) /@ cc && dd . cc == n, cc, Integers]]}]);

Remark At Will's request, the sequence of successive champions of the sequence $p(n)$ (that is, the $p(n)$ for which $p(n) > p(m)$ for all $m < n$) for $n \leq 1000$, together with their respective $n$, starts $$\begin{array}{cc} n & p(n)\\ \hline 1 & 0 \\ 6 & 1 \\ 12 & 2 \\ 24 & 5 \\ 36 & 7 \\ 48 & 10 \\ 60 & 34 \\ 120 & 278 \\ 180 & 751 \\ 240 & 2157 \\ 360 & 22208 \\ 720 & 676327 \\ 840 & 2225346 \\ \end{array}$$

Abundant numbers for which $p(n) = 0$ are precisely the weird numbers; the first of these are $$70, 836, 4030, 5830, 7192, 7912, \ldots,$$ and there are no odd weird numbers $< 10^{20}$.

As you point out, if $n$ is perfect then $p(n) = 1$, but the converse is not true; the first counterexamples are $$20, 78, 88, 102, 104, 114, \ldots$$ (see http://oeis.org/A064771 , which includes the perfect numbers too). Odd numbers of this type are rare: The first is $8925$, and there are only four such odd numbers $< 10^8$.

Answer 2

If you're asking for a formula for the function, I'm afraid there is no straightforward answer.

Sloane's OEIS tells us that A065205(n) = A033630(n) - 1. That leads me to $$p_\sigma(n) = f(n, n, 1) - 1$$ where $$f(n, m, k) = f(n, m, k + 1) + f(n, m - k, k + 1)0^{n \bmod k}$$ if $k \leq m$, else $0^m$.

(By the way, I think $p(n)$ is too plain and $p_{\sigma\tau}$ a bit excessive).