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Let $\sigma(n)$ be the sum of all divisors (including 1 and $n$) of $n$, and define the abundancy index of $n$ as $I(n) = \sigma(n)/n $. For example: $I(6)= \frac{1+2+3+6}{6} = 1/1+1/2 +1/3 +1/6 = 2$. Thus all perfect numbers have an abundancy index of 2. I wondered what the average abundancy of all numbers was, and it turned out to converge to $\pi^2/6 $.

I then calculated the abundancy index for certain subsets of integers, and more interesting relations emerged it seemed. Let $T= 1,3,6,..$ be set of the first million triangle numbers, then the average abundancy index of $T$ turned out to be $ 1.99990$ Furthermore I found for the numbers that are 1 or 2 more than a perfect square, $Q = 2,3,5,6,10,11,17,18,...$ that the median of the abundacy index for the first 1000000 numbers in $Q$ is 1.500000000991

For the first 150 numbers of the form $2^n+1$ I found the median 1.3333333406, and for the first 150 $2^n+2$ a median of 2.00000000139 Furthermore I found for the first 100 numbers of the form $3^n -1$ the average abundancy index of 2.00146

Anyone has some insights or knows some interesting papers regarding these values ?

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  • $\begingroup$ All of these questions concern estimating $\sum{\sigma(n)\over n}$, where the sum is over all numbers up to $N$ in some particular set, such as the triangular numbers. Analytic Number Theory has ways of arriving at such estimates. $\endgroup$
    – Gerry Myerson
    Commented Mar 14 at 11:33
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    $\begingroup$ Some of these are connected. $2^n+2=2\times (2^{n-1}+1)$ so $\sigma(2^n+2)=3\sigma(2^{n-1}+1)$ which implies that the abundancy of $2^n+1$ is $\frac 32$ times that of $2^{n-1}+1$, as you observed. $\endgroup$
    – lulu
    Commented Mar 14 at 11:35
  • $\begingroup$ You seem to switch between mean and median - is this just looking for roundish values? $\endgroup$
    – Henry
    Commented Mar 14 at 12:09

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