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A run of 2 or more consecutive integers is said to be perfect if the sum of its terms equals the sum of all the their proper divisors, adding common divisors as often as they occur. For example, 672, 673 is such a run (672 is a triperfect number whose proper divisors add up to 1344, while 673 is a prime, 1 being its only proper divisor).

Are there arbitrarily long perfect runs of consecutive numbers? Infinitely many of any length greater than 1?

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    $\begingroup$ If $\sigma$ denotes the sum of divisors, then we have the classical estimate $\sum_{m=1}^n \sigma(m) \sim \zeta(2) n^2/2$. Since $1 < \zeta(2)<2$, this means that runs of $k$ consecutive integers $(n-k+1,\ldots,n)$ where $k$ is ``large'' (I think $k \gg \log n$ suffices) cannot be perfect. $\endgroup$
    – François Brunault
    Commented Apr 24, 2019 at 20:38

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523776, 523777 is another example of a 2-run. Again, 523776 is a triperfect number.

An example of a 3-run is 5829840, 5829841, 5829842. No apparent pattern in these three numbers:

          Factorization         Sum of proper divisors
5829840   2^4 * 3^3 * 5 * 2699  14258160
5829841   29 * 53 * 3793        316439
5829842   2 * 2914921           2914924

Here are two more perfect 3-runs of consecutive numbers:

             Factorization                Sum of proper divisors
3414097920   2^16 * 3 * 5 * 23 * 151      8061430272
3414097921   13 * 31 * 41 * 206627        473814527
3414097922   2 * 1707048961               1707048964

39339578248  2^3 * 4917447281             34422130982
39339578249  29 * 4561 * 297421           1365596671
39339578250  2 * 3^2 * 5^3 * 7 * 2497751  82231007094

The first of these two, like the earlier example, were found by computer search with PARI/GP. I noticed that their middle odd terms turn up in A072188:

Numbers $n$ such that $n$ divides $\sigma(n-1)+\sigma(n)+\sigma(n+1)$, where $\sigma(n)$ is the sum of integer divisors of $n$.

The final of the above examples of perfect 3-runs comes from that sequence.

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    $\begingroup$ Are these the only examples below 6 million? Gerhard "Is Keeper Of Imperfect Information" Paseman, 2019.04.24. $\endgroup$
    – Gerhard Paseman
    Commented Apr 24, 2019 at 18:24
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    $\begingroup$ @GerhardPaseman: Yes, the above two examples, together with OP's (672, 673), are the only 2 and 3-runs below $10^8$. $\endgroup$
    – Freddy Barrera
    Commented Apr 24, 2019 at 19:28
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    $\begingroup$ Probably coincidental but your examples all fulfill $\tau(n)=2^m.d$ with $d$ a divisor of a prime. $\endgroup$
    – Sylvain JULIEN
    Commented Apr 25, 2019 at 16:15
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    $\begingroup$ My computer checked that there is no other perfect 2-run below $10^{10}$. An observation: in all the known perfect runs, the largest number is a large prime times some small prime factors. I don't know if this could help to find more runs. $\endgroup$
    – François Brunault
    Commented Apr 25, 2019 at 19:53
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    $\begingroup$ @FrançoisBrunault: I can confirm that there are no other perfect 2-runs and 3-runs below $1.8\times 10^{10}$, nor any perfect 4-runs. The closest for the latter is the run beginning at 113393279 that falls short by 4 of being perfect. $\endgroup$
    – Freddy Barrera
    Commented Apr 27, 2019 at 12:35

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