Sunday, 3 January 2021

Hemiperfect Numbers

The so-called hemiperfect numbers relate a concept called abundancy that I've dealt with in two previous posts:

I'll define the abundancy of a number \(n\) once again to be the ratio of the sum-of-divisors of \(n\) to \(n\) itself. It is given by the formula:$$ \frac{\sigma_1(n)}{n} \text{ where }\sigma_1(n) \text{ is the divisor function}$$Note that abundancy may also be defined as:$$\sigma_{-1}(n) \text{ where } \sigma_{-1}(n) \text{ represents the sum of the reciprocals of the divisors of } n$$A multiperfect (sometimes multiply perfect) number is a number whose abundancy is a whole number:$$\frac{\sigma_1(n)}{n}=k \text{ with } k \text{ an integer } \geq 2$$We can refer to such a number as \(k\)-perfect with 2-perfect numbers being the perfect numbers 6, 28, 496, 8128 etc.

Today I turned 26208 days old and discovered that this number is a member of OEIS A159907:


  A159907

Numbers \(n\) with half-integral abundancy index such that:                  $$\frac{\sigma_1(n)} {n} = k+\frac{1}{2} \text{ with integer }k$$


Numbers of this sort are termed hemiperfect. The sequence, up to 26208, consists of 2, 24, 4320, 4680, 26208 where:
$$\begin{align}
\frac{\sigma_1(2)} {2} &= \frac{3}{2}=1+\frac{1}{2}\\
\frac{\sigma_1(24)} {24} &= \frac{60}{24}=2+\frac{1}{2} \\
\frac{\sigma_1(4320)} {4320} &= \frac{15120}{4320}=3+\frac{1}{2} \\
\frac{\sigma_1(4680)} {4680} &= \frac{16380}{4680}=3+\frac{1}{2} \\
\frac{\sigma_1(26208)} {26208} &= \frac{91728}{26208}=3+\frac{1}{2}
\end{align}$$After this the numbers get bigger quickly. The next hemiperfect number is 8910720 which has an abundancy of 9/2 and is termed 9-hemiperfect. Similarly, 2 is 3-hemiperfect, 24 is 5-hemiperfect and 4320, 4680 and 26208 are 7-hemiperfect.

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