Why aren't 10 and 47 an amicable pair?

[ksnortum]

I am obviously missing something when it comes to amicable pairs.
If d(n) is a function that takes a natural number and returns a list of proper factors (excluding the number itself), then:
d(10) = 1, 2, 5, sum of 8; d(49) = 1, 7, sum of 8; I've looked it up and 10 and 47 are not an amicable pair. Why?

 

[mmm4444bot]

1 + 2 + 5 would need to equal 49 AND 1 + 7 would need to equal 10. That is why 10 and 49 are not an amicable pair of numbers.

I think that you misunderstand the definition of amicable pair.

HERE is an example. :cool:

 

[soroban]

Hello, ksnortum!

I am obviously missing something when it comes to amicable pairs.
If d(n) is a function that takes a natural number and returns the sum of the proper factors), then:
. . . d(10) = 1, 2, 5 . . . sum of 8.
. . . d(49) = 1, 7 . . . sum of 8.

I've looked it up and 10 and 47 are not an amicable pair. .Why?

You've misunderstood the definition of Amicable Numbers.

A pair of amicable numbers has this property:
. . the sum of the proper factors of one number equals the other number.


Such a pair is \(\displaystyle (220,\:284)\)

The factors of 220 are: .\(\displaystyle 1, 2, 4, 5, 10, 11,20,22,44,55,110\)
. . And: .\(\displaystyle 1+2+4+5+10+11 + 20+22+44+55+110 \;=\;284\)

The factors of 284 are: .\(\displaystyle 1, 2, 4, 71, 142\)
. . And: .\(\displaystyle 1+2+4+71+142 \;=\;220\)


There are many other such pairs:
. . \(\displaystyle (1184,\:1210),\;(2620,\:2924),\; (5020,\:5564),\;(6232,\:6368)\;\text{ . . .}\)


The generating procedure follows.

Take any power of 2, \(\displaystyle 2^n\text{ for }n > 1.\)

Then form these numbers: .\(\displaystyle \begin{Bmatrix}a &=& 3\!\cdot\!2^n-1 \\ b &=& 3\!\cdot\!2^{n-1} - 1 \\ c &=& 9\!\cdot\!2^{2n-1}-1 \end{Bmatrix}\)

If \(\displaystyle a,b,c\) are prime, then: .\(\displaystyle 2^nab\,\text{ and }\,2^nc\) are amicable.

 

[ksnortum]

Thank you.