Problems & Puzzles: Puzzles

Puzzle 1198 ssd(n)=s2(n)

On Nov 15, 2024, Fred Schneider sent the following nice puzzle:

Define ssd(n) = sum( sigma(each divisor of n)) and s2(n) = sigma(sigma(n)) where sigma(n) is the sum of the divisors of n

I noticed something recently: 1, 2 and 360 have the special property that ssd(n) = s2(n).

Let's call these Fred numbers

sigma(1) = sigma(sigma(1)) = 1 and
sigma(2) + sigma(1) = sigma(sigma(2)) = 4

The third solution 360 is a little more complex:

360 = 2^3 * 3^2 * 5

So, sigma(360) = (8 + 4 + 2 + 1)(9 + 3 + 1)(5 + 1) = 1170

Similarly, ssd(360) can be expressed elegantly as:

( sigma(8) + sigma(4) + sigma(2) + sigma(1) ) *
( sigma(9) + sigma(3) + sigma(1) ) *
( sigma(5) + sigma(1) )

= 3276

(This cross-product generates each divisor and its sigma equals the sum of each divisor's sigma)

sigma(360) = 1170 and sigma(1170) = 3276

Q1. Can you find any other "Fred" numbers?
Q2. If not, can you prove no others exist?


From Nov 23 to Nov 29 contributions came from JM Rebert, Jeff Heleen

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JM Rebert wrote.

I found :
n = 15360620700 = 2^2 * 3 * 5^2 * 3967 * 12907 
with the property that ssd(n) = s2(n) = 107082865890

 
div(4) = [1, 2, 4] -> sigma(4) =  7
div(3) = [1, 3] -> sigma( 3) = 4
div(25) = [1, 5, 25] -> sigma(25) = 31
div(3967) = [1, 3967] -> sigma(3967) = 3968
div(12907) = [1, 12907] -> sigma(12907) = 12908
and 4 * 3 *  25 * 3967 * 12907-= 15360620700

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Jeff wrote:

For n up to 1,000,000 I did not find any other solutions.
 
I noticed for n in the range 3 <= n <= 359 that
sigma(sigma(n)) > ssd(n) 322 times and
ssd(n) > sigma(sigma(n)) 35 times.
 
For n in the range 361 <= n <= 1,000,000
sigma(sigma(n)) > ssd(n) 988616 times and
ssd(n) > sigma(sigma(n)) 11024 times.

 

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