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?

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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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