My favourite summation
x ∈ S ∑ x − 1 1 = 3 1 + 7 1 + 8 1 + 1 5 1 + 2 4 1 + …
Find the value of the above summation when it's summed over all elements in set S .
Set S consists of all perfect powers excluding 1 and excluding duplicates.
S = { 4 , 8 , 9 , 1 6 , 2 5 , 2 7 , 3 2 , 3 6 , 4 9 , 6 4 , 8 1 , 1 0 0 , . . . }
• A perfect power is a number of the form m n where m and n are natural numbers and n = 1 .
The answer is 1.0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
@Isaac Buckley Yet another theorem I've learned from you today. :) My first attempt failed to avoid repetition, so I looked at the sequence of denominators on OEIS and found the connection to the G-E Theorem.
Log in to reply
Well you're actually one of the first people i found when coming to this website. You've taught me so much! That really made my day that i could repay you with some fun facts. Thanks a lot Brian, you inspire me.
I've actually just solved 1729 problems so it was already a good day. Made a fun small problem to celebrate. It's very famous but i searched and nobody had posted it yet. Hoping i can make it 3 little fun facts in one day.
Log in to reply
Thanks for your kind words. That is a fun new problem and a surprising result; definitely a third new fact for the day. :) Congrats on reaching the Hardy-Ramanujan "level"; I'm not sure what the next "famous" number is, but 1 7 7 1 is interesting in that it is the first tetrahedral palindrome, (after 0 , 1 and 4 , of course), and might be the only non-trivial one.
This is the Goldbach-Euler Theorem .
It's interesting to note that Goldbach's and Euler's proof involves assigning a "value" to the harmonic series, which is problematic, but later proofs have made this approach more rigorous.