S = n = 1 ∑ ∞ n 6 σ 2 ( n )
Let σ 2 ( n ) denote the sum of the squares of all the positive integer divisors of n . For example, σ 2 ( 6 ) = 1 2 + 2 2 + 3 2 + 6 2 = 5 0 .
Enter S π 1 0 as your answer.
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Wonderful! That's exactly what I had in mind! (+1)
One small complaint, though: You butchered the name of my beloved country man (Jakob) Bernoulli, a member of the greatest mathematical family in history. ;)
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Hm... I tried making a generating function but got stuck.
sorry for that....... i have corrected it.....
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read about dirichlet convolution (still under construction, if you want to help DM me at slack)
the defination of dirchlet series D G ( F ; s ) = n = 1 ∑ ∞ n s F ( n ) where F is an arithmetic function. we are to know D G ( f ∗ g ; s ) = D ( f ; s ) D ( g ; s ) where f ∗ g denotes the dirichlet convolution of f and g. define function f ( n ) = n 2 and u ( n ) = 1 . so we see that f ∗ u = d ∣ n ∑ ( f ( d ) u ( d n ) ) but u of anything is one so: f ∗ u = d ∣ n ∑ ( d 2 ) = σ 2 ( n ) so D G ( σ 2 ; 6 ) = D G ( f ; 6 ) D G ( u ; 6 ) = ( n = 1 ∑ ∞ n 6 n 2 ) ( n = 1 ∑ ∞ n 6 1 ) = ζ ( 4 ) ζ ( 6 ) we know we can calculate zeta of even integers using Bernoulli numbers.so, this equals after simplification 8 5 0 5 0 π 1 0