product of all multiplicative functions

n = 1 ϕ μ τ σ λ e ( n ) n 1729 = π a b a + b = ? \sum_{n=1}^\infty \dfrac{\phi\mu\tau\sigma\lambda e(n)}{n^{1729}}=\dfrac{\pi^a}{b}\\ a+b=? notations

all functions are listed in this wiki


The answer is 1.

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1 solution

Otto Bretscher
Feb 20, 2016

Not as bad as it looks!

Since e ( 1 ) = 1 e(1)=1 and e ( n ) = 0 e(n)=0 for n > 1 n>1 , the sum is 1 = π 0 1 1=\frac{\pi^0}{1} , and the answer is 1 \boxed{1} .

PS: You should tell us to enter a + b a+b as the answer! I assumed...

Thanks, i have edited the problem.

Aareyan Manzoor - 5 years, 3 months ago

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Wait what?

Julian Poon - 5 years, 3 months ago

Comrade @Aareyan Manzoor : I have a question on your recent roots of unity problem... you know I love those ;) Comparing the derivatives, I'm finding that the RHS is L i 2 ( x n ) n \frac{Li_2(x^n)}{n} . Am I on the right track? Maybe a calculus error?

Otto Bretscher - 5 years, 2 months ago

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I am sorry; i made an error. I had initially thought of n-1 but ended up writing n in the upper limit of the boundary of the summation.

Aareyan Manzoor - 5 years, 2 months ago

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@Aareyan Manzoor yes, that's what I thought... so you get a coefficient -1 on the other side. great problem though!

Otto Bretscher - 5 years, 2 months ago

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