Lambert series 2, double mersenne summation

Number Theory Level 5

∑ p prime ∑ k = 1 ∞ 1 2 p k − 1 = ∑ n = 1 ∞ f ( n ) 2 − n \large \displaystyle\sum_{p \ \text{prime}} \sum_{k=1}^\infty \dfrac{1}{2^{p^k}-1}= \sum_{n=1}^\infty f(n)2^{-n}

Let f ( n ) f(n) be an additive function satisfying the equation above.

Find f ( 1 0 1 2 3 ) \large f(10^{12^3}) .


The answer is 3456.

Aareyan Manzoor by Aareyan Manzoor , 18, USA

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1 solution

Alex Burgess
May 2, 2019

1 2 p i k − 1 = 2 − p i k 1 1 − 2 − p i k = 2 − p i k ( 1 + 2 − p i k + 2 − 2 p i k + . . . ) \frac{1}{2^{p_i^k} - 1} = 2^{-p_i^k}\frac{1}{1 - 2^{-p_i^k}} = 2^{-p_i^k}(1 + 2^{-p_i^k} + 2^{-2p_i^k} + ...) for some p i , k p_i,k .

This includes 2 − n 2^{-n} iff p i k ∣ n p_i^k | n .

Summing over all k k , we get a a amounts of 2 − n 2^{-n} where p i p_i divides into n n a a times.

Summing over all primes, for n = p 1 α 1 p 2 α 2 . . . n = p_1^{\alpha_1}p_2^{\alpha_2}... , we get α 1 + α 2 + . . . \alpha_1 + \alpha_2 + ... lots of 2 − n 2^{-n} .

Hence f ( n ) = Ω ( n ) f(n) = \Omega(n) the total number of prime factors of n n , which is trivially additive.

f ( 1 0 1 2 3 ) = 1 2 3 f ( 10 ) = 1 2 3 ( f ( 2 ) + f ( 5 ) ) = 2 ∗ 1 2 3 = 3456 f(10^{12^3}) = 12^3f(10) = 12^3 (f(2) + f(5)) = 2*12^3 = 3456 .

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Although, you only really need to determine f ( 2 ) f(2) and f ( 5 ) f(5) , and 2 − 2 , 2 − 5 2^{-2},2^{-5} are only found once in 1 2 − 2 1 , 1 2 − 5 1 \frac{1}{2^{-2^1}}, \frac{1}{2^{-5^1}} respectively.

Alex Burgess - 2 years, 1 month ago

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