Prime Trouble

For any positive integer n > 1 n > 1 , let P ( n ) P(n) denote the largest prime not exceeding n n . Let N ( n ) N(n) denote the next prime larger than P ( n ) P(n) ,

Then Evaluate: 1 P ( 2 ) N ( 2 ) + 1 P ( 3 ) N ( 3 ) + 1 P ( 4 ) N ( 4 ) + + 1 P ( 2016 ) N ( 2016 ) \large \frac{1}{P(2)N(2)}+\frac{1}{P(3)N(3)}+\frac{1}{P(4)N(4)}+\cdots+\frac{1}{P(2016)N(2016)}


Clarification: P ( 10 ) = 7 P(10) = 7 and N ( 10 ) = 11 N(10) = 11 , while P ( 11 ) = 11 P(11) = 11 and N ( 11 ) = 13 N(11) = 13 .

Note: If your answer is of the form a b \dfrac{a}{b} , where a , b a,b are coprime positive integers, then submit a + b a+b .


The answer is 6049.

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2 solutions

Kanad Pardeshi
Feb 15, 2018

Giorgos K.
Feb 6, 2018

Mathematica

Sum[1/((s=Prime@PrimePi@n)NextPrime@s),{n,2,2016}]

solution is 2015/4034

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