Bet you ran primes

Number Theory Level 3

A sequence defined as $a_n = p_{n+1}\bmod{p_n},$ where $p_n$ denotes the $n^\text{th}$ prime number.

What is the sum of the first 1000 terms $a_1+a_2+\cdots+ a_{1000}?$

Note: You can look up the values of any primes you want.

The answer is 7925.

1 solution

Alexander Gibson
Apr 22, 2018

Bertrand's postulate states that for any $n$ , there exists a prime number $p$ between $n$ and $2n$ .Applying this to $p_n$ , we get that $p_{n+1}$ lies between $p_n$ and $2p_n$ .Thus $p_{n+1}$ mod $p_{n}$ or $a_n$ is just $p_{n+1}-p_{n}$ .This therefore forms a telescoping sum and we get that the sum of the first 1000 terms is $p_{1000}-p_1$ $\equiv$ 7927-2 $\equiv$ 7925

Bet you ran primes

The answer is 7925.

1 solution

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