Extend Bertrand’s Postulate to Sums of Any Primes

In 1845, Bertrand conjectured what became known as Bertrand’s postulate: twice any prime strictly exceeds the next prime [1] . Tchebichef presented his proof of Bertrand’s postulate in 1850 and published it in 1852 [2] . It is now sometimes called the Bertrand-Chebyshev theorem. Surprisingly, a stronger statement seems not to be well known, but is elementary to prove: The sum of any two consecutive primes strictly exceeds the next prime, except for the only equality 2 + 3 = 5. After I conjectured and proved this statement independently, a very helpful referee pointed out that Ishikawa published this result in 1934 (with a different proof) [3] . This observation is a special case of a much more general result, Theorem 2, that is also elementary to prove (given the prime number theorem), and perhaps not previously noticed: If pn denotes the nth prime, n=1,2,3,⋯ with p1=2,p2=3,p3=5,⋯ and if c1,c2,⋯,cj are natural numbers (not necessarily distinct), and d1,d2,⋯,di are positive integers (not necessarily distinct), and then there exists a positive integer N such that pn−c1+pn−c2+⋯+pn−cj>pn+d1+pn+d2+⋯+pn+di for ll n≥N . We also have another result: If i<n and j are nonnegative integers, then there exists a large enough positive integer N such that, for all n≥N , pn+pn−i>pn+j . We give some numerical results.