cp's OEIS Frontend

Pi(n) = number of prime numbers <= n (A000720). Prime(n) = A000040(n).

Conjecture: For each n there is at least one prime p such that a(n) < p < a(n+1). From the conjecture follows that the prime gaps g(n) = p(n+1) - p(n) = O(sqrt(p(n))/log(p(n))). Legendre's hypothesis is that g(n) = O(sqrt(p(n))). - Thomas Ordowski, Aug 11 2012

Sequence elements A000040 of the form A128913(n) - 1.

Primes of the form pi(k)*pi(prime(k)) + e^(i*Pi), where pi(k) is the number of primes <= k, i is the imaginary unit, Pi = 3.14159...

I introduce this sequence which is A128913(n)/2 because it is closely related to the prime counting function Pi(n) and the sum of primes < n for large n.

This is, SumP(n) ~ n*Pi(n)/2. For n = 10^10 n*Pi(n)/2 = 2275262555000000000.

Sum primes < 10^n = 2220822432581729238. This has error 0.0245...For the largest known sum of primes, for sums < 10^20, we have n*Pi(n)/2 = 111040980128045942000000000000000000000. The sum of primes < 10^20 = 109778913483063648128485839045703833541. The error here is -0.01149... It converges quite slowly and better approximations have been found.

This relationship was derived by using the summation formula for an arithmetic progression. For the odd integers where n is even, let the first term = 1, the last term is n-1 and the number of terms is n/2. So the sum of the odd numbers < n is ((1 +n-1)*n/2)/2. If we let Pi(x) be the number of terms, we get the result n*Pi(n)/2. A closed formula, SumP(n) ~ n^2/(2*log(n)-1) is quite accurate. The best formula I have found is the remarkable SumP(n) ~ Pi(n^2).

This formula has an error of 6.162071097138 E-11 for the largest known sum of primes or sum < 10^20.

Proof: 2+3+..+prime(n) = A007504(n) ~ n^2 log n / 2 (Bach and Shallit, 1996). Let n = Pi(x) ~ x/log x. So A007504(n) ~ (x/log x)^2 log(x/log x) / 2 ~ x^2 / (2 log x) ~ Pi(x^2). QED. - Thomas Ordowski, Aug 12 2012

See the link Sum of Primes for derivations of these asymptotic formulas.