Bertrand theorem states that there is at least one prime in (x, 2x) for x > 1. Let Pn denote the n-th prime and take x = Pn. Then the theorem states that there is at least one prime in (Pn, 2Pn). It shows Pn+1 ？ Pn < Pn, which means that gap between primes is linearly controlled and supports infinitude of primes. Generalize the theorem into the prime index sequence {n}. Then the Bertrand-type theorem states there is at least one number n + k in (n, 2n) for n > 1 such that n + k is prime p and Pp is called double prime. It is obvious that the Bertrand-type theorem implies infinitude of double primes. An even number Ln is defined as the largest strong Goldbach number generated by Pn if every even number from 4 to Ln is the sum of two primes not greater than Pn but Ln + 2 is not such a sum. Since Ln ≤ Ln+1 for all n, Ln is a non-decreasing function and there exist growth points of Ln. If Ln？1 < Ln then n is a growth point of Ln and corresponding prime Pn is called a nontrivial prime to structure a growth of Ln. It is clear that the infinitude of nontrivial primes implies Goldbach conjecture. Comparing counted number of nontrivial primes with counted number of double primes, we can conjecture that there is at least one number n + k in (n, 2n) for n > 1 such that Pn+k is a nontrivial prime. The Bertrand-type conjecture has been verified up to n = 300,000. If it is proven then Goldbach conjecture is true. Comparing counted number of twin primes with counted number of double primes, we can conjecture that there is at least one number n + k in (n, 2n) for n > 1 such that Pn+k is a twin prime. The Bertrand-type conjecture has been verified up to
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