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The nth Prime Page will now find any of the first 2.623˙10^{15} primes or
π(x) for x up to 10^{17}. Just showing those entries submitted by 'Rivera': (Click here to show all) The least even number 2k such that SOP(2, P(k)) is divisible by 2k, i.e., SOP(2, P(k)) is congruent to 0 mod (2k). Note that SOP(2, P(k)) = the sum of primes from 2 to the prime P(k). [Rivera]
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