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When A342439(n) is the largest prime < 10^n obtained with the longest sum of the A342440(n) consecutive primes, then a(n) is the first prime of these A342440(n) consecutive primes.
4

%I #25 Mar 17 2021 15:40:32

%S 2,2,7,3,3,7,7,7,11,2,19,5,5,2,13,5,5,7,11

%N When A342439(n) is the largest prime < 10^n obtained with the longest sum of the A342440(n) consecutive primes, then a(n) is the first prime of these A342440(n) consecutive primes.

%C Inspired by the 50th problem of Project Euler (see link).

%C There must be at least two consecutive primes in the sum.

%C The terms a(4)-a(17) come from the Perl program and the results proposed by _Daniel Suteu_ in the link Archive.today.

%H Archive.today, <a href="https://archive.is/WNBa2">trizen / experimental-projects</a>.

%H Dreamshire, <a href="https://blog.dreamshire.com/project-euler-50-solution/">Project Euler 50 Solution</a>.

%H Project Euler, <a href="https://projecteuler.net/problem=50">Problem 50: Consecutive prime sum</a>.

%e A342439(2) = 41 = 2 + 3 + 5 + 7 + 11 + 13 hence a(2) = 2.

%Y Cf. A342439, A342440, A342443, A342444, A342454

%K nonn,more

%O 1,1

%A _Bernard Schott_, Mar 14 2021

%E a(4)-a(17) from _Daniel Suteu_, Mar 14 2021

%E a(18)-a(19) from _Martin Ehrenstein_, Mar 14 2021

%E a(7) and a(15) corrected by _Martin Ehrenstein_, Mar 14 2021