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%I #19 Mar 11 2023 00:13:22
%S 2,174,11010,877590,3576536040,7300395162060,8122095802580760,
%T 15497422946114018910,6949903578918639188850,
%U 482875127106370562524140180,2448313281623289989477792853630,20024982066721727911275778517919720,29200503600421680216708710172770859570
%N Least k such that the first n primes divide k and the next n primes divide k+1.
%H Samuel Harkness, <a href="/A360478/b360478.txt">Table of n, a(n) for n = 1..175</a>
%e a(3) = 11010 because the first 3 primes {2, 3, 5} divide 11010 and the next 3 primes {7, 11, 13} divide 11011.
%t K = {}; For[n = 1, n <= 13, n++, R = Prime[Range[2 n]];
%t A = R[[1 ;; n]]; t = Times @@ A; B = R[[n + 1 ;; 2 n]]; p = t + 1;
%t For[b = 1, b <= n, b++, While[Mod[p, Part[B, b]] != 0, p += t];
%t t *= Part[B, b]]; AppendTo[K, p - 1]]; Print[K]
%o (PARI) a(n)={my(p=primes(2*n)); lift(chinese(Mod(0, vecprod(p[1..n])), Mod(-1, vecprod(p[n+1..2*n]))))} \\ _Andrew Howroyd_, Feb 08 2023
%Y Cf. A069561.
%K nonn
%O 1,1
%A _Samuel Harkness_, Feb 08 2023
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