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Products m of k = 5 consecutive primes p_1..p_k, where only p_1 < m^(1/k).
3

%I #5 Sep 20 2024 06:01:18

%S 362469273063260281,390268963330916339,2501104163586622303,

%T 9136139450993677127,14786802713701223291,16175430816211360949,

%U 42275879149134880507,58976503686022007233,75786488186892877007,124796858811854774081,226284602311169194703,252607170708747107509

%N Products m of k = 5 consecutive primes p_1..p_k, where only p_1 < m^(1/k).

%C In other words, products m of k = 5 consecutive primes p_1..p_k, where floor(log_p_1 m) >= k but floor(log_p_j m) = k-1, j > 1.

%C a(n) = m is such that floor(log_p_1 m) = k but floor(log_p_j m) = k-1 for j > 1.

%C Contains neither A375263 nor A375264, since for m in either sequence, both p_1 and p_2 are smaller than m^(1/k).

%H Michael De Vlieger, <a href="/A376261/b376261.txt">Table of n, a(n) for n = 1..10000</a>

%t k = 5; s = {1}~Join~Prime[Range[k - 1]]; Reap[Do[s = Append[Rest[s], Prime[i + k - 1]]; r = Surd[Times @@ s, k]; If[Count[s, _?(# < r &)] == 1, Sow[Times @@ s] ], {i, 120}] ][[-1, 1]]

%Y Cf. A375263, A375264, A376136.

%K nonn

%O 1,1

%A _Michael De Vlieger_, Sep 17 2024