A376261
Products m of k = 5 consecutive primes p_1..p_k, where only p_1 < m^(1/k).
Original entry on oeis.org
362469273063260281, 390268963330916339, 2501104163586622303, 9136139450993677127, 14786802713701223291, 16175430816211360949, 42275879149134880507, 58976503686022007233, 75786488186892877007, 124796858811854774081, 226284602311169194703, 252607170708747107509
Offset: 1
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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]]
A374873
Smallest primes p_1 where products m of n consecutive primes p_1..p_n are such that only p_1 < m^(1/n).
Original entry on oeis.org
2, 3, 113, 3229, 15683, 736279, 8332427, 37305713, 4948884397, 6193302809, 316781230427
Offset: 2
a(2) = 2 since m = 2*3 = 6 and 3 > sqrt(6).
a(3) = 3 since m = 3*5*7 = 105 and 5 > 105^(1/3).
a(4) = 113 since m = 113 * 127 * 131 * 137 = 257557397 and 127 > 257557397^(1/4), etc.
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k = 1; Table[r = Range[0, n - 1]; While[(Set[{p, q, m}, {#[[1]], #[[2]], Times @@ #}]; q < Surd[m, n]) &[Prime[k + r]], k++]; p, {n, 2, 6}]
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a(n) = {my(ps = vector(n, k, prime(k))); forprime(p = prime(n+1), , if(ps[2]^n > vecprod(ps), return(ps[1])); ps = concat(vecextract(ps, "^1"), p));} \\ Amiram Eldar, Sep 23 2024
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