OFFSET
1,1
COMMENTS
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.
a(n) = m is such that floor(log_p_1 m) = k but floor(log_p_j m) = k-1 for j > 1.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
MATHEMATICA
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]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, Sep 17 2024
STATUS
approved