Charlie Neder, <a href="/A052202/b052202_1.txt">Table of n, a(n) for n = 1..1000</a>
Charlie Neder, <a href="/A052202/b052202_1.txt">Table of n, a(n) for n = 1..1000</a>
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Previous name was: If p^a(p,n) is highest power of p that divides n, then productProduct_(p=primes) [log(p)^a(p,n) ] > log(n).
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Previous name was: If p^a(p,n) is highest power of p that divides n, then product(p=primes)[log(p)^a(p,n) ] > log(n).
Old title: If p^a(p,n) is highest power of p that divides n, then product(p=primes)[log(p)^a(p,n) ] > log(n).
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If p^a(p,n) is highest power of p that divides n, then Product_(p=primes) log(p)^a(p,n) > log(n).
Numbers k such that the product of the logarithms of k's prime factors is greater than their sum.
Numbers n such that the product of the logs of n's prime factors is greater than their sum.
Primes are counted with multiplicity.
Old title: If p^a(p,n) is highest power of p that divides n, then product(p=primes)[log(p)^a(p,n) ] > log(n).
Title changed by Charlie Neder, Dec 04 2018
0,1,1
a(100000) = 189837189835. (End)
Charlie Neder, <a href="/A052202/b052202_1.txt">Table of n, a(n) for n = 01..9991000</a>
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(PARI) isok(n) = my(f=factor(n)); prod(k=1, #f~, log(f[k, 1])^f[k, 2]) > log(n); \\ Michel Marcus, Dec 04 2018
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