Conjecture 60. Generalization
of Legendre's Conjecture
T.D. Noe sent the following
conjecture:
Every fan of prime numbers knows
Legendre's conjecture:
For n>0, there is always
a prime between n^2 and (n+1)^2.
I propose that this is just a
special case of the following generalization:
For n>0 and k>=K, there
is always a prime between n^k and (n+1)^k, where
K=log(127)/log(16)=1.74717117169304146332...
Q. Prove it of
find a counterexample.

Contributions came from Patrick Capelle,
Luis Rodríguez & Enoch Haga.
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Patrick wrote:
This conjecture,
which could be improved by using K = sqrt(3), is just a special case of
the following generalization: CONJECTURE. Pi((n+1)^k) - Pi(n^k) >=
ceil(n^(k-2)) for n > 0, k >= K.
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Luis wrote:
The Conjecture 60 is
a weak case of Firoozbakht Conjecture:
P(n)^(n+1) > P(n+1)^n. It follows that
P(n)^(1+1/n) > P(n+1)
N must be > P(n), because if N = P(n) then we will have already a prime
in N.
If N^(1+1/n) > P(n+1) , we have at least the prime P(n+1 )between N and
N^(1+1/n) .
This is better than a prime between N^1.747 and (N+1)^1.747.
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Enoch wrote:
Since the
generalization comprises a narrower range than the special case, it is
only necessary to check whether there is always a prime greater than n^k
and less than (n+1)^k.
The generalization fails at 113 - 127 because the next prime greater
than 113 is 127, and we are looking for a prime within the range. The
word "between" does not include either the value at the beginning or the
end, but only those within the range.
Latter he added:
I think
K=log(127)/log(14) may work.
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T. D. Noe, replied:
Here is my response
to the comments from Patrick Capelle:
K=sqrt(3)=1.73205... does not work as a lower bound because for K in the
small range (log(113)/log(15),log(127)/log(16))=(1.74568..., 1.74717...)
the conjecture is false. I used exact interval arithmetic to find 29
intervals in which the conjecture appears true. The last interval is the
semi-infinite interval beginning with log(127)/log(16).
Here is my comment to Luis Rodríguez:
It is true that the Firoozbakht Conjecture implies this conjecture.
However, because the Firoozbakht Conjecture is stronger, it will
probably
be even more difficult to prove.
Here is my response to the comments from Enoch Haga:
If I change the conjecture to k>K, then there will be a prime between
15^k
and 16^k. It appears the the relatively large prime gap between 113 and
127 is the largest gap to overcome.
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