72 Jean-Marie De Koninck
320
the
12th
number n such that n! + 1 is prime (see the number 116).
323
the smallest Fibonacci pseudoprime : the only other one known is 377; for the
definition of a Fibonacci pseudoprime, see Ribenboim [169], pp. 126-127.
324
the
13th
number n such that n! 1 is prime (see the number 166).
325
the third pseudoprime in base 7: the ten smallest pseudoprimes in base 7 are
6, 25, 325, 561, 703, 817, 1 105, 1 825, 2 101 and 2 353;
the smallest (and possibly the only one) 3-hyperperfect number: we say that
a number n is 3-hyperperfect if it can be written as n = 1 + 3
d|n
1dn
d (which is
equivalent to the condition 3σ(n) = 4n + 2); see the number 21;
the smallest number which can be written as the sum of two squares in three
distinct ways: 325 = 12 + 182 = 62 + 172 = 102 + 152 (see the number 50).
330
the smallest number n such that π(n) = n/5; if, for each integer m 2, we set
Am = {n N : n/π(n) = m}, one can prove that #Am 1
for78
each m 2;
a clever computation provides the following table:
78Here
is how one can prove this result, first established in 1962 by S.W. Golomb [91]. Let as
usual p1, p2, . . . stand for the increasing sequence of prime numbers. Given an arbitrary integer
m 2, one needs to show that equation
n
π(n)
= m has at least one solution n. Since π(x) = o(x),
it follows that for all ε 0, there exists n0 = n0(ε) such that
π(n)
n
ε for all n n0. This is why
there exists a largest prime number pk such that
π(pk )
pk
=
k
pk

1
m
, so that (∗) pk mk. But then,
either mk pk+1 or else mk pk+1. This last case cannot occur, because one would then have
k+1
pk+1
k
pk+1

1
m
, contradicting the minimal choice of pk. Therefore, mk pk+1, an inequality
which coupled with (∗) yields pk mk pk+1 1, so that k = π(pk) π(mk) π(pk+1 1) = k,
which establishes that π(mk) = k. Setting n = mk, it follows that n/π(n) = m, thus providing
the required solution n. It is clear on the other hand that the result holds for an arbitrary set of
positive integers of zero density. In other words, one can also prove the following result:
Let {an} be an infinite sequence of positive integers, and let C(x) = #{an x}.
Assume that C(x) = o(x) when x ∞. Then there exists a positive integer η such that
the set of integer values taken by the quotient n/C(n) is equal to the set {m : m η}.
Moreover, η = 1 if and only if a1 = 1.
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