78 Jean-Marie De Koninck
362
the tenth number n such that n · 2n 1 is prime (see the number 115);
the third solution of σ(φ(n)) = σ(n) (see the number 87).
363
the fifth number n such that n ·
10n
+ 1 is
prime82:
the sequence of numbers
satisfying this property begins as follows: 1, 3, 9, 21, 363, 2161, 4839, . . . (see
the number 117).
364
the ninth number n such that f(n) f(m) for all numbers m n, where
f(n) :=
Σ∗
1
p
, with the star indicating that the sum runs over all prime numbers
p n which do not divide
(
2n
n
)
: here f(364) = 0.98744 . . .; the only numbers
n 20 000 satisfying this property are 3, 7, 9, 10, 27, 120, 121, 255, 364, 756,
3 160 and 3 250; Erd˝ os, Graham, Ruzsa & Straus [77] conjectured that there
exists a constant c such that f(n) c for all n 1 (see R.K. Guy [101], B33).
366
the number of digits in the decimal expansion of the
14th
perfect number
2606(2607
1).
367
the third prime number p such that p k! is composite for each positive integer
k such that k! p: the prime numbers smaller than 1 000 which satisfy this
property are 101, 211, 367, 409, 419, 461, 557, 673, 709, 937 and 967; it is not
known if there exist infinitely many prime numbers satisfying this property.
82One
can prove that at least one third of the numbers of the form
n10n
+ 1 are composite. In
fact, one can prove even more. Indeed, if S+(N) stands for the number of positive integers n N
such that rn :=
n10n
+ 1 is composite, then
(∗) S+(N)
N
3
+ π(N) (N 11).
To prove (∗), first observe that if n 2 (mod 3), then 3|rn; this follows from the fact that, in this
case, n = 3k + 2 for a certain integer k 0, so that
n10n
+ 1 = (3k +
2)103k+2
+ 1 200 ·
102k
+ 1 2 ·
12k
+ 1 0 (mod 3).
On the other hand, it follows from Fermat’s Little Theorem that if n + 1 is a prime number p 5,
then rn is a multiple of p, since in this case n = p 1, so that
n10n
+ 1 = (p
1)10p−1
+ 1 (p 1) + 1 0 (mod p).
From these two observations and the fact that S+(11) = 8, inequality (∗) follows.
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