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.