92 Jean-Marie De Koninck

533

• the common value of B(n), B(n + 1) and B(n + 2), when n = 417 162, namely

the only number n

109

such that B(n) = B(n + 1) = B(n + 2) (see the

number 417 162).

541

• the fifth horse number (see the number 13);

• the 100th prime number; if as usual, pk stands for the kth prime number, then

we have the following table:

k pk

10 29

100 541

1 000 7 919

10 000 104 729

100 000 1 299 709

1 000 000 15 485 863

k pk

107 179 424 673

108 2 038 074 743

109 22 801 763 489

1010

252 097 800 623

1011

2 760 727 302 517

1012

29 996 224 275 697

• the smallest prime number which is preceded by 17 consecutive composite num-

bers; indeed, there are no prime numbers between 523 and 541.

544

• the smallest number m for which inequality π(2x) ≥ π(x) +

π2(

√

x) holds for

all x ≥ m, a result

due96

to Panaitopol [159].

545

• the largest number x for which the diophantine equation

x2

+

32m

=

2yp

has a

solution in co-prime numbers x and y (Sz. Tengely,

[195])97;

here (x, y, m, p) =

(545, 53, 3, 3).

546 (= 2 · 3 · 7 · 13)

• the sixth ideal number (see the number 390);

• the 16th number n such that n! − 1 is prime (see the number 166).

96In that paper, Panaitopol also proves the following three inequalities:

(i) π(x + y) ≤ π(x) + π(y) +

1

22

π(x − y) for all numbers x ≥ y ≥ 2;

(ii) π(x + y) ≤ π(x) + π(y) + π(π(x − y)) for all numbers x ≥ y ≥ 2;

(iii) π(x + y) ≥

π2(

√

x) +

π2(

√

y) for all numbers x ≥ y ≥ 2, as long as x ≥ 3727.

97In

fact, Sz. Tengely proved that the only solutions of the diophantine equation

x2

+

32m

=

2yp

are (x, y, m, p) = (13, 5, 2, 3), (79,5,1,5), (545,53,3,3).