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).
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