16 Jean-Marie De Koninck
the smallest solution of σ(n) = σ(n + 20); the sequence of numbers satisfying
this equation begins as follows: 42, 51, 123, 141, 204, 371, 497, 708, 923, 992,
1034, 1343, 1391, 1484, 1595, 1691, 1826, 3266, 3317, 5015, 5152, 7367, 8003,
9132, 9287, 9494,. . .
43
the fourth prime number p such that 19p−1 1 (mod p2): the only prime
numbers p 232 satisfying this congruence are 3, 7, 13, 43, 137 and 63 061 489
(see Ribenboim [169], p. 347).
44
the second number n such that φ(x) = n has exactly three solutions: the
sequence of
numbers21
satisfying this property begins as follows: 2, 44, 56, 92,
104, 116, 140, . . . ;
the smallest number n which allows the sum
m≤n
ω(m)=2
1
m
to exceed 1; for each
k 2, if we let nk be the smallest integer n which allows the above sum to
exceed k, then the sequence (nk)k≥1 begins as follows: 44, 236, 1 853, 24 692,
627 208, 33 757 004,. . . 22.
45
the smallest number n such that τ (n) τ (n + 1) τ (n + 2): here 6 4 2;
if we denote by nk the smallest number n such that τ (n) τ (n + 1) . . .
τ (n + k), then n1 = 4, n2 = 45, n3 = 80, n4 = n5 = 28 974, n6 = 8 489 103 and
n7 = 80 314 575;
the smallest number n having at least two distinct prime factors and which
is such that p|n =⇒ p + 3|n + 3: the sequence of numbers which satisfy this
property begins as follows: 45, 147, 165, 357, 405, 567, 621, 637, 845, . . . (see
the number 399).
21See
the footnote tied to the number 24.
22One can estimate the size of the number nk with respect to k by using estimate π2(x) := #{n
x : ω(n) = 2} x(log log x)/(log x), which follows essentially from the Prime Number Theorem.
Indeed, by first writing the sum as a Stieltjes integral and then integrating by parts, one obtains
n≤x
ω(n)=2
1
n
=
x
6
1
t
d π2(t) =
π2(t)
t
x
6
+
x
6
π2(t)
t2
dt
x
6
log log t
t log t
dt
(log log x)2
2
.
This is why the number nk satisfies
(log log
nk)2
2
k as k ∞, which means that nk
ee

2k
.
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