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

.