42 Jean-Marie De Koninck

133 (= 7 · 19)

• the

100th

composite number; if we denote by nk the

kth

composite number,

then we have the

following53

table:

α n10α

1 18

2 133

3 1 197

4 11 374

5 110 487

6 1 084 605

7 10 708 555

α n10α

8 106 091 745

9 1 053 422 339

10 10 475 688 327

11 104 287 176 419

12 1 039 019 056 246

13 10 358 018 863 853

14 103 307 491 450 820

• the smallest solution of τ (n + 11) = τ (n) + 11; the sequence of numbers satis-

fying this equation begins as follows: 133, 2489, 3958, 4613, 5765, 8453, 9593,

13445, 16373, 21598, . . .

135

• the smallest number n such that n and n + 1 each have four prime factors

counting their multiplicity: 135 = 33 · 5 and 136 = 23 · 17; if we denote by

nk the smallest number n such that n and n + 1 each have k prime factors

counting their multiplicity (that is such that Ω(n) = Ω(n+1) = k), then n1 = 2,

n2 = 9, n3 = 27, n4 = 135, n5 = 944, n6 = 5 264, n7 = 29 888, n8 = 50 624,

n9 = 203 391, n10 = 3 290 624, n11 = 6 082 047 and n12 = 32 535 999 (see the

number 230 for the similar question with the ω(n) function);

• the smallest solution of σ(n) = σ(n + 23); the sequence of numbers satisfying

this equation begins as follows: 135, 231, 322, 682, 778, 1222, 1726, 1845, 5026,

22011, . . . ;

• the smallest odd number n 1 such that γ(n)|σ(n); the sequence of numbers

satisfying this property begins as follows: 135, 891, 1521, 3375, 5733, 10935,

11907, 41067, 43875, . . . ;

• the second solution of γ(n + 1) − γ(n) = 19 (see the number 98).

53We

obtained the values appearing in this table in the following manner. It is clear that, for

each k ≥ 2, the number nk satisfies relation (∗) nk = 1 + π(nk) + k. Using the Prime Number

Theorem in the form π(x) ∼ x/ log x+x/(log

x)2,

it follows from (∗) that nk ∼

nk

log nk

+

nk

log2

nk

+ k

(as k → ∞), so that nk 1 −

1

log nk

−

1

log2

nk

∼ k. It follows in particular that log nk ∼ log k.

Combining these last two estimates, one obtains nk ∼ k/ 1 −

1

log k

−

1

log2

k

. Setting k =

10α

provides a starting point for the first approximation of n10α . Indeed, using the approximation

n = k/ 1 −

1

log k

−

1

log2 k

and setting s = s(n) = 1 + π(n) + k − n, then, as long as a := s is not

equal to 0, one replaces n by n + a.