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.
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