Those Fascinating Numbers 55

197

• the smallest prime number p such that the numbers 2p + 1, 4p + 1, 8p + 1,

10p + 1, 14p + 1 and 16p + 1 are all

composite67;

• the seventh Keith number: a number n = d1d2 . . . dk with k ≥ 2 digits, is

called a Keith number if its associated sequence a1, a2, . . . contains the num-

ber n, where the sequence (aj )j≥1 is defined as follows: the k first terms are

d1, d2, . . . , dk, and the

jth

term, j ≥ k + 1, is the sum of the k preceding terms,

that is nj = nj−k + . . . + nj−1; thus the sequence associated with n = 197 is 1,

9, 7, 17, 33, 57, 107, 197, . . . ; the sequence of Keith numbers begins as follows:

14, 19, 28, 47, 61, 75, 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385, 7647,

7909, . . . ; there exist only 57 Keith numbers

1015;

it is not known if there

are infinitely many such numbers (see M. Keith [116], [117]).

198

• the second composite number n such that

σ(n) + φ(n) + γ(n)

n

is an integer:

20, 198 and 18 486 are the only composite numbers n

109

satisfying this

property.

199

• the prime number which appears the most often as the eighth prime factor of

an integer; if p0 = p0(k) stands for the prime number which appears the most

often as the kth prime factor of an integer, we have the following table68 (here,

by the rank of a prime number, we mean its rank in the increasing sequence of

prime numbers; thus, the rank of a prime number p is equal to π(p)):

67By

this approach, an important step was made in 1823 towards the proof of the first case of

Fermat’s Last Theorem, namely by showing that if the diophantine equation

xp

+

yp

=

zp

has

positive solutions in integers x, y, z, with p prime and not dividing xyz, then p ≥ 197.

68Here

is how one can obtain the values appearing in this table. First, given a positive integer k

and a prime number p0, let λk(p0) be the density of the set of numbers whose kth prime factor is

p0. One can then prove (see J.M. De Koninck & G. Tenenbaum [63]) that

(∗) λk (p0) =

1

p0

pp0

1 −

1

p

Sk−1(p0),

where

Sj (p0) =

m≥1

ω(m)=j

P (m)p0

1

m

(j = 1, 2, . . .).

Computing expression Sk−1(p0) in (∗) is made easier if one uses the recurrence relation (valid for

each fixed prime number p0)

kSk = P1Sk−1 − P2Sk−2 + P3Sk−3 − . . . +

(−1)kPk−1S1

+

(−1)k+1PkS0,

where S0 = 1 and

Pj = Pj (p0) =

pp0

1

(p −

1)j

(j = 1, 2, . . .).

Using these formulas, one easily finds the prime number p0 = p0(k) which appears the most often

as the kth prime factor of an integer; hence, the table.