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.
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