Those Fascinating Numbers 55
• the smallest prime number p such that the numbers 2p + 1, 4p + 1, 8p + 1,
10p + 1, 14p + 1 and 16p + 1 are all
• 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
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
it is not known if there
are infinitely many such numbers (see M. Keith , ).
• the second composite number n such that
σ(n) + φ(n) + γ(n)
is an integer:
20, 198 and 18 486 are the only composite numbers n
• 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)):
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
positive solutions in integers x, y, z, with p prime and not dividing xyz, then p ≥ 197.
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 ) that
(∗) λk (p0) =
Sj (p0) =
(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 − . . . +
where S0 = 1 and
Pj = Pj (p0) =
(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.