Those Fascinating Numbers 85
410
the smallest solution of
φ(n)
n
=
16
41
; the sequence of numbers satisfying this
equation begins as follows: 410, 820, 1640, 2050, 3280, 4100, 6560, 8200, . . .
411
the smallest number n for which the Moebius function µ takes successively,
starting with n, the values 1,0,1,0,1,0: it is also the smallest number n for
which the Moebius function µ takes successively, starting with n, the values
1,0,1,0,1,0,1 as well as 1,0,1,0,1,0,1,0 (see the number 3 647).
412
the smallest number not divisible by 8 and which is not the sum of four non
zero distinct squares (F.Halter-Koch): see R.K. Guy [101], C20.
417
the smallest number n such that ω(n), ω(n + 1), ω(n + 2), ω(n + 3) are all
distinct, namely in this case with the values 2, 3, 1 and 4: if nk stands for
the first of the k (k 2) smallest consecutive
numbers87
at which the ω(n)
function takes k distinct values, then n2 = 5, n3 = 28, n4 = 417, n5 = 14 322,
n6 = 461 890, n7 = 46 908 264 and n8 = 7 362 724 275 (see the number 726 for
the same matter, but this time with the Ω(n) function).
419
the
100th
number of the form

, where p is prime and β a positive integer; if
we denote by nk the kth prime power, then we have the following table88:
α n10α
1 16
2 419
3 7 517
4 103 511
α n10α
5 1 295 953
6 15 474 787
7 179 390 821
8 2 037 968 761
α n10α
9 22 801 415 981
10 252 096 675 073
11 2 760 723 662 941
12 29 996 212 395 727
(see also the table appearing at number 455).
87One
can establish the size of nk for all k 2. First, it is clear that nk
kk
; indeed, using the
Prime Number Theorem, nk p1p2 . . . pk
e(1+o(1))pk

e(1+o(1))k log k kk.
As for the upper
bound, J.M. De Koninck, J. Friedlander and F. Luca [49] proved that nk =
exp{O(k2 log2
k)}.
88Here is how one can obtain the values appearing in the table. It is clear that, for each k 2,
the number nk satisfies relation (∗)
r
j=1
π(n
1/j
k
) = k, where r = [log nk/ log 2]. Using the Prime
Number Theorem in the form π(x) x/ log x, it follows from (∗) that
nk
log nk
k (as k ∞), so
that nk k log nk k log k. Setting k =
10α
provides a starting point a first approximation of
n10α . Indeed, starting with the approximation n = [k · log k], set s =
∑for
r
j=1
π(n1/j
), and thereafter
as long as a := k s is not equal to 0, replace n by n + a.
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