Those Fascinating Numbers 333
624 041 002
the smallest number n such that P (n) P (n + 1) . . . P (n + 11):
here 312070501 208047001 156035251 124828201 104023501
89163001 39008813 4079353 48647 3079 2719 431 (see the
number 1 851).
635 318 657
the smallest number which can be written as196 the sum of two co-prime fourth
powers in two distinct ways: 635 318 657 =
594
+
1584
=
1334
+
1344;
the
sequence of numbers satisfying this property begins as follows: 635 318 657,
3 262 811 042, 8 657 437 697, 68 899 596 497, 86 409 838 577, 160 961 094 577,
2 094 447 251 857, 4 231 525 221 377, . . . ; it is known that this sequence is infi-
nite197;
on the other hand, the sequence of numbers with two representations
as the sum of two fifth powers is most likely
finite198.
683 441 871
the first of the smallest ten consecutive numbers at which the Ω(n) function
takes distinct values, namely here the values 6, 8, 1, 11, 7, 5, 4, 3, 2, 10 (see
the number 726).
699 117 024
the 22nd number n such that φ(n) + σ(n) = 3n (see the number 312).
196Euler
was the first to make this observation, that is some 150 years before Ramanujan (see the
number 1 729) informed G.H. Hardy that the number 1 729 was a very interesting number since it
was expressible as the sum of two cubes in two distinct ways. Hardy claims he then asked Ramanujan
if he knew of any number which could be written as the sum of two fourth powers in two distinct
ways, a problem for which Ramanujan had no solution.
197Indeed,
consider the expression x(η, a, b) :=
a7
+
a5b2

2a3b4
+
3ηa2b5
+
ab6.
Checking that
x4(1,
a, b) +
x4(−1,
b, a) =
x4(−1,
a, b) +
x4(1,
b, a), and setting 1 a b, one can then generate
infinitely many such representations. This result is essentially due to Euler, as was pointed out by
Jean-Marc Deshouillers.
198In
fact, no one knows of any number which can be written in two distinct ways as the sum of two
fifth powers. By a heuristic argument, one can show that such a representation is unlikely. Indeed,
let x be a large number. It is clear that the number of numbers n =
a5
+
b5
x is approximately
(x1/5)2
=
x2/5.
Hence, the probability that a number n chosen at random is of the form n =
a5
+
b5
is of the order of
n2/5
n
=
1
n3/5
. Therefore, it follows that
Prob[n =
a5
+
b5
=
c5
+
d5]

1
n3/5
2
=
1
n6/5
.
This is why the expected number of numbers n [r, x] having two distinct representations as the
sum of two fifth powers is of the order of
r≤n≤x
1
n6/5

x
r
1
t6/5
dt =
−5t−1/5
x
r
=
5
r1/5

5
x1/5
5
r1/5
,
a quantity which tends to 0 as r ∞.
Previous Page Next Page