62 Jean-Marie De Koninck
239
one of the two numbers (the other one is 23) which cannot be written as the
sum of eight cubes (in non negative integers) or less: 239 = 2 · 43 +4 · 33 +3 · 13;
the largest number n such that P (n2 + 1) 17, where P (m) stands for the
largest prime factor of m: Maurice Mignotte proved that P (n2 + 1) 17 if
n 240, while 2392 + 1 = 2 · 134 (R.K. Guy [101], A1); see the number 137;
the only solution n 1 of the diophantine equation n2 + 1 = 2m4: in 1942,
W. Ljunggren proved that the only two positive integer solutions of n2+1 = 2m4
are (n, m) = (1, 1) and (239,13) (R.K. Guy [101], D6);
240
the 12th highly composite number (see the number 180);
the smallest number m for which equation σ(x) = m has exactly seven solutions,
namely 114, 135, 158, 177, 203, 209 and 239;
the fifth number n 1 such that φ(σ(n)) = n (see the number 128).
241
the smallest number n such that τ (n) τ (n + 1) . . . τ (n + 4) (as well as
the smallest such that τ (n) τ (n + 1) . . . τ (n + 5)): here 2 6 6 6
6 8; if we denote by nk the smallest number n such that τ (n) τ (n + 1)
. . . τ (n + k 1), then we have the following table:
k 2 3 4 5 6 7 8
nk 9 13 13 241 241 12 853 12 853
k 9 10 11 12
nk 234 613 376 741 78 312 721 125 938 261
242
the smallest number n such that n, n + 1, n + 2 and n + 3 are all divisible by
a square 1: here 242 = 2 · 112, 243 = 35, 244 = 22 · 61 and 245 = 5 · 72; if
we denote by nk the smallest number n such72 that µ(n) = µ(n + 1) = . . . =
µ(n + k 1) = 0, then n2 = 8, n3 = 48, n4 = 242, n5 = 844, n6 = 22 020,
n7 = 217 070, n8 = 1 092 747, n9 = 8 870 024, n10 = n11 = 221 167 422 and
n12 = 47 255 689 915; if we denote by mk, the smallest number m such that
72It
follows from the Chinese Remainder Theorem that nk exists for each k 2 and moreover
that nk
k
i=1
pi
2
; similarly, mk, exists for each k 2 and 2, and we also have that
mk,
k
i=1
pi .
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