254 Jean-Marie De Koninck
324 900
the
12th
number n such that σ(n) and σ2(n) have the same prime factors,
namely the prime numbers 3, 7, 13, 31 and 127; indeed,
σ(n) = 3 · 7 · 13 · 31 · 127,
σ2(n) =
33
·
76
· 13 · 31 · 127;
see the number 180.
326 981 (= 79 · 4139)
the value of 9! 8! + 7! 6! + 5! 4! + 3! 2! + 1!; setting ak := k! (k
1)! + (k 2)! + . . . +
(−1)k2!
+
(−1)k+11!,
then a1 = a2 = 1, a3 = 5, a4 = 19,
a5 = 101, a6 = 619, a7 = 4 421, a8 = 35 899 and a9 = 326 981.
331 777
the seventh prime number of the form
n4
+1, here with n = 24 (see the number
1 297).
332 928
the seventh powerful number n such that n + 1 is also powerful: here 332 928 =
27
·
32
·
172
and 332 929 =
5772
(see the number 288);
the seventh solution w + s to the aligned houses problem (see the number 35).
332 929
the eight number n such that the binomial coefficient
(
n
2
)
is a perfect square:
here
(
332 929
2
)
=
2354162
(see the number 289); it is easy to prove that this
sequence of numbers is
infinite177.
333 667
the largest prime factor of the number 12345678987654321 =
1111111112,
whose complete factorization is given by
12345678987654321 =
34
·
372
·
3336672,
a perfect
square178.
177Indeed one only needs to prove that equation
n(n−1)
2
= 2 has infinitely many solutions. But
this equation is equivalent to the quadratic equation
n2
n 2
2
= 0, whose positive solutions are
given by n =
1+

1+8
2
2
; but then such a solution n is an integer if and only if 1 + 8
2
=
x2
for
some integer x. Now this equation can be written in the form x2 8 2 = 1, that is a Fermat-Pell
equation which, as is well known, has infinitely many solutions.
178It is interesting to observe that when the number 12345678987654321 is placed after itself, one
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