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