106 Jean-Marie De Koninck

779

• the smallest number n such that φ(n) = 6!; if nk stands for the smallest number

n such that φ(n) = k!, then n1 = 1, n2 = 3, n3 = 7, n4 = 35, n5 = 143, n6 =

779, n7 = 5 183, n8 = 40 723, n9 = 364 087, n10 = 3 632 617, n11 = 39 916 801

and n12 = 479 045 521; Erd˝ os observed that equation φ(n) = k! has a solution

n for each number k ≥ 1, while this was less obvious for σ(n) = k! for each

k ≥ 3 (see the number 1 560).

780

• the smallest solution of σ(n) = 3n + 12: this equation has only four solutions

smaller than

109,

namely 780, 2 352, 430 272 and 184 773 312.

782

• the second number n such that σ(n) = σ(n + 13) (see the number 182).

787

• one of the only two prime numbers with three digits (the other is 101) whose

digits are consecutive (see the number 67).

823

• the smallest number n such that φ(n) φ(n + 1) φ(n + 2) φ(n + 3): here

822 408 400 348 (see the numbers 313 and 1 484).

828

• the smallest number n which allows the sum

m≤n

1

φ(m)

to exceed 13 (see the

number 177);

• the smallest number n such that h(n) = 6, where h(n) stands for the number

of divisors d of n such that d + n/d is a perfect square; thus h(1) = 1, h(2) = 0

and h(3) = 2, and obviously, for n 1, h(n) is either 0 or an even positive

integer; if n2k stands for the smallest number n such that h(n) = 2k, then we

have the

following111

table:

2k 2 4 6 8 10 12 14 16 18

n2k 3 48 828 8 820 26 100 417 600 2 114 100 3 009 600 43 864 128

111If

one could prove that for any positive integer , there exists a number n such that h(n) = 2 ,

one would automatically obtain a proof of the existence of an elliptic curve of rank .