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 .
Previous Page Next Page