98 Jean-Marie De Koninck

629

• the smallest number n satisfying φ(n) = 4φ(n + 1); the sequence of numbers

satisfying this property begins as follows: 629, 1469, 85139, 100889, 139859,

154979, 168149, 304079, 396899, 838199, 1107413, 1323449, 1465463, 2088839,

2160899, 2505879, 2684879, 2693249, 2800181, 3404609, . . . ; if nk stands for

the smallest number n such that φ(n) = kφ(n + 1), then n1 = 1, n2 = 5,

n3 = 119, n4 = 629 and n5 = 17 907 119 (for the analogue problem with

equation φ(n + 1) = kφ(n) for an arbitrary number k, see the number 1 260).

641

• the smallest prime factor of the Fermat101 number F5 =

225

+1 = 641·6 700 417;

• the smallest number n such that

φ9(n)

= 2, where

φ9(n)

stands for the ninth

iteration of the φ function (see the number 137).

644

• the smallest number n such that ω(n) = ω(n + 1) = ω(n + 2) = 3: indeed,

644 =

22

· 7 · 23, 645 = 3 · 5 · 43, 646 = 2 · 17 · 19; the sequence of numbers

satisfying this property begins as follows: 644, 740, 804, 986, 1034, 1064, 1104,

1220, . . . ; see the number 37 960 to discover the smallest number n such that

ω(n) = ω(n + 1) = . . . = ω(n + k − 1) = for any given k ≥ 2 and ≥ 2.

645

• the largest number n such that the sum of the numbers ≤ n is equal to the

sum of the first m squares for a certain number m: in other words, 645 is the

largest number n such that

(∗) 1 + 2 + 3 + . . . + n =

12

+

22

+

32

+ . . . +

m2

for a certain number m;

the only solutions of (∗) are (n, m) = (1, 1), (10,5), (13,6) and (645,85); see

R. Finkelstein & H. London [81] as well as the number 1 291 and its footnote;

• the smallest solution of σ2(n) = σ2(n + 6): the sequence of numbers satisfy-

ing this equation begins as follows: 645, 3 237, 2 842 557, 4 086 717, 6 690 813,

44 819 997, 59 160 021, . . .

102

101This

factorization was obtained by Euler, who had noticed that any prime factor of a Fermat

number Fk is necessarily of the form r · 2k+2 + 1 for a certain positive integer r, here with r = 5.

102One

can prove that if Hypothesis H is true (see page xvii for its statement), this equation has

infinitely many solutions; in fact, it is proved in J.M. De Koninck [45] that if Hypothesis H is true,

then equation σ2(n) = σ2(n + 2) has infinitely many solutions and the same holds for equation

σ2(n) = σ2(n + ), where ≥ 4 is an even number.