88 Jean-Marie De Koninck

462

• the

12th

number n such that n ·

2n

− 1 is prime (see the number 115);

• the smallest number which cannot be written as the sum of eight non zero

distinct squares (R.K. Guy [101], C20).

463

• the first term of the smallest sequence of five consecutive prime numbers all of

the form 4n + 3 (as well as six or seven consecutive prime numbers all of the

form 4n + 3); if we denote by qk the first term of the smallest sequence of k

consecutive prime numbers all of the form 4n + 3, we have the following table:

k qk

1 3

2 3

3 199

4 199

5 463

k qk

6 463

7 463

8 36 551

9 39 607

10 183 091

k qk

11 241 603

12 241 603

13 241 603

14 9 177 431

15 9 177 431

k qk

16 95 949 311

17 105 639 091

18 341 118 307

19 727 334 879

20 727 334 879

(see the number 2 593 for the similar question with 4n + 1);

• the prime number which allows one to write the number 6 as the difference

of two powerful numbers: 6 =

54

·

73

−

4632

= 214 375 − 214 369, a repre-

sentation discovered by W. Narkiewicz and that S.W. Golomb thought to be

impossible; Mollin & Walsh [141] proved that each integer k 0 has infinitely

many representations as the difference of two powerful numbers.

464

• the third solution of σ(n) = 2n + 2: the list of solutions of this equation begins

as follows: 20, 104, 464, 650, 1 952, 130 304, 522 752, 8 382 464,. . .

91

467

• the prime number which appears the most often as the tenth prime factor of

an integer (see the number 199).

469

• the

15th

number n such that n! − 1 is prime (see the number 166).

91It is easy to show that each number n = 2α·p, where α is a positive integer such that p = 2α+1−3

is prime, is a solution of σ(n) = 2n + 2: it is the case when α = 2, 3, 4, 5, 8, 9, 11, 13, 19, 21, 23, 28, 93;

the solutions corresponding to α = 2, 3, 4, 5, 8, 9, 11 are included in the above list.