Those Fascinating Numbers 395
2106(2107
1)
the
11th
perfect number, a 65 digit number.
69 113 789 582 492 712 943 486 800 506 462 734 562 847 413 501 952 000
000 000 000 001
the largest known prime of the form 1! · 2! · 3! · . . . · r! + 1, here with r = 14 (see
the number 125 411 328 001).
52! + 52 + 1
the largest known prime of the form n! + n + 1, in this case a 68 digit number;
the only known numbers n such that n! + n + 1 is prime are n = 2, 4, 6, 10 and
52 (see the number 52).
37 032 592 805 942 775 592 027 297 064 629 098 681 015 432 812 873 314
981 288 646 788 880 382 401
the largest prime factor of 7979 + 1, whose complete factorization213 is given
by
7979
+ 1 =
24
· 5 · 34919 · 188021 · 45780868646549
·918150595356645610443476284621975471619075398615249· P74.
6 086 555 670 238 378 989 670 371 734 243 169 622 657 830 773 351 885
970 528 324 860 512 791 691 264
the largest known sublime number: this number is sublime because n =
2126(261−
1)(231

1)(219

1)(27

1)(25

1)(23
1), and therefore one can easily check
that τ (n) =
26(27
1) and σ(n) =
2126(2127
1) are both perfect: the only
other known sublime number is 12.
2126(2127 1)
the
12th
perfect number, a 77 digit number.
213To
obtain this factorization, one can use a technique that allows to factor numbers of the form
nn
+ 1 when n 3 (mod 4), which is indeed the case for n = 79. Indeed, using Mathematica
and the feature Factor[((n x2)n + 1)/(n x2 + 1)], one obtains two polynomials (in x) both of
degree n 1. Then, setting x = 1, two non trivial factors of
nn
+ 1 appear. In the case n = 79, the
second of these factors is precisely the largest prime factor of 7979 + 1. When n 2 (mod 4), the
same approach works, but this time by writing Factor[((n
x2)n
+
1)/(n2

x4
+ 1)], in which case
one gets two polynomials of degree n 2.
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