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 79 79 + 1 = 2 4 · 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. 213 To 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.
Previous Page Next Page