Those Fascinating Numbers 209
30 618
possibly the largest number n such that n(n + 1)(n + 2) . . . (n + 6) and (n +
1)(n + 2) . . . (n + 7) have the same prime factors: here
30618 · 30619 · . . . · 30624 =
29
·
39
· 5 · 7 · 11 · 29 · 59 · 61
·67 · 113 · 173 · 251 · 271 · 457 · 1531,
30619 · 30620 · . . . · 30625 =
28
·
32
·
55
·
72
· 11 · 29 · 59 · 61
·67 · 113 · 173 · 251 · 271 · 457 · 1531;
if nk, for k 2, stands for the largest number n such that n(n + 1)(n +
2) . . . (n + k 1) and (n + 1)(n + 2) . . . (n + k) have the same prime factors,
then the conjectured values of the first nk’s (assuming the abc
Conjecture160)
are n2 = 2, n3 = 24, n4 = 32, n5 = 400, n6 = 480, n7 = 30 618, n8 = 34 992,
n9 = 39 366, n10 = 43 740 and n11 = 107 800.
30 693
the ninth number which is not a palindrome, but whose square is a palindrome
(see the number 26).
30 784
the tenth number whose square can be written as the sum of three fourth
powers: 30 7842 = 964 + 1204 + 1604 (see the number 481).
31 469
the smallest prime number which is preceded by 71 composite numbers; indeed,
there are no primes between 31 397 and 31 469.
31 907
the smallest prime number p such that p + 50 is prime and such that each
number between p and p + 50 is composite (see the number 139).
160Indeed, it is easy to show that if the abc Conjecture is true, then each number nk is well defined.
Indeed, let k 2 be fixed. First observe that P (n) k, since otherwise p|n for some prime p k,
in which case p cannot divide any of the numbers n + i for i = 1, 2, . . . , k, thereby contradicting the
fact that n(n + 1) . . . (n + k 1) and (n + 1) . . . (n + k) have the same prime divisors. By the same
argument, we also have P (n + k) k. Thus, applying the abc Conjecture (with a = n, b = k and
c = n + k), we have
n + k γ(n · k · (n +
k))1+ε
p≤k
p
1+ε
,
an inequality which cannot hold if n is sufficiently large.
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