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A373209
Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.
1
68, 112, 128, 162, 200, 212, 252, 294, 318, 336, 338, 372, 448, 450, 498, 502, 542, 578, 592, 598, 612, 648, 672, 678, 708, 752, 762, 808, 812, 852, 878, 888, 938, 952, 992, 996, 1012, 1038, 1098, 1102, 1116, 1122, 1188, 1202, 1212, 1248, 1258, 1328, 1362, 1380
OFFSET
1,1
COMMENTS
Among the first 10000 terms (from a(1) = 68 through a(10000) = 697578), k^2 - 1 and k^2 + 1 are each the product of three distinct primes, except for
125 terms for which k^2 + 1 = 5^3 times a prime
6 terms for which k^2 + 1 = 13^3 times a prime
1 terms for which k^2 + 1 = 17^3 times a prime
1 terms for which k^2 + 1 = 29^3 times a prime, and
4 terms for which k^2 - 1 = p^3 * (p^3 +/- 2) (with p = 19, 29, 37, 83, respectively).
The first term for which both k^2 - 1 and k^2 + 1 are of the form p^3 * q is k = 41457661182: k^2 - 1 = 3461^3 * 41457661183, while k^2 + 1 = 5^3 * 13749901365452077097.
FORMULA
{ k : tau(k^2 - 1) = tau(k^2 + 1) = 8}, where tau() is the number of divisors function, A000005.
EXAMPLE
68 is a term: both 68^2 - 1 = 4623 = 3 * 23 * 67 and 68^2 + 1 = 4625 = 5^3 * 37 have 8 divisors.
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jun 21 2024
STATUS
approved