cp's OEIS Frontend

Related to 46th IMO 2005 Problem 4, which asks to find all positive integers k such that gcd(2^n + 3^n + 6^n - 1, k) = 1 for all n. The answer is k = 1. The main part of the solution is to show that for all primes p >= 5 satisfies p divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1. By Fermat's little theorem, 2^(p-1) == 3^(p-1) == 6^(p-1) == 1 (mod p), so 2^(p-2) + 3^(p-2) + 6^(p-2) - 1 == 2^(-1) + 3^(-1) + 6^(-1) - 1 == 0 (mod p). Here a^(-1) is the multiplicative inverse of a modulo p.

All terms are multiples of 10 but not divisible by 4. If prime(n) !== 3 (mod 8) then a(n) is divisible by 50; if n > 3 and prime(n) == 1 (mod 4) then a(n) is divisible by 250; if prime(n) == 1, 13, 17 (mod 20) then a(n) is divisible by 1250.

Note that a(1) = 1 is also an integer, while a(2) = 10/3 is not.

The H function is 0 for the first three primes. The term 61 comes from Schwindt's paper. The terms 1680023 and 7308036881 come from Dobson's paper.

Let q_2 = (2^(p-1) - 1)/p and q_3 = (3^(p-1) - 1)/p. Then, as proved by Emma Lehmer, H(floor(p/6)) == -2*q_2 - (3/2)*q_3 (mod p) when p > 3. This congruence provides an efficient means of detecting when H(floor(p/6)) vanishes mod p. - John Blythe Dobson, Mar 01 2014

Also {2, 3} union {primes p : p^2 divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1}. Except for the term 3, p is a term of this sequence if and only if p^2 is in A318761. There are no more terms up to 7*10^10. - Jianing Song, Dec 26 2018

More generally (see Lehmer's paper, p. 352, eq. 13), if p^2 divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1, then it also divides 2^(k(p-1)-1) + 3^(k(p-1)-1) + 6^(k(p-1)-1) - 1 for any natural number k. Also, since H(floor(p/6)) == -2*q_2 - (3/2)*q_3 == -q_4 - (1/2)q_27 == -(1/2)(q_16 + q_27) == -(1/2)q_432 (mod p), the terms of this sequence greater than 3 coincide with the values of p that divide q_432, and can be found in Richard Fischer's list of vanishing Fermat quotients, which extended to 1.31*10^14 at the last revision of 19 December 2020. - John Blythe Dobson, Jan 02 2021