OFFSET
3,2
COMMENTS
7^p - 6^p - 1 is divisible by 1806p = 6*7*43*p where p prime > 3 (see the proof with the general case).
The sequence is generalizable with the form a(n) = ((k^p - (k-1)^p - 1)) /(k*(k-1)*p*q) where p = prime(n), k integer such that q = k*(k-1) + 1 prime (q = A002383(n) with k = A055494(n)).
k*(k-1)*p*q divides k^p - (k-1)^p - 1, proof :
(1) p divides k^p - (k-1)^p - 1 (Fermat’s theorem)
(2) k*(k-1) divides k^p - (k-1)^p - 1
(3) q = k*(k-1) + 1 divides k^p - (k-1)^p - 1. Suppose k^p - (k-1)^p - 1 ==r (mod q). Then ((k-1)^p)*k^p - ((k-1)^p)*(k-1)^p - (k-1)^p ==r*(k-1)^p (mod q). But the first term is congruent to -1 (mod q), the second term is congruent to k^p (mod q) and the last term is congruent to (k-1)^p (mod q). We obtain r (mod q) = r*(k-1)^p (mod q) => r = 0.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 3..100
Peter Vandendriessche and Hojoo Lee, Problems in elementary number theory, Problem A43.
MAPLE
with(numtheory): for n from 3 to 25 do:p:=ithprime(n):x:=(7^p - 6^p - 1)/(1806*p): printf(`%d, `, x):od:
PROG
(PARI) a(n)={my(p=prime(n)); (7^p - 6^p - 1)/(1806*p)} \\ Andrew Howroyd, Feb 25 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 27 2012
STATUS
approved