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A212738 a(n) = (7^p - 6^p - 1)/(1806p) where p is the n-th prime.

%I A212738 #18 Feb 25 2018 22:48:42
%S A212738 1,43,81271,3570505,7025726485,314435374639,639872336584027,
%T A212738 60775577624897675065,2794429652350970000851,
%U A212738 276858360603194024261113585,600808083611945729624598396925,28083738921571587634894783049047,61728002094732427074308383210511683
%N A212738 a(n) = (7^p - 6^p -  1)/(1806p) where p is the n-th prime.
%C A212738 7^p - 6^p - 1 is divisible by 1806p = 6*7*43*p where p prime > 3 (see the proof with the general case).
%C A212738 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)).
%C A212738 k*(k-1)*p*q divides k^p - (k-1)^p - 1, proof :
%C A212738 (1)   p divides k^p - (k-1)^p - 1 (Fermat’s theorem)
%C A212738 (2)   k*(k-1) divides k^p - (k-1)^p - 1
%C A212738 (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.
%H A212738 Andrew Howroyd, <a href="/A212738/b212738.txt">Table of n, a(n) for n = 3..100</a>
%H A212738 Peter Vandendriessche and Hojoo Lee, <a href="http://www.scribd.com/doc/24487088/Hojoo-Lee-Peter-Vandendriessche-Problems-in-Elementary-Number-Theory">Problems in elementary number theory</a>, Problem A43.
%p A212738 with(numtheory): for n from 3 to 25 do:p:=ithprime(n):x:=(7^p - 6^p - 1)/(1806*p): printf(`%d, `, x):od:
%o A212738 (PARI) a(n)={my(p=prime(n)); (7^p - 6^p - 1)/(1806*p)} \\ _Andrew Howroyd_, Feb 25 2018
%Y A212738 Cf. A002383, A055494.
%K A212738 nonn
%O A212738 3,2
%A A212738 _Michel Lagneau_, May 27 2012