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

1, 43, 81271, 3570505, 7025726485, 314435374639, 639872336584027, 60775577624897675065, 2794429652350970000851, 276858360603194024261113585, 600808083611945729624598396925, 28083738921571587634894783049047, 61728002094732427074308383210511683

Offset: 3

Author

Michel Lagneau, May 27 2012

Keywords

• nonn

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.

Crossrefs

Cf. A002383, A055494.

Programs

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:

PARI

a(n)={my(p=prime(n)); (7^p - 6^p - 1)/(1806*p)} \\ Andrew Howroyd, Feb 25 2018