A284761 - OEIS

%I #8 Apr 05 2017 08:07:17

%S 1,1,1,1,1,1,1,6,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,3,1,1,1,1,2,2,1,1,2,2,1,1,1,1,1,1,3,1,1,1,1,1,2,6,1,1,1,1,1,

%U 1,1,1,1,1,2,2,1,1,1,1,4,2,1,1,1,1,1,3

%N a(n) = gcd(A279513(n), A279513(n+1)).

%C Two consecutive numbers, say n and n+1, cannot share a prime factor (gcd(n, n+1)=1). However, their prime tower factorizations can share some prime numbers; this is the case iff a(n)>1 (see A182318 for the definition of the prime tower factorization of a number).

%C If p is prime, then a(p-1) = a(p) = 1.

%C If p is an odd prime, then a(p^2) = 2.

%C This sequence contains a multiple of p for any prime p:

%C - let m = A074792(p)^p-1,

%C - m is a multiple of p, hence p divides A279513(m),

%C - m+1 = A074792(p)^p, hence p divides A279513(m+1),

%C - hence p divides gcd(A279513(m), A279513(m+1)) = a(m).

%C This sequence contains infinitely many distinct values; see A284821 for these distinct values in order of appearance, and A284822 for the corresponding indexes.

%H Rémy Sigrist, <a href="/A284761/b284761.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A284761/a284761.pdf">Illustration of the first terms</a>

%e a(8) = gcd(A279513(8), A279513(9)) = gcd(A279513(2^3), A279513(3^2)) = gcd(2*3, 3*2) = 6.

%Y Cf. A074792, A182318, A284821, A284822.

%K nonn

%O 1,8

%A _Rémy Sigrist_, Apr 02 2017