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A284761 a(n) = gcd(A279513(n), A279513(n+1)).

%I A284761 #8 Apr 05 2017 08:07:17
%S A284761 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 A284761 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 A284761 1,1,1,1,1,2,2,1,1,1,1,4,2,1,1,1,1,1,3
%N A284761 a(n) = gcd(A279513(n), A279513(n+1)).
%C A284761 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 A284761 If p is prime, then a(p-1) = a(p) = 1.
%C A284761 If p is an odd prime, then a(p^2) = 2.
%C A284761 This sequence contains a multiple of p for any prime p:
%C A284761 - let m = A074792(p)^p-1,
%C A284761 - m is a multiple of p, hence p divides A279513(m),
%C A284761 - m+1 = A074792(p)^p, hence p divides A279513(m+1),
%C A284761 - hence p divides gcd(A279513(m), A279513(m+1)) = a(m).
%C A284761 This sequence contains infinitely many distinct values; see A284821 for these distinct values in order of appearance, and A284822 for the corresponding indexes.
%H A284761 Rémy Sigrist, <a href="/A284761/b284761.txt">Table of n, a(n) for n = 1..10000</a>
%H A284761 Rémy Sigrist, <a href="/A284761/a284761.pdf">Illustration of the first terms</a>
%e A284761 a(8) = gcd(A279513(8), A279513(9)) = gcd(A279513(2^3), A279513(3^2)) = gcd(2*3, 3*2) = 6.
%Y A284761 Cf. A074792, A182318, A284821, A284822.
%K A284761 nonn
%O A284761 1,8
%A A284761 _Rémy Sigrist_, Apr 02 2017