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Greatest common divisor of consecutive terms in A227113.
3

%I #5 Jul 05 2013 07:58:49

%S 1,2,1,3,1,5,1,7,2,4,3,3,1,11,1,13,2,2,1,17,1,19,2,4,3,3,1,23,1,5,2,4,

%T 1,29,1,31,1,3,1,5,1,37,1,3,1,41,1,43,2,4,3,5,1,47,1,7,1,3,2,4,1,53,1,

%U 5,1,3,1,59,1,61,2,2,1,67,2,2,1,3,1,71,1

%N Greatest common divisor of consecutive terms in A227113.

%C a(n) = GCD(A227113(n),A227113(n+1));

%C a(2*n) > 1 by definition of A227113;

%C 9 is the smallest odd number m with a(m) > 1: a(9) = GCD(A227113(9),A227113(10)) = GCD(14,8) = 2;

%C a(A227289(n)) = n and a(m) <> n for m < A227289(n).

%H Reinhard Zumkeller, <a href="/A227288/b227288.txt">Table of n, a(n) for n = 1..10000</a>

%o (Haskell)

%o a227288 n = a227288_list !! (n-1)

%o a227288_list = zipWith gcd (tail a227113_list) a227113_list

%K nonn

%O 1,2

%A _Reinhard Zumkeller_, Jul 05 2013