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Greatest common divisors of consecutive terms of A091780.
0

%I #2 Mar 30 2012 17:37:53

%S 3,3,3,3,5,5,5,5,5,3,7,7,7,7,5,5,9,9,3,3,11,11,11,11,5,5,13,13,13,13,

%T 3,3,17,17,17,17,5,3,19,19,19,19,5,7,7,21,3,3,23,23,23,23,5,5,3,27,3,

%U 11,7,7,13,3,29,29,29,29,5,3,31,31,31,31,5,5,3,9,3,3,37,37,37,37,5,11,11,11

%N Greatest common divisors of consecutive terms of A091780.

%C It appears that each prime p>2 first appears in this sequence in a run of length 4, except for prime(3)=5, which first occurs in a run of length 5. This has been confirmed through prime(135)=761.

%e The fifth and sixth terms of A091780 are 15 and 10, respectively, so a(5)=5.

%Y Cf. A091780.

%K nonn

%O 0,1

%A _John W. Layman_, Nov 21 2005