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Group the natural numbers so that every 2n-th group product is divisible by the single number in the next group. (1), (2,3,4,5), (6), (7,8,9,10,11), (12), (13,14,15,16,17,18,19),(20), (21,22,23,24,25,26,27),(28),... Sequence contains the number of terms in the 2n-th group.
3

%I #5 Nov 04 2018 13:13:30

%S 4,5,7,7,7,8,10,13,9,9,13,7,7,11,7,13,13,11,11,15,11,13,15,13,15,16,

%T 14,11,11,13,9,17,17,19,15,15,13,17,14,8,15,18,20,15,17,17,17,9,19,11,

%U 20,20,13,19,19,15,15,9,14,14,17,16,15,18,13,20,14,14,14,17,15,15,15,21,13

%N Group the natural numbers so that every 2n-th group product is divisible by the single number in the next group. (1), (2,3,4,5), (6), (7,8,9,10,11), (12), (13,14,15,16,17,18,19),(20), (21,22,23,24,25,26,27),(28),... Sequence contains the number of terms in the 2n-th group.

%C a(n) = A109895(x+1)-A109895(x)-1 - Simon Nickerson (simonn(AT)maths.bham.ac.uk), Jul 15 2005

%Y Cf. A079759, A109895, A109897.

%K nonn

%O 1,1

%A _Amarnath Murthy_, Jul 13 2005

%E More terms from Simon Nickerson (simonn(AT)maths.bham.ac.uk), Jul 15 2005