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%I #4 Dec 05 2013 19:55:34
%S 1,2,3,5,4,6,7,8,15,9,10,11,12,19,13,23,14,16,17,18,27,20,30,31,24,21,
%T 22,36,25,26,37,28,47,39,29,33,32,51,44,34,35,56,38,40,41,42,63,48
%N a(1) = 1; for n>1, a(n) is the smallest previously unused natural number such that a(n) divides the sum of next a(n) terms.
%C 4 comes after 5 hence the next terms after 5 are so chosen that the property holds for 4 as well.
%C The next term a(49) does not exist, so there are no more terms. This is because a(25)=24 and a(27)=22 both must divide a(49). Assuming that a(49)=x, this leads to 867+x=24a and 824+x=22b for some integers a and b. Subtracting, we get 43=24a-22b, which is a contradiction. - _John W. Layman_, Jul 29 2003
%e 5 divides the sum (4 + 6 +7 +8 + 15) of the next five terms and 4 does so for (6+7+8+15).
%Y Cf. A074137, A074138.
%K fini,full,nonn
%O 1,2
%A _Amarnath Murthy_, Aug 28 2002
%E More terms from _John W. Layman_, Jul 29 2003
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