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%I #3 Dec 15 2017 17:36:56
%S 1,1,1,1,1,1,1,1,1,1,3,1,2,3,3,4,3,1,3,1,1,2,3,1,5,3,1,5,3,1,2,6,3,5,
%T 3,1,3,3,3,1,3,1,3,3,1,4,3,1,3,1,3,5,3,1,2,3,4,2,3,1,5,5,1,3,3,2,2,3,
%U 3,1,3,1,3,3,3,2,3,5,3,1,1,5,3,1,2,3,3,5,3,1,4,9,4,3,3,9,2,5,3,1,3,1,2,2,6
%N Number of consecutive integers starting with n needed to sum to a Niven number.
%e a(25)=5 because 25 + 26 + 27 + 28 + 29 = 135 and 135/9 = 15, 5 consecutive
%e integers are needed.
%Y Cf. A005349, A110029, A110031.
%K base,easy,nonn
%O 1,11
%A _Jason Earls_, Jul 08 2005
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