%I #3 Mar 31 2012 10:30:58
%S 0,0,0,0,0,0,0,0,0,5,5,10,11,13,17,25,22,25,30,29,32,30,29,33,36,34,
%T 36,42,37,41,37,40,35,37,37,42,38,45,46,46,46,50,44,43,40,34,31,30,25,
%U 25,28,31,31,32,30,26,24,26,30,28,35,35,37,39,48,50,47,50,45,42,36,40,33
%N Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 5.
%C Variation on Angelini's A112395. The sequence cycles at a(28)=42 and the loop has 312 terms. Computed by Gilles Sadowski.
%e a(28)=42 because 2+9 + 3+3 + 3+6 + 3+4 + 3+6 = 42
%Y Cf. A112395, A004207, A065075, A001370.
%K base,easy,nonn
%O 1,10
%A _Alexandre Wajnberg_, Dec 11 2005