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A094405 a(1) = 1; a(n) = (sum of previous terms) mod n.

%I A094405 #7 Mar 30 2012 17:38:57
%S A094405 1,1,2,0,4,2,3,5,0,8,4,6,10,4,5,7,11,1,17,11,18,10,15,1,21,11,16,26,
%T A094405 17,27,16,24,7,5,1,29,13,17,25,1,33,15,20,30,5,45,33,7,2,42,22,32,52,
%U A094405 38,8,2,47,23,32,50,25,35,55,31,46,10,3,57,29,41,65,41,64,36,53,11,2,62,26
%N A094405 a(1) = 1; a(n) = (sum of previous terms) mod n.
%C A094405 Theorem. For all values of n>=397, a(n)=97. Proof. Let s(n) denote Sum[a(i), i=1..n-1]. Calculation shows that s(397)=38606=397*97+97. Thus a(397)=397*97+97 mod 397=97. Then s(398)=s(397)+97=398*97+97, giving a(398)=97. A simple inductive argument shows that a(397+k)=97 for all integers k>=0. - _John W. Layman_, Jun 07 2004
%C A094405 Conjecture: For any seed a(1) the sequence "a(n) = (sum of previous terms) mod n" ends with repeating constant. This is true for a(1) = 1,...,941. - _Zak Seidov_, Feb 24 2006
%C A094405 Essentially the same as A066910. [From _R. J. Mathar_, Sep 05 2008]
%e A094405 a(4) = 0 because the previous terms 1, 1, 2 sum to 4 and 4 mod 4 is 0. a(5) = 4 because the previous terms 1, 1, 2, 0 sum to 4 and 4 mod 5 is 4.
%p A094405 L := [1]; s := 1; p := 2; while (nops(L) < 90) do; if 1>0 then; t := s mod p; L := [op(L),t]; s := s+t; p := p+1; fi; od; L;
%K A094405 nonn
%O A094405 1,3
%A A094405 Chuck Seggelin (seqfan(AT)plastereddragon.com), Jun 03 2004