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a(n) = n + (sum of preceding terms) mod n; a(0) = 0.
0

%I #3 Mar 30 2012 18:51:00

%S 0,1,3,4,4,7,7,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,

%T 48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,

%U 94,96,98,100,102,104,106,108,110,112,114,116,118,120,122,124,126,128,130

%N a(n) = n + (sum of preceding terms) mod n; a(0) = 0.

%C a(n) = 2*(n-1) for n > 6.

%Y A000217.

%K nonn

%O 0,3

%A _Reinhard Zumkeller_, Jan 09 2009