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

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, 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, 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

Offset: 1

Author

Chuck Seggelin (seqfan(AT)plastereddragon.com), Jun 03 2004

Keywords

• nonn

Comments

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

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

Essentially the same as A066910. [From R. J. Mathar, Sep 05 2008]

Examples

			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.
		

Programs

Maple

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;