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

1,3

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]

Links

Table of n, a(n) for n=1..79.

Example

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.

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;

Crossrefs

Sequence in context: A351912 A136265 A066910 * A155984 A028609 A107490

Adjacent sequences: A094402 A094403 A094404 * A094406 A094407 A094408

Keyword

nonn

Author

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

Status

approved