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]
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
KEYWORD
nonn
AUTHOR
Chuck Seggelin (seqfan(AT)plastereddragon.com), Jun 03 2004
STATUS
approved