Re A094405: John W. Layman's Theorem & Conjection

[zak seidov]

Dear SeqFans,

1. My conjection:

For any seed a(1) sequence 
"a(n) = (sum of previous terms) mod n"
ends with repeating constant.

2. I checked it for a(1) = 1,3,5,..,941.

2a. Seq's with a(1) = 2m-1 & a(1) = 2m merge
starting with a(3) hence no need to consider 
cases of  even a(1)'s.
 
3. Some examples:

3a. Sequences with a(1) =
{9,33,37,39,43,45,47,49,53,55,59,93,97,99,101}
all have a(n>1241) = 316.

3b. Sequences with a(1)=
{449, 451, 455, 457, 471, 473, 475, 479}
all have a(n>19207) = 4788 (record values1).

3c. Sequences with a(1) =
 {1, 51, 103, 107, 109, 129, 133, 137, 165}
(* including original A094405 *)
all have a(n>397) = 97.

Any ideas for proof/check/contra-example?
Zak 

%S A094405
1,1,2,0,4,2,3,5,0,8,4,6,10,4,5,7,11,1,17,11,18,10,15,1,21,11,16,26,17,
%T A094405
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,
%U A094405
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
(layman(AT)math.vt.edu), Jun 07 2004
%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;
%Y A094405 Sequence in context: A112824 A001100
A066910 this_sequence A028609 A107490 A079534
%Y A094405 Adjacent sequences: A094402 A094403 A094404
this_sequence A094406 A094407 A094408
%K A094405 nonn
%O A094405 1,3
%A A094405 Chuck Seggelin
(seqfan(AT)plastereddragon.com), Jun 03 2004



__________________________________________________
Do You Yahoo!?
Tired of spam?  Yahoo! Mail has the best spam protection around 
http://mail.yahoo.com