OFFSET
1,3
COMMENTS
Conjecture (seems provable): More generally let a and b(1) be integers. If b(n+1) = b(n) + b(n) (mod(n+a)) there is an integer x(a,b(1)) such that b(n+1) = b(n) + x(a,b(1)) for n sufficiently large. We have x(0,1) = x(1,1) = x(2,1) = 97, x(3,1) = 1, x(4,1) = 3, x(5,1) = 3, x(6,1) = 6, ..., x(97,1) = 43, x(0,11) = 2, etc. - Benoit Cloitre, Aug 20 2002
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..1000
EXAMPLE
a(397) = 38606 = 2*97*199 = (2*199)*97 = 398*97 = (397+1)*97; a(397) mod 397 = (397*97 + 97) mod 397 = 97, a(398) = a(397) + a(397) mod 397 = (397+1)*97 + 97 = (398+1)*97, etc.: a(n+1) = a(n) + 97 for n >= 397.
MATHEMATICA
s=1; lst={s}; Do[s+=Mod[s, n]; AppendTo[lst, s], {n, 1, 6!, 1}]; lst (* Vladimir Joseph Stephan Orlovsky, Nov 07 2008 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Aug 19 2002
STATUS
approved