

A309869


Let b(0)=n, b(1)=n+1, b(2)=n+2, and b(k)=(b(k3)*b(k2)*b(k3)) mod (b(k3)+b(k2)+b(k1)) for k>=3. Continue until either b(k)=0, in which case a(n)=k, or the ordered triple [b(k2),b(k1),b(k)] has appeared before, in which case a(n)=k. If neither of these ever occur, then a(n)=0.


1



3, 50, 3, 3, 10, 3, 3, 17, 3, 3, 11, 3, 3, 5, 3, 3, 6, 3, 3, 5, 3, 3, 120, 3, 3, 6, 3, 3, 121, 3, 3, 5, 3, 3, 13, 3, 3, 1688, 3, 3, 24, 3, 3, 7, 3, 3, 15, 3, 3, 21, 3, 3, 32, 3, 3, 191, 3, 3, 19, 3, 3, 12, 3, 3, 36, 3, 3, 14, 3, 3, 194, 3, 3, 17, 3, 3, 16, 3, 3, 22, 3, 3
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OFFSET

1,1


COMMENTS

If b(k)=0 we get b(k+1)=0, b(k+2)=0, and the sequence ends because 0 mod 0 is undefined. If [b(k2),b(k1),b(k)] has appeared before, the sequence has entered a cycle.
a(n)=3 unless n==2 (mod 3).
Can a(n)=0?


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


FORMULA

a(266*(1)^k+132*k)=5. Conjecture: the only other n for which a(n)=5 are 14 and 32.  Robert Israel, Aug 21 2019


EXAMPLE

a(14)=5 because b(3)=(14*15*16) mod (14+15+16)=30, b(4)=(15*16*30) mod (15+16+30)=2, and b(5)=(16*30*2) mod (16+30+2)=0.
a(8)=17 because (b(15),b(16),b(17))=(21,18,18)=(b(11),b(12),b(13)).


MAPLE

f:= proc(m) local a, b, c, t, n, s, S;
a:= m; b:= m+1; c:= m+2;
S:= {[m, m+1, m+2]};
for n from 3 to 10^6 do
t:= (a*b*c) mod (a+b+c);
if t = 0 then return n fi;
a:= b; b:= c; c:= t;
s:= [a, b, c];
if member(s, S) then return n fi;
S:= S union {s};
od;
FAIL
end proc:
map(f, [$1..100]);


CROSSREFS

Sequence in context: A063893 A291707 A145572 * A012857 A246221 A249249
Adjacent sequences: A309866 A309867 A309868 * A309870 A309871 A309872


KEYWORD

sign


AUTHOR

J. M. Bergot and Robert Israel, Aug 20 2019


STATUS

approved



