OFFSET
1,1
COMMENTS
For each d>=1, (10^(3*d)-4)/3+10^(2*d) (the concatenation of x+2, x+1 and x where x = (10^d-4)/3) is in the sequence, being divisible by x+1 and x+3. Thus the sequence is infinite.
It appears that a(n) is of the form (10^(3*d)-4)/3+10^(2*d) for n >= 11. - Chai Wah Wu, Jun 19 2019
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..56
EXAMPLE
232231230 is the concatenation of 232, 231 and 230, and is divisible by 231 and 230, so it is in the sequence.
MAPLE
f:= proc(x)
local t1, t2, q, a, b;
t1:= 10^length(x);
t2:= t1*10^length(x+1);
q:= x*(1+t1+t2)+2*t2+t1;
a:= (q/x)::integer;
b:= (q/(x+1))::integer;
if a and b then return q elif not(a) and not(b) then return NULL fi;
if (q/(x+2))::integer then q else NULL fi
end proc:
map(f, [$1..10^8]);
CROSSREFS
KEYWORD
nonn,base
AUTHOR
J. M. Bergot and Robert Israel, Jun 05 2019
STATUS
approved