

A306643


Numbers that, for some x, are the concatenation of x, x+1 and x+2 and are divisible by at least two of x, x+1 and x+2.


1




OFFSET

1,1


COMMENTS

012 is not included because leading 0's are not allowed.
If x = 10^k  2, then xx+1x+2 (with  denoting concatenation) will be congruent to 22 modulo x, 9 modulo x+1, and 0 modulo x+2.
If x = 10^k  1, then xx+1x+2 will be congruent to 12 modulo x and 1 modulo x+1.
Therefore, the only term such that x and x+2 have different lengths is 8910.
By reducing modulo x + {0,1,2} it can be shown that if at least two of x10^k+2, x+110^2k1, and x+22*10^2k10^k are true  presuming x and x+2 are the same length  then xx+1x+2 is in this sequence. No further terms corresponding to x < 10^18. (End)
No further terms corresponding to x < 10^50.  Chai Wah Wu, Jun 19 2019


LINKS



EXAMPLE

230231232 is the concatenation of 230, 231 and 232 and is divisible by 231 and 232.


MAPLE

cat3:= proc(x)
local t;
t:= 10^length(x+2);
x*(1 + t*(1+10^length(x+1)))+t+2
end proc:
f:= proc(x) local q, a, b;
q:= cat3(x);
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..1000]);


PROG

(Python)
for k in range(1, 8):
..for x in range(10**(k1), 10**k2): # will not find 8910
....if sum([not (10**k+2)%x, not (10**(2*k)1)%(x+1), \
....not (2*10**(2*k)+10**k)%(x+2)]) >= 2:
......print(str(x)+str(x+1)+str(x+2)) # Charlie Neder, Jun 05 2019


CROSSREFS



KEYWORD

nonn,base,more


AUTHOR



STATUS

approved



