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.

123, 234, 345, 456, 8910, 101112, 230231232

Offset: 1

Author

J. M. Bergot and Robert Israel, Jun 03 2019

Keywords

• nonn
• base
• more

Comments

012 is not included because leading 0's are not allowed.

From Charlie Neder, Jun 05 2019: (Start)

If x = 10^k - 2, then x|x+1|x+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 x|x+1|x+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 x|10^k+2, x+1|10^2k-1, and x+2|2*10^2k-10^k are true - presuming x and x+2 are the same length - then x|x+1|x+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

Examples

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

Crossrefs

Subsequence of A001703.

Cf. A308527.

Programs

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]);

Python

for k in range(1,8):
  for x in range(10**(k-1),10**k-2): # 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