[seqfan] Digit "a" divides the concatenation "bc"

Eric Angelini Eric.Angelini at kntv.be
Sun Oct 18 18:25:06 CEST 2015

Hello SeqFans,

--let's call a, b and c any three consecutive digits
   of S (for instance 1, 2 and 3);
--S was build such that  "a" always divides the
   concatenation "bc" (here, 1 divides 23)

--two remarks:
(1) we accept the triplet (4,0,8), for instance,
    as we accept that < "4" divides "08" >;
(2) we never divide by zero (when "a" = 0 we
    simply ignore it).

Could the integer "100" (or 2000, or  700006,
etc., that is, integers having  2 or more consecutive
zeros) be part of S? Yes, as "1" (or "2" or "7") _do_
divide the concatenation "00".

--last point:
S was started with a(1)=1 and a(2)=2,  then
extended with the smallest integer not yet
present in S and not leading to a contradiction.

S = 1,2,3,4,5,6,54,52,50,51,53,57,56,540,8,

Why is "9" there ("9" is the last term of S computed
here)? Because after "210", the smallest available
integer not leading to a contradiction is 9:
--"2" divides "10"
--"1" divides any concatenation that start with "0x"
-- "0" is skipped
... We thus have a large choice for the integer
following "210" (and "9" will be itself followed by "18",
as "18" is the smallest available integer divisible by "9").

There is an infinite amount of integers that won't enter S.
I guess they form the seq T hereunder:

T=201,203,205,207,209,211,213,215,... ,301,302,
304,305,307,308,310,311,... ,401,402 403,405, etc.


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