Additions using 0 (zero)

Today is self-description time again!

We have decided here that the digit 0 has two meanings:

a) zero

b) add the number to my left to the number to my right.

X = 109, 108, 208, 107, 1019, 206, 307, 106, 406, 1018, 2018, 105, 1029, 205, 505, 204, 1039, 303, 604, 1017, 3017, 2016, 703, 104, 802, 101, 207, 4016, 203, 1049, 302, 5015, 103, 901, 102, 306, 2028, 201, 1059, 202, 2099, 304, 3027, 1016, ...

The first term of X must be read like this:

109 == "add 1 to 9" (and we get the term 10);

The second term of X reads:

108 == "add 1 to 8" (and we get the term 9);

Now:

208 == "add 2 to 8" (and we get the term 10);

107 == "add 1 to 7" (and we get the term 8);

1019 == "add 1 to 19" (and we get the term 20);

206 == "add 2 to 6" (and we get the term 8);

307 == "add 3 to 7" (and we get the term 10);

106 == "add 1 to 6" (and we get the term 7);

...

The successive yellow terms on the right (seq Y) spell the original sequence X. Comparison:

X = 109, 108, 208, 107, 1019, 206, 307, 106,...

Y = 10  9  10  8  20  8  10  7 ...

Designing X

We wanted X to be the lexicographically earliest sequence of distinct positive terms such that the successive additions (whose results form Y) reproduce X.

A note

We see (immediately above) three 10s in Y; as each 10 is the result of an addition (when 0 is replaced by a "+" sign in X), and because of the command "the sequence X is made of distinct positive terms", we must find distinct additions to produce those 10s (like 109, 208, 307, ... 901).

Ok – but what will happen if we have to produce another 10 in Y, after having used 901? Well, we don't produce 10... Instead, we will produce "10a", with "a" being the next available digit of X.

We have an example here, at the end of X, with 2099:

X = 109, 108, 208, 107, 1019, 206, 307, 106, 406, 1018, 2018, 105, 1029, 205, 505, 204, 1039, 303, 604, 1017, 3017, 2016, 703, 104, 802, 101, 207, 4016, 203, 1049, 302, 5015, 103, 901, 102, 306, 2028, 201, 1059, 202, 2099, 304, 3027, 1016, ...

This 2099 is transformed in 2+99 and 2+99 = 101.

The last "1" of "101" is the "a" we were speaking of.

 

What else can we say? All terms of X will contain exactly one zero – not at the beginning nor at the end of the integer. This is why the first term of X is 109: there would be a contradiction if a(1) = 100, or 101, or 102, ... or 108.

No doubt

X is infinite, no doubt, as it will always be possible to use the "10a" technique to extend X.

The "Tom Clancy variant" is here.

Best,

É.