A self-binary array T

 

Hello Math Fun,
Could someone compute a few more binary terms? I am stuck with S(21).
Here is the idea:
We build an array T and associate to T a sequence S of binary terms that is just T read by its successive antidiagonals [in short, the term S(k) is the binary representation of the base-10 T(k)]. It turns out that this sequence S is also the upper row of T.
You will understand very quickly how T was lexicographically build by distinct nonnegative terms. We start T with 0:
 
..0

T is always extended by its successive antidiagonals with the smallest term not yet in T that doesn’t lead to a contradiction.
..0.....1.....a
..2.....b
..c
 

Now, as we want the upper row to represent T in binary, we must assign to “a” the binary value of ‘2’, which is 10 [this 10 is the 3rd term of the upper row of the array and must thus represent T(3) in binary; as 10 is 2 in binary, we assign this 10 to the letter a]:

..0.....1.....10
..2.....b
..c
 
The only constraint on the above “b” is to be the smallest integer not yet used, thus b=3 (and c=4);
..0.....1....10.....e
..2.....3.....f
..4.....g
..h
 
The next number we must find is “e”. But as this “e” is the 4th horizontal term (of the first row, by construction) it must represent in binary the 4th term of T (read by antidiagonals); so “e” is simply the binary representation of T(4), which is the 10 already present in T; as “10” is represented by the binary 1010, we have “e”=1010.

..0.....1....10..1010.....i
..2.....3.....f.....j
..4.....g.....k
..h.....l
..m
 
The terms f, g and h are easy to find (the smallest unused, etc.):
..0.....1....10..1010.....i
..2.....3.....5.....j
..4.....6.....k
..7.....l
..m
 
The binary number “i” that we must find is the 5th term of the upper row – it must then represent in binary the 5th term of T when T is read by its successive antidiagonals; as T(5)=3 we have i=11 (which is 3 in binary):

..0.....1....10..1010....11
..2.....3.....5.....8
..4.....6.....9
..7.....l
..m
 
To extend further T, we cannot re-use 10 for “l”, neither 11 for “m” (as they are already in T); we take instead l=12 and m=13:
..0.....1....10..1010....11.....n
..2.....3.....5.....8.....o
..4.....6.....9.....p
..7....12.....q
.13.....r
..s
 
Again, “n” being the 6th term of the upper row, we want it to represent T(6) in binary; as T(6)=4, we have n=100 (the other letters of the present antidiagonal are computed easily):
..0.....1....10..1010....11...100.....t
..2.....3.....5.....8....14.....u
..4.....6.....9....15.....v
..7....12....16.....w
.13....17.....x
.18.....y
..z
 
Ok, you’ve got the idea.
But self-binary representations become quickly quite big. And as the letter “t” above stands for the base-10 integer 1111110010 I am stuck with the binary representation of this 1111110010.
 
If I am not wrong (I usually am), the binary sequence S begins like this:
S = 0,1,10,1010,11,100,1111110010,101,110,111,1011,1000,1001,1100,1101,1100100,1110,10000,10001,10010,.?.,
Best,
É.

___________________________________

Michael Branicky was quick to correct and extend S on Math Fun:

> I get the next term is 1000010001110100011000101111010

(and also that 1111 is missing between 1110 and 10000)

We have thus so far for (the corrected) S:

S = 0,1,10,1010,11,100,1111110010,101,110,111,1011,1000,1001,1100,1101,1100100,1110,10000,10001,10010,1000010001110100011000101111010,...

Best (merci Michael!)
É.