ecl1 - More Powerful Single-Sequence BMS System?

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ecl1 - More Powerful Single-Sequence BMS System?
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A sequence is written as (x_1,x_2,x_3,x_4,...,x_n)[a] where a>0 is an integer and x_i >= 0 are integers.

If the last number is 0, the normal BMS rule applies (remove it, square the number in the [] brackets).
If the last number is not 0, then do these:
---- Let the last number be N.
---- Find the rightmost number strictly less than N, and call it K.
---- The part of the sequence including K and things to the left of K is the "left" part of the sequence.
---- The part of the sequence strictly to the right of K is the "right" part of the sequence.
---- Decrease the last number (N) by 1.  Make this the new value of N.
---- Find the difference between N and K, call this D.
(Note: if the ordinal is less than e0, D will always be equal to 0.)
---- Take the left part of the sequence, and concatenate sequences S_0, S_1, ... S_(a-1) to it.
---- S_i is defined as the right part of the sequence, where every term in it is incremented by D\*i.

Example #1:
(0,2,1,3)
Left part = (0,2,1)
Right part = (2) (after decrementing from 3)
N = 2
K = 1
D = 2-1 = 1
S_0 = (2) incremented by D*0 = 0
S_1 = (2) incremented by D*1 = 1, which is (3).
S_2 = (2) incremented by D*2 = 2, which is (4).
etc etc...
So the result is the left part (0,2,1) concatenated with (2),(3),(4),(5),.... etc
which is (0,2,1,2,3,4,5,6,7,8,....)

Example #2:
(0,2,2)
Left part = (0)
Right part = (2,1) after decrementing
N = 1
K = 0
D = 1-0 = 1
S_0 = (2,1) incremented by D*0 = 0
S_1 = (2,1) incremented by D*1 = 1, which is (3,2)
S_2 = (2,1) incremented by D*2 = 2, which is (4,3)
etc etc...
So the result is (0,2,1,3,2,4,3,5,4,6,5....)

Example #3:
(0,4,7)
Left part = (0,4)
Right part = (6) after decrementing
N = 6
K = 4
D = 6-4 = 2
S_0 = (6) incremented by D*0 = 0
S_1 = (6) incremented by D*1 = 2, which is (8)
S_2 = (6) incremented by D*2 = 4, which is (10)
etc etc...
So the result is (0,4,6,8,10,12,14,......)