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COMMENT
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This sequence is a loop of 12 steps repeated indefinitely from the 13th term.
The sequence starting with any pair of successive terms of the sequence runs also in the same loop.
Note that the parity of the terms of the loop has a periodicity of 3 :odd,odd,even,odd,odd,even . . . then one at least of the digits of any pair of successive terms of the loop id odd;*
The same rule applied from the fisrt terms 0 an 1 gives A003893 (=truncanacci(0,1)) which is a loop of 60 steps. That is true of course also for any of the 59 other pairs (A003893(i),A003893(1+i))
Another loop of 3 steps is given for the 3 last pairs of digits, one of them at least being odd : Truncanacci(0,5) =0,5,5,0,5,5
Among the pairs of two even digits, 20 of them generate the same loop of 20 steps :Truncanacci(0,2)=0,2,2,4,6,0,6,6,2,8,0,8,8,6,4,0,4,4,8,2,0,2,2,4,6,0,6,6,2,8,0,8,8,6,4,0,4,4,8,2
4 more pairs generate a loop of this size : Truncanacci(4,6)= 2,6,8,4,2,6,8,4
The last pair of even digits (0,0) gives the constant term 0 or a loop of one step.
As we have studied 60+12+3=75 starting pairs with at least an odd term and 20+4+1=25 pairs of even digits, all possible pairs from (0,0) to (9,9) are exhausted
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