OFFSET
1,2
COMMENTS
Start with a(1) = 1 and a(2) = 2; read the sequence digit by digit starting from the left:
when the read digit is smaller than the next one, double the last integer of the sequence and extend the sequence with the result rearranged (smallest digits first and leading zeros erased);
when the read digit is bigger than the next one, double the last integer of the sequence and extend the sequence with the result rearranged (biggest digits first, zeros at the end);
when both digits are equal, do a circular permutation of the last integer of the sequence and extend the sequence with the result (this will erase a few zeros in some cases, as 100023850 becomes 238501).
LINKS
Jean-Marc Falcoz, Table of n, a(n) for n = 1..3002
EXAMPLE
As the only digit of a(1) = 1 is smaller than 2 (the next digit), we extend the sequence with 4 (that is 2 times 2);
as the only digit of a(2) = 2 is smaller than 4 (the next digit), we extend the sequence with 8 (that is 2 times 4);
as the only digit of a(3) = 4 is smaller than 8 (the next digit), we extend the sequence with 16 (that is 2 times 8 -- with 1 coming before 6);
as the only digit of a(4) = 8 is bigger than 1 (the next digit), we extend the sequence with 32 (that is 2 times 16 -- with 3 coming before 2);
as the first digit of a(5) = 1 is smaller than 6 (the next digit), we extend the sequence with 46 (that is 2 times 32 = 64 that is rearranged in 46);
as the last digit of a(5) = 6 is bigger than 3 (the next digit), we extend the sequence with 92 (that is 2 times 46 = 92 rearranged in 29);
. . .
as the last digit of a(9) = 1 is equal to 1 (the next digit), we extend the sequence with 554219 (this is the circular permutation of the previous term, 955421);
etc.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Jean-Marc Falcoz and Eric Angelini, Jul 12 2018
STATUS
approved