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A375243
Infinite variant of A375232.
0
0, 10, 20, 100, 30, 102, 40, 101, 203, 105, 60, 1024, 300, 107, 200, 150, 304, 1026, 80, 109, 230, 10457, 0, 120, 306, 110, 204, 1058, 303, 10279, 0, 1046, 302, 501, 0, 201, 3048, 170, 206, 1059, 330, 1042, 0, 1000, 320, 105678, 400, 210, 3000, 190, 202, 1045, 360, 1027, 800, 1001, 2034, 510, 0, 10269
OFFSET
1,2
COMMENTS
Instead of stopping the sequence when no integer is available, we extend it with 0 and go on (0 being the only term allowed to be repeated whenever nothing else works). This method seems to work ad infinitum.
Around 90% of the terms are not equal to 0.
For the first 1000 terms, the largest chunk between two successive 0 is the 24-integer long serie [101101, 2630, 8015, 40044, 10122, 333330, 1907, 200222, 10456, 10200, 80008, 101110, 3204, 5107, 6660, 9012, 1400, 202000, 8051, 10276, 40400, 101111, 20223, 5109].
EXAMPLE
The finite sequence A375232 ends with 80, 109, 230, 10457. If we extend it with a(23) = 0, we can compute a(24) = 120, a(25) = 306 then 110, 204, 1058, 303 and 10279. No more integers are available at that stage. But, again, we can extend the sequence with a(31) = 0, then a(32) = 1046 and 302, 501, 0, 201, 3048, 170, etc.
A repeated single 0 is counted as a term of the sequence.
CROSSREFS
Sequence in context: A215878 A200985 A375232 * A166641 A101244 A260743
KEYWORD
nonn,base
AUTHOR
STATUS
approved