OFFSET
1,2
COMMENTS
The sequence S starts with a(1) = 1 and a(2) = 3. S is extended by duplicating the first term A among the not yet duplicated terms, under the condition that the concatenation [last term Z of the sequence;A] is divisible by A. If this is not the case, we then extend S with the smallest integer X not yet present in S such that the concatenation [last term Z of the sequence;X] is not divisible by X. This is the lexicographically first sequence with this property. The terms 2 and 5 will never appear.
LINKS
Jean-Marc Falcoz, Table of n, a(n) for n = 1..10002
EXAMPLE
S starts with a(1) = 1 and a(2) = 3
Can we duplicate a(1) to form a(3)? Yes, as [31] is divisible by 1, of course; thus a(3) = 1.
Can we duplicate a(2) to form a(4)? No, as [13] is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [a(3);X] is not divisible by X. We get X = 4 and thus a(4) = 4.
Can we duplicate a(2) to form a(5)? No, as [43] is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [a(4);X] is not divisible by X. We get X = 6 and thus a(5) = 6.
Can we duplicate a(2) to form a(6)? Yes, as [63] is divisible by 3, of course; thus a(6) = 3.
Can we duplicate a(3) to form a(7)? Yes, as [31] is divisible by 1, of course; thus a(7) = 1.
Can we duplicate a(4) to form a(8)? No, as [14] is not divisible by 4; we thus extend S with the smallest integer X not yet in S such that [a(7);X] is not divisible by X. We get X = 7 and thus a(8) = 7.
Can we duplicate a(4) to form a(9)? No, as [74] is not divisible by 4; we thus extend S with the smallest integer X not yet in S such that [a(8);X] is not divisible by X. We get X = 8 and thus a(9) = 8.
Can we duplicate a(4) to form a(10)? Yes, as [84] is divisible by 4, of course; thus a(10) = 4.
Etc.
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Eric Angelini and Jean-Marc Falcoz, Mar 11 2019
STATUS
approved