OFFSET
1,1
COMMENTS
Unlike A304237, there are no consecutive numbers in the sequence up to n=82.
Could have started sequence with offset a(0)=0 and it would be the same sequence.
EXAMPLE
a(8)=112 since 112 is the smallest positive number not yet in the sequence that is obtained when the first and last digits of a(7)=21 are exchanged and used adjacent in that order (without the adjacency condition, we would have a(8)=102 as in A304237).
PROG
(PARI) firstTerms(n)={my(Seq=vector(n), a=[1, 0], c, y, k, h=Vecsmall(0, 1000*n)); print1("10, "); Seq[1]=10; h[11]=1; for(i=2, n, for(t=11, oo, if(!h[t+1], c=digits(t); y=1; while((y<#c)&&(!c[y]), y++); for(u=y, #c-1, if(k=([c[u], c[u+1]]==[a[#a], a[1]]), break)); if(k, Seq[i]=t; print1(t", "); a=c; h[t+1]=1; break)))); return(Seq)} \\ R. J. Cano, May 11 2018
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Enrique Navarrete, May 08 2018
STATUS
approved