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%I #19 Nov 26 2018 06:57:02
%S 1,2,3,4,5,6,7,8,9,11,21,31,41,51,61,71,81,91,92,93,94,95,96,97,98,99,
%T 101,201,301,401,501,601,701,801,901,911,921,931,941,951,961,971,981,
%U 991,992,993,994,995,996,997,998,999,1001,2001,3001,4001,5001,6001,7001
%N Positive integers read backwards, but omit a number if it is <= an earlier number.
%C a(n+1) - a(n) is a power of 10 or 2 if a(n) is of the form 10^m - 1. - _David A. Corneth_, Nov 26 2018
%H David A. Corneth, <a href="/A076643/b076643.txt">Table of n, a(n) for n = 1..10000</a>
%F a((9*m^2 + 7*m + 2) / 2) = 10^m - 1, m > 0. - _David A. Corneth_, Nov 25 2018
%e The beginning list is 1,2,3,4,5,6,7,8,9,1,11; but 1 < 9 so omit it.
%p d := N->`if`(N=0,[0],ListTools[Reverse](convert(N,base,10))); a := N->sum(d(N)[n]*10^(n-1),n=1..nops(d(N))); dropsort := proc(S)::list; description "Forms a strictly increasing list by dropping terms."; local T,V; T := k->max(seq(S[j],j=1..k)); V := ListTools[MakeUnique]([seq(T(j),j=1..nops(S))]); return(V); end proc; dropsort([seq(a(n),n=1..2000)]);
%t Union@ FoldList[Max, IntegerReverse@ Range@ 1007] (* _Michael De Vlieger_, Nov 25 2018 *)
%o (PARI) nxt(n) = {my(d = digits(n), i = 1); while(i <= #d && d[i] == 9, i++); if(i <= #d, n+10^(#d-i), n+2)}
%o first(n) = my(res = vector(n)); res[1] = 1; for(i = 2, n, res[i] = nxt(res[i-1])); res \\ _David A. Corneth_, Nov 25 2018
%Y Cf. A004086.
%K nonn,easy,look,base
%O 1,2
%A Francois Jooste (phukraut(AT)hotmail.com), Oct 23 2002