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A300000 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, with a(1) = 1.

%I A300000 #22 Jul 08 2022 12:19:01
%S A300000 1,10,99,999,9990,99900,999000,9990000,99900000,999000000,9990000000,
%T A300000 99899999991,998999999919,9989999999190,99899999991900,
%U A300000 998999999918991,9989999999189910,99899999991899109,998999999918991090,9989999999189910900,99899999991899108991,998999999918991089910,9989999999189910899100
%N A300000 The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, with a(1) = 1.
%C A300000 The sequence starts with a(1) = 1 and is always extended with the smallest integer not yet present in the sequence and not leading to a contradiction.
%C A300000 By definition, Sum_{k=1..n} a(k) = c(n) = concatenation of the first n digits of the sequence, therefore a(n) = c(n) - c(n-1). For n > 2, this defines a(n) recursively, without the need for solving an implicit equation, as the definition might suggest. - _M. F. Hasler_, Feb 22 2018
%C A300000 From _Michael S. Branicky_, Dec 07 2020: (Start)
%C A300000 Each digit 0-9 eventually appears.  The digit 1 appears first in a(1), 0 in a(2), 9 in a(3), 8 in a(12), 2 in a(68), 7 in a(71), 3 in a(2280), 6 in a(2283), 5 in a(2417), and 4 in a(4280).
%C A300000 All ten digits appear in each of a(4280) through a(121000).
%C A300000 Conjecture: a(n) contains all ten digits for n >= 4280. (End)
%C A300000 Conjecture holds through a(169000). - _Michael S. Branicky_, Jul 08 2022
%H A300000 Michael S. Branicky, <a href="/A300000/b300000.txt">Table of n, a(n) for n = 1..1001</a> (terms 1..300 from Jean-Marc Falcoz)
%F A300000 a(n) = c(n) - c(n-1), where c(n) is the concatenation of the first n digits. c(n) ~ 1.1*10^(n-1), and a(n) ~ 0.999*10^(n-1). - _M. F. Hasler_, Feb 22 2018
%e A300000 1 + 10 = 11 which is the concatenation of 1 and 1.
%e A300000 1 + 10 + 99 = 110 which is the concatenation of 1, 1 and 0.
%e A300000 1 + 10 + 99 + 999 = 1109 which is the concatenation of 1, 1, 0 and 9.
%e A300000 Otherwise said:
%e A300000 a(3) = concat(1,1,0) - (1 + 10) = 110 - 11 = 99,
%e A300000 a(4) = concat(1,1,0,9) - (11 + 99) = 1109 - 110 = 999,
%e A300000 a(5) = concat(1,1,0,9,9) - 1109 = 11099 - 1109 = 9990,
%e A300000 a(6) = concat(1,1,0,9,9,9) - 11099 = 99900, etc. - _M. F. Hasler_, Feb 22 2018
%t A300000 a[1]=1;a[2]=10;a[n_]:=a[n]=FromDigits[Flatten[IntegerDigits/@Table[a[k],{k,n-1}]][[;;n]]]-Total@Table[a[m],{m,n-1}];
%t A300000 Table[a[l],{l,30}]  (* _Giorgos Kalogeropoulos_, May 20 2019 *)
%o A300000 (PARI) a(n, show=1, a=1, c=a, d=[c])={for(n=2, n, show&&print1(a","); a=-c+c=c*10+d[1]; d=concat(d[^1],if(n>2,digits(a)))); a} \\ _M. F. Hasler_, Feb 22 2018
%o A300000 (Python)
%o A300000 def a(n):
%o A300000     alist, c, ckm1 = [1, 10], "110", 11
%o A300000     for k in range(3, n+1):
%o A300000         ck = 10*ckm1 + int(c[k-1])
%o A300000         ak, ckm1 = ck - ckm1, ck
%o A300000         c += str(ak)
%o A300000         alist.append(ak)
%o A300000     return alist[n-1]
%o A300000 print([a(n) for n in range(1, 24)]) # _Michael S. Branicky_, Dec 07 2020
%Y A300000 A299865, A299866, A299867, A299868, A299869, A299870, A299871 and A299872 show the same type of sequence but with a different start.
%Y A300000 The partial sums (the sequence c(n) mentioned in the Comments) is A299301.
%K A300000 nonn,base,nice,easy
%O A300000 1,2
%A A300000 _Eric Angelini_, Feb 10 2018