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A309872 For each term, take the last digit of the previous term and count all the appearances of that digit up to and including the previous term; the first term is 1.

%I A309872 #25 Feb 19 2020 01:33:46
%S A309872 1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,1,12,2,3,2,4,2,5,2,6,2,7,2,8,
%T A309872 2,9,2,10,2,11,16,3,3,4,3,5,3,6,4,4,5,4,6,5,5,6,6,7,3,7,4,7,5,7,6,8,3,
%U A309872 8,4,8,5,8,6,9,3,9,4,9,5,9,6,10,3,10,4,10,5,10,6,11,23,11
%N A309872 For each term, take the last digit of the previous term and count all the appearances of that digit up to and including the previous term; the first term is 1.
%H A309872 Robert Israel, <a href="/A309872/b309872.txt">Table of n, a(n) for n = 1..10000</a>
%e A309872 To get the eleventh term, you need to get the last digit of the tenth term, which is 1, and then count all the 1's already in the sequence: 1, 1, 2, 1, 3, 1, 4, 1, 5, 1; there are six 1's, so the eleventh term is 6.
%p A309872 Q:= Array(0..9):
%p A309872 A:= Vector(100):
%p A309872 Q[1]:= 1:
%p A309872 A[1]:= 1:
%p A309872 for n from 2 to 100 do
%p A309872   d:= A[n-1] mod 10;
%p A309872   A[n]:= Q[d];
%p A309872   L:= convert(%,base,10);
%p A309872   for i in L do Q[i]:= Q[i]+1 od
%p A309872 od:
%p A309872 convert(A,list); # _Robert Israel_, Feb 18 2020
%o A309872 (PARI) f = vector(base=10); for (n=1, 91, v = if (n==1, 1, f[1+(v%base)]); apply (d -> f[1+d]++, if (v, digits(v, base), [0])); print1 (v ", ")) \\ _Rémy Sigrist_, Aug 21 2019
%o A309872 (Python)
%o A309872 s, a, n = "1", [1], 1
%o A309872 while n < 100:
%o A309872     n = n+1
%o A309872     d = s[len(s)-1]
%o A309872     i, aa = 0, 0
%o A309872     while i < len(s):
%o A309872         if s[i] == d:
%o A309872             aa = aa+1
%o A309872         i = i+1
%o A309872     s, a = s+str(aa), a+[aa]
%o A309872 for n in range(1, 92): print(a[n-1], end=', ') # _A.H.M. Smeets_, Aug 22 2019
%Y A309872 Cf. A248034 (first term is 0).
%K A309872 base,nonn,look
%O A309872 1,3
%A A309872 _Maxim Skorohodov_, Aug 21 2019