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At step n the sequence lists the number of occurrences of digit (n mod k), with k>0, in all the numbers from 1 to n. Case k=9.
9

%I #11 Jan 01 2022 17:56:19

%S 1,1,1,1,1,1,1,1,0,2,1,1,1,1,1,1,1,1,12,3,2,2,2,2,2,2,2,13,13,4,3,3,3,

%T 3,3,3,14,14,14,5,4,4,4,4,4,15,15,15,15,6,5,5,5,5,16,16,16,16,16,7,6,

%U 6,6,17,17,17,17,17,17,8,7,7,18,18,18,18,18,18,18,9,8,19,19,19,19,19,19,19

%N At step n the sequence lists the number of occurrences of digit (n mod k), with k>0, in all the numbers from 1 to n. Case k=9.

%H Harvey P. Dale, <a href="/A136713/b136713.txt">Table of n, a(n) for n = 0..1000</a>

%e For n=19 we have 12 because the digit (19 mod 9)=1 is present 12 times: 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19.

%e For n=20 we have 3 because the digit (20 mod 9)=2 is present 3 times: 2, 12, 20.

%p P:=proc(n,m) local a,b,c,d,i,v; v:=array(1..m); for i from 1 to m-1 do v[i]:=1; print(1); od; if m=10 then v[m]:=1; print(1); else v[m]:=0; print(0); fi; for i from m+1 by 1 to n do a:=(i mod m); for b from i-m+1 by 1 to i do d:=b; while d>0 do c:=d-(trunc(d/10)*10); d:=trunc(d/10); if c=a then if a=0 then v[m]:=v[m]+1; else v[a]:=v[a]+1; fi; fi; od; od; if a=0 then print(v[m]); else print(v[a]); fi; od; end: P(101,9);

%t Table[Count[Flatten[IntegerDigits/@Range[n]],Mod[n,9]],{n,100}] (* _Harvey P. Dale_, Jan 01 2022 *)

%Y Cf. A136706, A136707, A136708, A136709, A136710, A136711, A136712, A136714.

%K easy,base,nonn

%O 0,10

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jan 18 2008