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%I #27 Nov 20 2020 09:31:39
%S 0,1,2,3,4,5,6,7,8,9,0,1,1,2,2,3,3,4,4,0,1,1,2,2,3,3,4,4,0,1,1,2,2,3,
%T 3,4,4,0,1,1,2,2,3,3,4,4,0,1,1,2,2,3,3,4,4,0,1,1,2,2,3,3,4,4,0,1,1,2,
%U 2,3,3,4,4,0,1,1,2,2,3,3,4,4,0,1,1,2,2
%N Floor of (digital root of n) divided by (number of digits in n).
%H Fernando Villanueva, <a href="/A338379/b338379.txt">Table of n, a(n) for n = 0..9999</a>
%e For n = 12345 the length of n is 5 (because 12345 has 5 digits) and its digital root is 6 (1+2+3+4+5=15 then 1+5=6). Then [6/5]=1 (where [x] is the integer part of x). Therefore a(12345)=0.
%t Array[Floor[If[#1 == 0, 0, (#2 /. 0 -> 9)/IntegerLength[#1]]] & @@ {#, Mod[#, 9]} &, 105, 0] (* _Michael De Vlieger_, Nov 05 2020 *)
%o (Python)
%o #Defines digital root
%o def recsum(n):
%o nstr = str(n)
%o if len(nstr) == 1:
%o return int(nstr)
%o else:
%o nl = sum([int(j) for j in nstr])
%o return recsum(nl)
%o #Prints first 1000 terms
%o for i in range(1000):
%o print(recsum(i)//len(str(i)))
%o (PARI) a(n) = if (n, (n-1)%9+1, 0) \ #Str(n); \\ _Michel Marcus_, Nov 05 2020
%Y Cf. A010888 (digital root of n), A055642 (length of n).
%K nonn,base
%O 0,3
%A _Fernando Villanueva_, Oct 23 2020
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