A340490 - OEIS

%I #48 Sep 13 2021 08:52:36

%S 9,13,14,15,16,17,18,19,20,2,20,1,1,2,15,2,20,2,20,2,20,2,20,2,20,2,

%T 20,5,20,2,20,2,20,1,1,2,13,2,20,2,20,2,20,2,20,2,20,7,20,2,20,2,20,2,

%U 20,1,1,2,11,2,20,2,20,2,20,2,20,9,20,2,20,2,20,2

%N a(n) is the number of digits after the n-th digit of the Champernowne constant A033307 until the next appearance of that digit.

%C In typing the positive integers without leaving spaces between them, it is interesting to see how many places later we press the same number key on the keyboard. This sequence answers that question.

%H Md. Towhidul Islam, <a href="/A340490/b340490.txt">Table of n, a(n) for n = 1..31968</a>

%F A033307(n - 1 + a(n)) = A033307(n - 1).

%e In concatenating the positive integers, we get 1 first. The next occurrence of 1 is in 10. So 1 occurs 9 places later, which gives a(1)=9. The second digit 2 occurs again in writing 12. So 2 occurs 13 places later and a(2) is 13.

%o (PARI) C(nn) = {my(list = List()); for (n=1, nn, my(d=digits(n)); for (k=1, #d, listput(list, d[k]););); list;} \\ A033307

%o posi(list, i) = {for (j=i+1, #list, if (list[i] == list[j], return (j-i)););}

%o lista(nn) = {my(list = C(nn)); my(listp = List()); for (i=1, #list, my(pos = posi(list, i)); if (! pos, break); listput(listp, pos);); Vec(listp);} \\ _Michel Marcus_, Jan 11 2021

%o (Python)

%o def aupton(terms):

%o     alst, chapnk, k = [], [1], 1

%o     for n in range(1, terms+1):

%o         chapn = chapnk.pop(0)

%o         while chapn not in chapnk:

%o             k += 1

%o             chapnk.extend(list(map(int, str(k))))

%o         alst.append(chapnk.index(chapn) + 1)

%o     return alst

%o print(aupton(74)) # _Michael S. Branicky_, Sep 13 2021

%Y Cf. A033307 (the parent sequence).

%K nonn,base

%O 1,1

%A _Md. Towhidul Islam_, Jan 10 2021