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A340490
a(n) is the number of digits after the n-th digit of the Champernowne constant A033307 until the next appearance of that digit.
2
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, 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, 20, 1, 1, 2, 11, 2, 20, 2, 20, 2, 20, 2, 20, 9, 20, 2, 20, 2, 20, 2
OFFSET
1,1
COMMENTS
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.
LINKS
FORMULA
A033307(n - 1 + a(n)) = A033307(n - 1).
EXAMPLE
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.
PROG
(PARI) C(nn) = {my(list = List()); for (n=1, nn, my(d=digits(n)); for (k=1, #d, listput(list, d[k]); ); ); list; } \\ A033307
posi(list, i) = {for (j=i+1, #list, if (list[i] == list[j], return (j-i)); ); }
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
(Python)
def aupton(terms):
alst, chapnk, k = [], [1], 1
for n in range(1, terms+1):
chapn = chapnk.pop(0)
while chapn not in chapnk:
k += 1
chapnk.extend(list(map(int, str(k))))
alst.append(chapnk.index(chapn) + 1)
return alst
print(aupton(74)) # Michael S. Branicky, Sep 13 2021
CROSSREFS
Cf. A033307 (the parent sequence).
Sequence in context: A050236 A193995 A070607 * A335662 A191891 A135439
KEYWORD
nonn,base
AUTHOR
Md. Towhidul Islam, Jan 10 2021
STATUS
approved