This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
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.
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
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
a(0) = 11 as the string '0' first appears at position 1 and again at position 12, giving a difference of 11. a(10) = 180 as the string '10' first appears at position 11 and again at position 191 (where it forms the start of the number 100), giving a difference of 180. a(12) = 13 as the string '12' first appears at position 2 (the first number to appear before its natural position) and again at position 15 (its natural position), giving a difference of 13. This is the first term to differ from A337227.
from itertools import count def A342162(n): s1, s2, m = tuple(int(d) for d in str(n)), tuple(), -1 l = len(s1) for i, s in enumerate(int(d) for k in count(0) for d in str(k)): s2 = (s2+(s,))[-l:] if s2 == s1: if m >= 0: return i-m m = i # Chai Wah Wu, Feb 18 2022
The sequence of digits of the integers is 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0... The 15th term, for instance, is a 1. Thus 1+15=16 is the 15th term of this sequence. Next one is 2 (the digit) + 16 (the rank) = 18
dgs = Flatten[IntegerDigits/@Range[0, 50]]; Plus@@@Partition[Riffle[dgs, Range[Length[dgs]]], 2] (* Harvey P. Dale, Jan 15 2011 *)
See Links section.
def aupton(terms): alst, A033307, last, used, n, an = [], '1', 1, set(), 0, 0 while n <= terms: while an in used: an += 1 while len(A033307) <= n + an: last += 1; A033307 += str(last) if A033307[n + an] == A033307[n]: alst += [an]; used.add(an); n += 1; an = 0 else: an += 1 return alst print(aupton(63)) # Michael S. Branicky, Jan 30 2021
8.10000006707603361330731967383416787753583647347857972252510...
RealDigits[1/ChampernowneNumber[10] , 10, 120][[1]]
0.123456789101112...= 1/9+1/(9*10)+1/(9*10*10)+1/(9*10*10*10)+... .
a(4) = 8, then 8 positions along in sequence A033307 reaches digit 1 (first 1 of 13) so a(5) = 1. a(5) = 1, then 1 position along in sequence A033307 reaches digit 3 (of number 13) so a(6) = 3. The terms of A033307 and those which become the terms here begin 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 ... ^ ^ ^ ^ ^ ^ ^ ^
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