|
|
A340835
|
|
a(n) is the least k such that the digit reversal of k is greater than or equal to n.
|
|
2
|
|
|
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
A004086 gives the digit reversal of a number.
All positive terms belong to A070698.
This sequence is nondecreasing.
|
|
LINKS
|
|
|
FORMULA
|
a(n) <= n + 1 with equality iff n = 10^k for some k > 0.
|
|
EXAMPLE
|
For n = 1000:
- A004086(k) < 1000 for any k <= 1000,
- so a(1000) = 1001.
|
|
PROG
|
(PARI) { base = 10; k = 0; r = 0; for (n=0, 67, while (r<n, r = fromdigits(Vecrev(digits(k++, base)), base); ); print1 (k ", ")) }
(Python)
if n == 0:
return 0
s = str(n)
for i, x in enumerate(s):
if x != '9':
break
else:
return n
s1, s2 = s[:i+1], s[i+1:]
if s2 == '':
if s1[-1] == '0':
return int(str(n+1)[::-1])
else:
return int(s[::-1])
if int(s2) <= 1:
return int('1'+s2[-2::-1]+s1[::-1])
else:
return int('1'+'0'*(len(s2)-1)+str(int(s1)+1)[::-1]) # Chai Wah Wu, Mar 14 2021
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|