A340835 a(n) is the least k such that the digit reversal of k is greater than or equal to n.
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
Offset: 0
Examples
For n = 1000: - A004086(k) < 1000 for any k <= 1000, - A004086(1001) = 1001 >= 1000, - so a(1000) = 1001.
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
Programs
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PARI
{ base = 10; k = 0; r = 0; for (n=0, 67, while (r -
Python
def A340835(n): 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
Formula
a(n) <= n + 1 with equality iff n = 10^k for some k > 0.
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