A261423 Largest palindrome <= n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66, 66, 66, 66, 66, 66, 66, 77, 77

Offset: 0

Author

N. J. A. Sloane, Aug 28 2015

Keywords

• nonn
• base

Comments

Might be called the palindromic floor function.

Let P(n) = n with the second half of its digits replaced by the first half of the digits in reverse order. If P(n) <= n, then a(n) = P(n), else if n=10^k then a(n) = n-1, else a(n) = P(n-10^floor(d/2)), where d is the number of digits of n. - M. F. Hasler, Sep 08 2015

The largest differences of n - a(n) occur for n = m*R(2k) - 1, where 1 <= m <= 9 and R(k)=(10^k-1)/9. In this case, n - a(n) = 1.1*10^k - 1. - M. F. Hasler, Sep 05 2018

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Palindromic Number
• Wikipedia, Palindromic number
• Index entries for sequences related to palindromes

Crossrefs

Cf. A002113, A261424, A261914 (previous palindrome).

Cf. A262038.

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

A262257(n) = Levenshtein distance between n and a(n). - Reinhard Zumkeller, Sep 16 2015

Programs

Haskell

a261423 n = a261423_list !! n
a261423_list = tail a261914_list  -- Reinhard Zumkeller, Sep 16 2015

Maple

# P has list of palindromes
palfloor:=proc(n) global P; local i;
for i from 1 to nops(P) do
   if P[i]=n then return(n); fi;
   if P[i]>n then return(P[i-1]); fi;
od:
end;

Mathematica

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Table[k = n;
While[Nand[palQ@ k, k > -1], k--]; k, {n, 0, 78}] (* Michael De Vlieger, Sep 09 2015 *)

PARI

A261423(n,d=digits(n),m=sum(k=1,#d\2,d[k]*10^(k-1)))={if( n%10^(#d\2)M. F. Hasler, Sep 08 2015, minor edit on Sep 05 2018

Python

def P(n):
    s = str(n); h = s[:(len(s)+1)//2]; return int(h + h[-1-len(s)%2::-1])
def a(n):
    s = str(n)
    if s == '1'+'0'*(len(s)-1) and n > 1: return n - 1
    Pn = P(n)
    return Pn if Pn <= n else P(n - 10**(len(s)//2))
print([a(n) for n in range(79)]) # Michael S. Branicky, Jun 25 2021

Formula

n - a(n) < 1.1*10^floor(d/2), where d = floor(log_10(n)) + 1 is the number of digits of n. - M. F. Hasler, Sep 05 2018