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 A261423 Largest palindrome <= n. 87
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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 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 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 *) PROG (PARI) A261423(n, d=digits(n), m=sum(k=1, #d\2, d[k]*10^(k-1)))={if( n%10^(#d\2)

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Last modified June 5 01:27 EDT 2020. Contains 334828 sequences. (Running on oeis4.)