A262038 -id:A262038 - cp's OEIS frontend

A002113 Palindromes in base 10.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515

Offset: 1

N. J. A. Sloane

n is a palindrome (i.e., a(k) = n for some k) if and only if n = A004086(n). - Reinhard Zumkeller, Mar 10 2002
It seems that if n*reversal(n) is in the sequence then n = 3 or all digits of n are less than 3. - Farideh Firoozbakht, Nov 02 2014
The position of a palindrome within the sequence can be determined almost without calculation: If the palindrome has an even number of digits, prepend a 1 to the front half of the palindrome's digits. If the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit. Examples: 98766789 = a(19876), 515 = a(61), 8206028 = a(9206), 9230329 = a(10230). - Hugo Pfoertner, Aug 14 2015
This sequence is an additive basis of order at most 49, see Banks link. - Charles R Greathouse IV, Aug 23 2015
The order has been reduced from 49 to 3; see the Cilleruelo-Luca and Cilleruelo-Luca-Baxter links. - Jonathan Sondow, Nov 27 2017
See A262038 for the "next palindrome" and A261423 for the "preceding palindrome" functions. - M. F. Hasler, Sep 09 2015
The number of palindromes with d digits is 10 if d = 1, and otherwise it is 9 * 10^(floor((d - 1)/2)). - N. J. A. Sloane, Dec 06 2015
Sequence A033665 tells how many iterations of the Reverse-then-add function A056964 are needed to reach a palindrome; numbers for which this will never happen are Lychrel numbers (A088753) or rather Kin numbers (A023108). - M. F. Hasler, Apr 13 2019
This sequence is an additive basis of order 3, see Cilleruelo, Luca, & Baxter and Sigg. - Charles R Greathouse IV, Apr 08 2025

Karl G. Kröber, "Palindrome, Perioden und Chaoten: 66 Streifzüge durch die palindromischen Gefilde" (1997, Deutsch-Taschenbücher; Bd. 99) ISBN 3-8171-1522-9.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 50-52.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 120.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, List of first 19999 palindromes: Table of n, a(n) for n = 1..19999
Hunki Baek, Sejeong Bang, Dongseok Kim, and Jaeun Lee, A bijection between aperiodic palindromes and connected circulant graphs, arXiv:1412.2426 [math.CO], 2014.
William D. Banks, Derrick N. Hart, and Mayumi Sakata, Almost all palindromes are composite, Math. Res. Lett., Vol. 11, No. 5-6 (2004), pp. 853-868.
William D. Banks, Every natural number is the sum of forty-nine palindromes, arXiv:1508.04721 [math.NT], 2015; Integers, 16 (2016), article A3.
Javier Cilleruelo, Florian Luca and Lewis Baxter, Every positive integer is a sum of three palindromes, Mathematics of Computation, Vol. 87, No. 314 (2018), pp. 3023-3055, arXiv preprint, arXiv:1602.06208 [math.NT], 2017.
Patrick De Geest, World of Numbers.
Kritkhajohn Onphaeng, Tammatada Khemaratchatakumthorn, Phakhinkon Napp Phunphayap, and Prapanpong Pongsriiam, Exact Formulas for the Number of Palindromes in Certain Arithmetic Progressions, Journal of Integer Sequences, Vol. 27 (2024), Article 24.4.8. See p. 2.
Phakhinkon Phunphayap and Prapanpong Pongsriiam, Reciprocal sum of palindromes, arXiv:1803.00161 [math.CA], 2018.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Prapanpong Pongsriiam and Kittipong Subwattanachai, Exact Formulas for the Number of Palindromes up to a Given Positive Integer, Intl. J. of Math. Comp. Sci. (2019) 14:1, 27-46.
E. A. Schmidt, Positive Integer Palindromes. [Cached copy at the Wayback Machine]
Markus Sigg, On a conjecture of John Hoffman regarding sums of palindromic numbers, arXiv preprint (2015). arXiv:1510.07507 [math.NT]
Eric Weisstein's World of Mathematics, Palindromic Number.
Wikipedia, Palindromic number.
Index entries for sequences that are an additive basis, order 3.
Index entries for sequences related to palindromes
Index entries for "core" sequences

Subsequence of A061917 and A221221.
A110745 is a subsequence.
Union of A056524 and A056525.
Palindromes in bases 2 through 11: A006995 and A057148, A014190 and A118594, A014192 and A118595, A029952 and A118596, A029953 and A118597, A029954 and A118598, A029803 and A118599, A029955 and A118600, this sequence, A029956. Also A262065 (base 60), A262069 (subsequence).
Palindromic primes: A002385. Palindromic nonprimes: A032350.
Palindromic-pi: A136687.
Cf. A029742 (complement), A086862 (first differences).
Palindromic floor function: A261423, also A261424. Palindromic ceiling: A262038.
Cf. A004086 (read n backwards), A064834, A118031, A136522 (characteristic function), A178788.
Ways to write n as a sum of three palindromes: A261132, A261422.
Minimal number of palindromes that add to n using greedy algorithm: A088601.
Minimal number of palindromes that add to n: A261675.

Filtered([0..550],n->ListOfDigits(n)=Reversed(ListOfDigits(n))); # Muniru A Asiru, Mar 08 2019

a002113 n = a002113_list !! (n-1)
  a002113_list = filter ((== 1) . a136522) [1..] -- Reinhard Zumkeller, Oct 09 2011

import Data.List.Ordered (union)
  a002113_list = union a056524_list a056525_list -- Reinhard Zumkeller, Jul 29 2015, Dec 28 2011

[n: n in [0..600] | Intseq(n, 10) eq Reverse(Intseq(n, 10))]; // Vincenzo Librandi, Nov 03 2014

read transforms; t0:=[]; for n from 0 to 2000 do if digrev(n) = n then t0:=[op(t0),n]; fi; od: t0;
# Alternatively, to get all palindromes with <= N digits in the list "Res":
N:=5;
Res:= $0..9:
for d from 2 to N do
  if d::even then
    m:= d/2;
    Res:= Res, seq(n*10^m + digrev(n),n=10^(m-1)..10^m-1);
  else
    m:= (d-1)/2;
    Res:= Res, seq(seq(n*10^(m+1)+y*10^m+digrev(n),y=0..9),n=10^(m-1)..10^m-1);
  fi
od: Res:=[Res]: # Robert Israel, Aug 10 2014
# A variant: Gets all base-10 palindromes with exactly d digits, in the list "Res"
d:=4:
if d=1 then Res:= [$0..9]:
elif d::even then
    m:= d/2:
    Res:= [seq(n*10^m + digrev(n), n=10^(m-1)..10^m-1)]:
else
    m:= (d-1)/2:
    Res:= [seq(seq(n*10^(m+1)+y*10^m+digrev(n), y=0..9), n=10^(m-1)..10^m-1)]:
fi:
Res; # N. J. A. Sloane, Oct 18 2015
isA002113 := proc(n)
    simplify(digrev(n) = n) ;
end proc: # R. J. Mathar, Sep 09 2015

palQ[n_Integer, base_Integer] := Module[{idn = IntegerDigits[n, base]}, idn == Reverse[idn]]; (* then to generate any base-b sequence for 1 < b < 37, replace the 10 in the following instruction with b: *) Select[Range[0, 1000], palQ[#, 10] &]
base10Pals = {0}; r = 2; Do[Do[AppendTo[base10Pals, n * 10^(IntegerLength[n] - 1) + FromDigits@Rest@Reverse@IntegerDigits[n]], {n, 10^(e - 1), 10^e - 1}]; Do[AppendTo[base10Pals, n * 10^IntegerLength[n] + FromDigits@Reverse@IntegerDigits[n]], {n, 10^(e - 1), 10^e - 1}], {e, r}]; base10Pals (* Arkadiusz Wesolowski, May 04 2012 *)
nthPalindromeBase[n_, b_] := Block[{q = n + 1 - b^Floor[Log[b, n + 1 - b^Floor[Log[b, n/b]]]], c = Sum[Floor[Floor[n/((b + 1) b^(k - 1) - 1)]/(Floor[n/((b + 1) b^(k - 1) - 1)] - 1/b)] - Floor[Floor[n/(2 b^k - 1)]/(Floor[n/(2 b^k - 1)] - 1/ b)], {k, Floor[Log[b, n]]}]}, Mod[q, b] (b + 1)^c * b^Floor[Log[b, q]] + Sum[Floor[Mod[q, b^(k + 1)]/b^k] b^(Floor[Log[b, q]] - k) (b^(2 k + c) + 1), {k, Floor[Log[b, q]]}]] (* after the work of Eric A. Schmidt, works for all integer bases b > 2 *)
Array[nthPalindromeBase[#, 10] &, 61, 0] (* please note that Schmidt uses a different, a more natural and intuitive offset, that of a(1) = 1. - Robert G. Wilson v, Sep 22 2014 and modified Nov 28 2014 *)
Select[Range[10^3], PalindromeQ] (* Michael De Vlieger, Nov 27 2017 *)
nLP[cn_Integer]:=Module[{s,len,half,left,pal,fdpal},s=IntegerDigits[cn]; len=Length[s]; half=Ceiling[len/2]; left=Take[s,half]; pal=Join[left,Reverse[ Take[left,Floor[len/2]]]]; fdpal=FromDigits[pal]; Which[cn==9,11,fdpal>cn,fdpal,True,left=IntegerDigits[ FromDigits[left]+1]; pal=Join[left,Reverse[Take[left,Floor[len/2]]]]; FromDigits[pal]]]; NestList[nLP,0,100] (* Harvey P. Dale, Dec 10 2024 *)

is_A002113(n)=Vecrev(n=digits(n))==n \\ M. F. Hasler, Nov 17 2008, updated Apr 26 2014, Jun 19 2018

is(n)=n=digits(n);for(i=1,#n\2,if(n[i]!=n[#n+1-i],return(0))); 1 \\ Charles R Greathouse IV, Jan 04 2013

a(n)={my(d,i,r);r=vector(#digits(n-10^(#digits(n\11)))+#digits(n\11));n=n-10^(#digits(n\11));d=digits(n);for(i=1,#d,r[i]=d[i];r[#r+1-i]=d[i]);sum(i=1,#r,10^(#r-i)*r[i])} \\ David A. Corneth, Jun 06 2014

\\ recursive--feed an element a(n) and it gives a(n+1)
nxt(n)=my(d=digits(n));i=(#d+1)\2;while(i&&d[i]==9,d[i]=0;d[#d+1-i]=0;i--);if(i,d[i]++;d[#d+1-i]=d[i],d=vector(#d+1);d[1]=d[#d]=1);sum(i=1,#d,10^(#d-i)*d[i]) \\ David A. Corneth, Jun 06 2014

\\ feed a(n), returns n.
inv(n)={my(d=digits(n));q=ceil(#d/2);sum(i=1,q,10^(q-i)*d[i])+10^floor(#d/2)} \\ David A. Corneth, Jun 18 2014

inv_A002113(P)={P\(P=10^(logint(P+!P,10)\/2))+P} \\ index n of palindrome P = a(n), much faster than above: no sum is needed. - M. F. Hasler, Sep 09 2018

A002113(n,L=logint(n,10))=(n-=L=10^max(L-(n<11*10^(L-1)),0))*L+fromdigits(Vecrev(digits(if(nM. F. Hasler, Sep 11 2018

# edited by M. F. Hasler, Jun 19 2018
def A002113_list(nMax):
  mlist=[]
  for n in range(nMax+1):
     mstr=str(n)
     if mstr==mstr[::-1]:
        mlist.append(n)
  return mlist # Bill McEachen, Dec 17 2010

from itertools import chain
A002113 = sorted(chain(map(lambda x:int(str(x)+str(x)[::-1]),range(1,10**3)),map(lambda x:int(str(x)+str(x)[-2::-1]), range(10**3)))) # Chai Wah Wu, Aug 09 2014

from itertools import chain, count
A002113 = chain(k for k in count(0) if str(k) == str(k)[::-1])
print([next(A002113) for k in range(60)]) # Jan P. Hartkopf, Apr 10 2021

is_A002113 = lambda n: (s:=str(n))[::-1]==s # M. F. Hasler, May 23 2024

from math import log10, floor
def A002113(n):
  if n < 2: return 0
  P = 10**floor(log10(n//2)); M = 11*P
  s = str(n - (P if n < M else M-P))
  return int(s + s[-2 if n < M else -1::-1]) # M. F. Hasler, Jun 06 2024

[n for n in (0..515) if Word(n.digits()).is_palindrome()] # Peter Luschny, Sep 13 2018

def palQ(n: Int, b: Int = 10): Boolean = n - Integer.parseInt(n.toString.reverse) == 0
(0 to 999).filter(palQ()) // _Alonso del Arte, Nov 10 2019

A136522(a(n)) = 1.
A178788(a(n)) = 0 for n > 9. - Reinhard Zumkeller, Jun 30 2010
A064834(a(n)) = 0. - Reinhard Zumkeller, Sep 18 2013
a(n+1) = A262038(a(n)+1). - M. F. Hasler, Sep 09 2015
Sum_{n>=2} 1/a(n) = A118031. - Amiram Eldar, Oct 17 2020
a(n) = (floor(d(n)/(c(n)*9 + 1)))*10^A055642(d(n)) + A004086(d(n)) where b(n, k) = ceiling(log((n + 1)/k)/log(10)), c(n) = b(n, 2) - b(n, 11) and d(n) = (n - A086573(b(n*(2 - c(n)), 2) - 1)/2 - 1). - Alan Michael Gómez Calderón, Mar 11 2025

A261423 Largest palindrome <= n.

Original entry on oeis.org

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

N. J. A. Sloane, Aug 28 2015

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

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

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

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

# 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;

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 *)

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

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

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

A206913 Greatest binary palindrome <= n; the binary palindrome floor function.

Original entry on oeis.org

0, 1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 9, 9, 9, 9, 15, 15, 17, 17, 17, 17, 21, 21, 21, 21, 21, 21, 27, 27, 27, 27, 31, 31, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 45, 45, 45, 45, 45, 45, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 63, 63, 65, 65, 65, 65

Offset: 0

Hieronymus Fischer, Feb 13 2012

Also the greatest binary palindrome < n + 1;

For n > 0, a(n-1) is the greatest binary palindrome < n.

			a(0) = 0 since 0 is the greatest binary palindrome <= 0;
a(1) = 1 since 1 is the greatest binary palindrome <= 1;
a(2) = 1 since 1 is the greatest binary palindrome <= 2;
a(3) = 3 since 3 is the greatest binary palindrome <= 3.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A006995, A206915, A206920, A030301.

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

a206913 n = last $ takeWhile (<= n) a006995_list
-- Reinhard Zumkeller, Feb 27 2012

Let n > 2, p = 1 + 2*floor((n-1)/2), m = floor(log_2(p)), q = floor((m+1)/2), s = floor(log_2(p-2^q)),
F(x, r) = floor(x/2^q)*2^q + Sum_{k = 0...q - 1} (floor(x/2^(r-k)) mod 2)*2^k;
If F(p, m) <= n then a(n) = F(p, m), otherwise a(n) = F(p-2^q, s).
By definition: F(p, m) = floor(p/2^q)*2^q + A030101(p) mod 2^q; also: F(p-2^q, s) = floor((p-2^q)/2^q)*2^q + A030101(p-2^q) mod 2^q; [Edited and corrected by Hieronymus Fischer, Sep 08 2018]
a(n) = A006995(A206915(n));
a(n) = A006995(A206915(A206914(n+1))-1);
a(n) = A006995(A206916(A206914(n+1))-1).

A175298 Smallest number >=n whose binary representation is palindromic and has a 1 whenever the binary representation of n has a 1.

Original entry on oeis.org

0, 1, 3, 3, 5, 5, 7, 7, 9, 9, 15, 15, 15, 15, 15, 15, 17, 17, 27, 27, 21, 21, 31, 31, 27, 27, 27, 27, 31, 31, 31, 31, 33, 33, 51, 51, 45, 45, 63, 63, 45, 45, 63, 63, 45, 45, 63, 63, 51, 51, 51, 51, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 65, 65, 99, 99, 85, 85, 119, 119, 73

Offset: 0

Leroy Quet, Mar 24 2010

Old name: "Convert n to binary. OR each respective digit of binary n and binary A030101(n), where A030101(n) is the reversal of the order of the digits in the binary representation of n (given in decimal). a(n) is the decimal value of the result."
By "respective" digits of binary n and binary A030101(n), the rightmost digit of A030101(n) ( which is a 1) is OR'ed with the rightmost digit of n. A030101(n) is represented with the appropriate number of leading 0's.
This is the binary next-palindrome function, the base-2 analog of A262038. - N. J. A. Sloane, Dec 08 2015

			20 in binary is 10100. The reversal of the binary digits is 00101. So, from leftmost to rightmost respective digits, we OR 10100 and 00101: 1 OR 0 = 1. 0 OR 0 = 0. 1 OR 1 = 1. 0 OR 0 = 0. And 0 OR 1 = 1. So, 10100 OR 00101 is 10101, which is 21 in decimal. So a(20) = 21.

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Cf. A006995, A030101, A175297, A262038.

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

Table[f = IntegerDigits[x, 2]; f = f + Reverse[f]; FromDigits[ Table[If[Positive[f[[r]]], 1, 0], {r, 1, Length[f]}], 2], {x, STARTPOINT, ENDPOINT}] (* Dylan Hamilton, Oct 15 2010 *)
f[n_] := Block[{id = IntegerDigits[n, 2]}, FromDigits[ BitOr[ id, Reverse@id], 2]]; Array[f, 72] (* Robert G. Wilson v, Nov 07 2010 *)

Extended, with redundant initial entries included, by Dylan Hamilton, Oct 15 2010

Edited with new name and offset by N. J. A. Sloane, Dec 08 2015

A206914 Least binary palindrome >= n; the binary palindrome ceiling function.

Original entry on oeis.org

0, 1, 3, 3, 5, 5, 7, 7, 9, 9, 15, 15, 15, 15, 15, 15, 17, 17, 21, 21, 21, 21, 27, 27, 27, 27, 27, 27, 31, 31, 31, 31, 33, 33, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 51, 51, 51, 51, 51, 51, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 65, 65, 73, 73

Offset: 0

Hieronymus Fischer, Feb 15 2012

For n > 0 also the least binary palindrome > n - 1;

a(n+1) is the least binary palindrome > n

			a(0) = 0 since 0 is the least binary palindrome >= 0;
a(1) = 1 since 1 is the least binary palindrome >= 1;
a(2) = 3 since 3 is the least binary palindrome >= 2;
a(5) = 5 since 5 is the least binary palindrome >= 5;

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A206915, A206920, A006995.

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

a206914 n = head $ dropWhile (< n) a006995_list
-- Reinhard Zumkeller, Feb 27 2012

a(n) = A006995(A206916(n));
a(n) = A006995(A206916(A206913(n-1))+1);
a(n) = A006995(A206915(A206913(n-1))+1);

A265509 a(n) = largest base-2 palindrome m <= 2n+1 such that every base-2 digit of m is <= the corresponding digit of 2n+1; m is written in base 10.

Original entry on oeis.org

1, 3, 5, 7, 9, 9, 9, 15, 17, 17, 21, 21, 17, 27, 21, 31, 33, 33, 33, 33, 33, 33, 45, 45, 33, 51, 33, 51, 33, 51, 45, 63, 65, 65, 65, 65, 73, 73, 73, 73, 65, 65, 85, 85, 73, 73, 93, 93, 65, 99, 65, 99, 73, 107, 73, 107, 65, 99, 85, 119, 73, 107, 93, 127, 129, 129, 129, 129, 129, 129, 129, 129, 129, 129, 129, 129, 153

Offset: 0

N. J. A. Sloane, Dec 09 2015

A007088(a(n)) = A265510(n). - Reinhard Zumkeller, Dec 11 2015

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

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

Cf. A007088, A030308.

a265509 n = a265509_list !! n
a265509_list = f (tail a030308_tabf) [[]] where
   f (bs:_:bss) pss = y : f bss pss' where
     y = foldr (\d v -> 2 * v + d) 0 ys
     (ys:_) = dropWhile (\ps -> not $ and $ zipWith (<=) ps bs) pss'
     pss' = if bs /= reverse bs then pss else bs : pss
-- Reinhard Zumkeller, Dec 11 2015

ispal := proc(n) # test for base-b palindrome
local L, Ln, i;
global b;
    L := convert(n, base, b);
    Ln := nops(L);
    for i to floor(1/2*Ln) do
        if L[i] <> L[Ln + 1 - i] then return false end if
    end do;
    return true
end proc
# find max pal <= n and in base-b shadow of n, write in base 10
under10:=proc(n) global b;
local t1,t2,i,m,sw1,L2;
if n mod b = 0 then return(0); fi;
t1:=convert(n,base,b);
for m from n by -1 to 0 do
if ispal(m) then
   t2:=convert(m,base,b);
   L2:=nops(t2);
   sw1:=1;
   for i from 1 to L2 do
      if t2[i] > t1[i] then sw1:=-1; break; fi;
                      od:
   if sw1=1 then return(m); fi;
fi;
                       od;
end proc;
b:=2; [seq(under10(2*n+1),n=0..144)]; # Gives A265509
# find max pal <= n and in base-b shadow of n, write in base b
underb:=proc(n) global b;
local t1,t2,i,m,mb,sw1,L2;
if n mod b = 0 then return(0); fi;
t1:=convert(n,base,b);
for m from n by -1 to 0 do
if ispal(m) then
   t2:=convert(m,base,b);
   L2:=nops(t2);
   sw1:=1;
   for i from 1 to L2 do
      if t2[i] > t1[i] then sw1:=-1; break; fi;
                      od:
   if sw1=1 then mb:=add(t2[i]*10^(i-1), i=1..L2); return(mb); fi;
fi;
                       od;
end proc;
b:=2; [seq(underb(2*n+1),n=0..144)]; # Gives A265510

A265509 = FromDigits[Min /@ Transpose[{#, Reverse@#}], 2] &@IntegerDigits[2 # + 1, 2] & (* JungHwan Min, Aug 22 2016 *)

A262037 Replace the second half of digits of n with the first half in reverse order.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Sep 08 2015

This is related to the "palindromic floor" and "palindromic ceiling" functions A261423 and A262038: a(n) = A261423(n) iff the first half of digits reversed does not exceed the second half of digits (without considering the middle digit in case of an odd number of digits), and a(n) = A262038(n) in the opposite case, i.e., first half of digits reversed is greater than or equal to second half of digits.
a(n) is either equal to the palindromic floor A261423(n) (e.g., a(1234) = 1221), or to the palindromic ceiling A262038(n) (= next larger palindrome, e.g., a(1324)=1331). In this sense it can be seen as a "palindromic round" function. However, it does not always yield the closest of the two (a(1900) = 1991 but 1881 would be closer to 1900). The sequence A262040 which has this property would better merit the name of "palindromic round function".
This simple function can be used to construct the next larger or next smaller palindrome, A261423 and A262038: indeed, if a(n) has the required property (less than or greater than n) then it is already the desired result, otherwise the result is given by a(n +- 10^k), where k is half the number of digits of n.

			a(31) = 33 since the second half ("1") gets replaced by the first half ("3").
a(314) = 313 since the second half ("4") is replaced by the first half ("3"), the middle "1" being untouched.
a(3141) = 3113 since the second half (41) is replaced by the first half (31), reversed (13).
a(31415) = 31413 as above, the middle 4 being left untouched.
a(314156) = 314413. This is the first instance in these examples where a(n) differs from A261423(n), which would yield 313313 here.

Dominic McCarty, Table of n, a(n) for n = 0..10000

Cf. A002113, A261423, A262038, A262039, A262040.

f[n_] := Block[{d = IntegerDigits@ n}, FromDigits[Take[d, Ceiling[Length[d]/2]]~Join~Reverse@ Take[d, Floor[Length[d]/2]]]]; Table[f@ n, {n, 0, 120}] (* Michael De Vlieger, Sep 09 2015 *)

a(n,d=digits(n),m=sum(k=1,#d\2,d[k]*10^(k-1)))=n+m-n%10^(#d\2)

def A262037(n):
    s = str(n); h = s[:(len(s)+1)//2]; return int(h + h[-1-len(s)%2::-1])
print([A262037(n) for n in range(77)]) # Michael S. Branicky, Sep 15 2022

A262039 Nearest palindrome to n; in case of a tie choose the larger palindrome.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 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

M. F. Hasler, Sep 08 2015

In analogy to the numerical "round" function, we "round up" to the next larger palindrome A262038(n) if it is at the same distance or closer, else we "round down" to the next smaller palindrome A261423(n). See A262040 for a variant where the next smaller palindrome is chosen in case of equal distance.

			a(10) = 11 since we round up if the next smaller palindrome (here 9) is at the same distance, both 9 and 11 are here at distance 1 from n = 10.
a(16) = 11 since |16 - 11| = 5 is smaller than |16 - 22| = 6.
a(17) = 22 since |17 - 22| = 5 is smaller than |17 - 11| = 6.
a(27) = 22 since |22 - 27| = 5 is smaller than |27 - 33| = 6.
a(28) = 33 since |33 - 28| = 5 is smaller than |22 - 28| = 6, and so on.
a(100) = 101 because we round up again in this case, where 99 and 101 both are at distance 1 from n = 100.

Cf. A002113, A261423, A262037, A262038, A262040.

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d];
f[n_] := Block[{k = n}, While[Nand[palQ@ k, k > -1], k--]; k];
g[n_] := Block[{k = n}, While[! palQ@ k, k++]; k];
h[n_] := Block[{a = f@ n, b = g@ n}, Which[palQ@ n, n, (b - n) - (n - a) > 0, a, (b - n) - (n - a) <= 0, b]]; Table[h@ n, {n, 0, 73}] (* Michael De Vlieger, Sep 09 2015 *)

A262040 Nearest palindrome to n; in case of a tie choose the smaller palindrome.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 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

M. F. Hasler, Sep 08 2015

In contrast to A262039, here we "round down" to the next smaller palindrome A261423(n) if it is at the same distance or closer, else we "round up" to the next larger palindrome A262038(n).

			a(10) = 9 since we round down if the next larger palindrome (here 11) is at the same distance, both 9 and 11 are here at distance 1 from n = 10.
a(16) = 11 since |16 - 11| = 5 is smaller than |16 - 22| = 6.
a(17) = 22 since |17 - 22| = 5 is smaller than |17 - 11| = 6.
a(27) = 22 since |22 - 27| = 5 is smaller than |27 - 33| = 6.
a(28) = 33 since |33 - 28| = 5 is smaller than |22 - 28| = 6, and so on.
a(100) = 99 because we round down in this case, where 99 and 101 both are at distance 1 from n = 100.

Cf. A002113, A261423, A262037, A262038, A262039.

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d];
f[n_] := Block[{k = n}, While[Nand[palQ@ k, k > -1], k--]; k];
g[n_] := Block[{k = n}, While[! palQ@ k, k++]; k];
h[n_] := Block[{a = f@ n, b = g@ n}, Which[palQ@ n, n, (b - n) - (n - a) >= 0, a, (b - n) - (n - a) < 0, b]]; Table[h@ n, {n, 0, 73}] (* Michael De Vlieger, Sep 09 2015 *)

A265510 a(n) = largest base-2 palindrome m <= 2n+1 such that every base-2 digit of m is <= the corresponding digit of 2n+1; m is written in base 2.

Original entry on oeis.org

1, 11, 101, 111, 1001, 1001, 1001, 1111, 10001, 10001, 10101, 10101, 10001, 11011, 10101, 11111, 100001, 100001, 100001, 100001, 100001, 100001, 101101, 101101, 100001, 110011, 100001, 110011, 100001, 110011, 101101, 111111, 1000001, 1000001, 1000001, 1000001, 1001001, 1001001, 1001001, 1001001, 1000001, 1000001

Offset: 0

N. J. A. Sloane, Dec 09 2015

a(n) = A007088(A265509(n)). - Reinhard Zumkeller, Dec 11 2015

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

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

Cf. A007088.

a265510 = a007088 . a265509  -- Reinhard Zumkeller, Dec 11 2015

cp's OEIS Frontend

A002113 Palindromes in base 10.

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A261423 Largest palindrome <= n.

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A206913 Greatest binary palindrome <= n; the binary palindrome floor function.

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A175298 Smallest number >=n whose binary representation is palindromic and has a 1 whenever the binary representation of n has a 1.

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A206914 Least binary palindrome >= n; the binary palindrome ceiling function.

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A265509 a(n) = largest base-2 palindrome m <= 2n+1 such that every base-2 digit of m is <= the corresponding digit of 2n+1; m is written in base 10.

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