A261424 -id:A261424 - cp's OEIS frontend

A088601 Number of steps to reach 0 when iterating A261424(x) = x - (the largest palindrome less than x), starting at n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2

Offset: 1

Amarnath Murthy, Oct 13 2003

The sequence "minimum number of palindromes that sum up to n", A261675, coincides with this sequence up to a(301). But then a(302) = 3 since 302 = 292 + 9 + 1, whereas 302 = 111 + 191.
While it has been conjectured [proved by Cilleruelo & Luca, 2016 -Ed.] that every number can be represented as a sum of at most 3 palindromes, the terms of this sequence, which correspond to a greedy representation, can be larger than 3 (see A109326). For example, 1022 can be represented as 33 + 989, but a(1022) = 4, because the greedy decomposition gives 1022 = 1001 + 11 + 9 + 1. - Giovanni Resta, Aug 20 2015
Presumably this sequence is unbounded (compare A109326). - N. J. A. Sloane, Sep 02 2015
This sequence is unbounded. Let n(1) := 1. To construct n(j+1), take a natural number m with 10^m > n(j) and set n(j+1) := 10^(2m) + 1 + n(j). Then a(n(j)) = j. - Markus Sigg, Oct 26 2015
In A109326 an explicit formula for a smaller (conjectured sharp) upper bound was already given earlier. - M. F. Hasler, Sep 09 2018

			a(10) = 2: f(10) = 10-9 = 1, f(1) = 1-1 = 0, two steps.

Giovanni Resta, Table of n, a(n) for n = 1..10000
William D. Banks, Every natural number is the sum of forty-nine palindromes, arXiv:1508.04721 [math.NT], 2015 and INTEGERS 16 (2016) A3
Javier Cilleruelo, Florian Luca and Lewis Baxter, Every positive integer is a sum of three palindromes, arXiv:1602.06208 [math.NT], 2016-2017.
M. F. Hasler, Sum of palindromes, OEIS wiki, Sept. 2015
Markus Sigg, On a conjecture of John Hoffman regarding sums of palindromic numbers, arXiv:1510.07507 [math.NT], 2015.

Cf. A109326 gives index of first occurrence of n in this sequence ("greedy inverse").

Cf. A002113, A261132, A261422, A261423, A261424.

# From N. J. A. Sloane, Aug 28 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;
GA:=proc(n) global P,palfloor; local a,i,k;
a:=1; k:=n;
for i from 1 to 30 do
  if k-palfloor(k)=0 then return(a);
  else k:=k-palfloor(k); a:=a+1; fi;
od; end;
[seq(GA(n),n=0..200)];

Length@ NestWhileList[f, #, # > 0 &] & /@ Range@ 105 - 1 (* Michael De Vlieger, Oct 26 2015 *)

ispal(n) = my(d=digits(n)); Vecrev(d)==d;
fp(n) = {while(!ispal(n), n--); n;}
a(n) = {nb = 0; while (n, n -= fp(n); nb++); nb;} \\ Michel Marcus, Aug 20 2015
/* The above fp() is extremely inefficient already for mid-sized numbers. The PARI function A261423 should be preferred.*/

A088601(n)=for(i=1,oo,(n-=A261423(n))||return(i)) \\ M. F. Hasler, Sep 09 2018

def P(n):
    s = str(n); h = s[:(len(s)+1)//2]; return int(h + h[-1-len(s)%2::-1])
def A261423(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))
def A088601(n): return 0 if n == 0 else 1 + A088601(n - A261423(n))
print([A088601(n) for n in range(1, 106)]) # Michael S. Branicky, Jul 12 2021

a(n) < log_2(log_10(n)) + 3. - M. F. Hasler, Sep 09 2018

More terms from David Wasserman, Aug 11 2005

A002113 Palindromes in base 10.

Original entry on oeis.org

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

cp's OEIS Frontend

A088601 Number of steps to reach 0 when iterating A261424(x) = x - (the largest palindrome less than x), starting at n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A002113 Palindromes in base 10.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Haskell

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

PARI

PARI

PARI

Python

Python

Python

Python

Python

SageMath

Scala

Formula

A261423 Largest palindrome <= n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

A109326 Smallest positive number that requires n steps to be represented as a sum of palindromes using the greedy algorithm.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Python

Formula

Extensions

A303534 Amount by which n exceeds the largest binary palindrome less than or equal to n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs