A206913 -id:A206913 - cp's OEIS frontend

A006995 Binary palindromes: numbers whose binary expansion is palindromic.

0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 85, 93, 99, 107, 119, 127, 129, 153, 165, 189, 195, 219, 231, 255, 257, 273, 297, 313, 325, 341, 365, 381, 387, 403, 427, 443, 455, 471, 495, 511, 513, 561, 585, 633, 645, 693, 717, 765, 771, 819, 843

Offset: 1

N. J. A. Sloane, Robert G. Wilson v, Mira Bernstein, L. J. Upton

If b > 1 is a binary palindrome then both (2^(m+1) + 1)*b and 2^(m+1) + 2^m - b are also, where m = floor(log_2(b)). - Hieronymus Fischer, Feb 18 2012
Floor and ceiling: If d > 0 is any natural number, then A206913(d) is the greatest binary palindrome <= d and A206914(d) is the least binary palindrome >= d. - Hieronymus Fischer, Feb 18 2012
The greatest binary palindrome <= the n-th non-binary-palindrome is that binary palindrome with number A154809(n)-n+1. The corresponding formula identity is: A206913(A154809(n)) = A006995(A154809(n)-n+1). - Hieronymus Fischer, Mar 18 2012
From Hieronymus Fischer, Jan 23 2013: (Start)
The number of binary digits of a(n) is A070939(a(n)) = 1 + floor(log_2(n)) + floor(log_2(n/3)), for n > 1.
Also: A070939(a(n)) = A070939(n) + A070939(floor(n/3)) - 1, for n <> 2. (End)
Rajasekaran, Shallit, & Smith show that this is an additive basis of order 4. - Charles R Greathouse IV, Nov 06 2018

			a(3) = 3, since 3 = 11_2 is the 3rd symmetric binary number;
a(6) = 9, since 9 = 1001_2 is the 6th symmetric binary number.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, arXiv:2202.13694 [math.NT], 2022.
William D. Banks and Igor E. Shparlinski, Average Value of the Euler Function on Binary Palindromes, Bulletin Polish Acad. Sci. Math., Vol. 54 (2006), pp. 95-101, alternative link.
Manfred Madritsch and Stephan Wagner, A central limit theorem for integer partitions, Monatshefte für Mathematik, Vol. 161, No. 1 (2010), pp. 85-114, alternative link. doi:10.1007/s00605-009-0126-y.
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.
Aayush Rajasekaran, Using Automata Theory to Solve Problems in Additive Number Theory, MS thesis, University of Waterloo, 2018.
Aayush Rajasekaran, Jeffrey Shallit and Tim Smith, Sums of Palindromes: an Approach via Nested-Word Automata, preprint arXiv:1706.10206 [cs.FL], Jun 30 2017.
Aayush Rajasekaran, Jeffrey Shallit and Tim Smith, Additive Number Theory via Automata Theory, Theory of Computing Systems, Vol. 64 (2020), pp. 542-567.
L. J. Upton, Letter to N. J. A. Sloane with attachments, Oct. 1993.
Index entries for sequences that are an additive basis, order 4.
Index entries for sequences related to binary expansion of n
Index entries for sequences related to palindromes

See A057148 for the binary representations.
Cf. A178225, A005408, A164126, A154809 (complement).
Cf. A002113, A048700, A048701, A206913, A206914, A244162.
Even numbers that are not the sum of two terms: A241491, A261678, A262556.
Cf. A145799.
Primes: A016041 and A117697.
Cf. A000051 (a subsequence).

a006995 n = a006995_list !! (n-1)
a006995_list = 0 : filter ((== 1) . a178225) a005408_list
-- Reinhard Zumkeller, Oct 21 2011

[n: n in [0..850] | Intseq(n,2) eq Reverse(Intseq(n,2))];  // Bruno Berselli, Aug 29 2011

dmax:= 15; # to get all terms with at most dmax binary digits
revdigs:= proc(n)
  local L, Ln, i;
  L:= convert(n,base,2);
  Ln:= nops(L);
  add(L[i]*2^(Ln-i),i=1..Ln);
end proc;
A:= {0,1}:
for d from 2 to dmax do
  if d::even then
    A:= A union {seq(2^(d/2)*x + revdigs(x),x=2^(d/2-1)..2^(d/2)-1)}
  else
    m:= (d-1)/2;
    B:={seq(2^(m+1)*x + revdigs(x),x=2^(m-1)..2^m-1)};
    A:= A union B union map(`+`,B,2^m)
  fi
od:
A;  # Robert Israel, Aug 17 2014

palQ[n_Integer, base_Integer] := Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], palQ[ #, 2]&]
Select[ Range[0, 1000], # == IntegerReverse[#, 2] &] (* Robert G. Wilson v, Feb 24 2018 *)
Select[Range[0, 1000], PalindromeQ[IntegerDigits[#, 2]]&] (* Jean-François Alcover, Mar 01 2018 *)

for(n=0,999,n-subst(Polrev(binary(n)),x,2)||print1(n,",")) \\ Thomas Buchholz, Aug 16 2014

for(n=0,10^3, my(d=digits(n,2)); if(d==Vecrev(d), print1(n,", "))); \\ Joerg Arndt, Aug 17 2014

is_A006995(n)=Vecrev(n=binary(n))==n \\ M. F. Hasler, Feb 23 2018

A006995(n,m=logint(n,2),c=1<<(m-1),a,d)={if(n>=3*c,a=n-3*c;d=2*c^2,a=n-2*c;n%2*c+d=c^2)+sum(k=1,m-2^(n<3*c),if(bittest(a,m-1-k),1<>k))+(n>2)} \\ Based on Fischer's smalltalk program. - M. F. Hasler, Feb 23 2018

from itertools import count, islice, product
def bin_pals(): # generator of binary palindromes in base 10
    yield from [0, 1]
    digits, midrange = 2, [[""], ["0", "1"]]
    for digits in count(2):
        for p in product("01", repeat=digits//2-1):
            left = "1"+"".join(p)
            for middle in midrange[digits%2]:
                yield int(left + middle + left[::-1], 2)
print(list(islice(bin_pals(), 58))) # Michael S. Branicky, Jan 09 2023

def A006995(n):
    if n == 1: return 0
    a = 1<<(l:=n.bit_length()-2)
    m = a|(n&a-1)
    return (m<Chai Wah Wu, Jun 10 2024

def palgenbase2(): # generator of palindromes in base 2
    yield 0
    x, n, n2 = 1, 1, 2
    while True:
        for y in range(n,n2):
            s = format(y,'b')
            yield int(s+s[-2::-1],2)
        for y in range(n,n2):
            s = format(y,'b')
            yield int(s+s[::-1],2)
        x += 1
        n *= 2
        n2 *= 2 # Chai Wah Wu, Jan 07 2015

[n for n in (0..843) if Word(n.digits(2)).is_palindrome()] # Peter Luschny, Sep 13 2018

A006995
"Answer the n-th binary palindrome
(nonrecursive implementation)"
| m n a b c d k2 |
n := self.
n = 1 ifTrue: [^0].
n = 2 ifTrue: [^1].
m := n integerFloorLog: 2.
c := 2 raisedToInteger: m - 1.
n >= (3 * c)
  ifTrue:
   [a := n - (3 * c).
   d := 2 * c * c.
   b := d + 1.
   k2 := 1.
   1 to: m - 1
    do:
     [:k |
     k2 := 2 * k2.
     b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]]
  ifFalse:
   [a := n - (2 * c).
   d := c * c.
   b := d + 1 + (n \\ 2 * c).
   k2 := 1.
   1 to: m - 2
    do:
     [:k |
     k2 := 2 * k2.
     b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]].
^b // by Hieronymus Fischer, Feb 15 2013

A178225(a(n)) = 1; union of A048700 and A048701. - Reinhard Zumkeller, Oct 21 2011
From Hieronymus Fischer, Dec 31 2008, Jan 10 2012, Feb 18 2012: (Start)
Written as a decimal, a(10^n) has 2*n digits. For n > 1, the decimal expansion of a(10^n) starts with 22..., 23... or 24...:
a(1000) = 249903,
a(10^4) = 24183069,
a(10^5) = 2258634081,
a(10^6) = 249410097687,
a(10^7) = 24350854001805,
a(10^8) = 2229543293296319,
a(10^9) = 248640535848971067,
a(10^10)= 24502928886295666773.
Inequality: (2/9)*n^2 < a(n) < (1/4)*(n+1)^2, if n > 1.
lim sup_{n -> oo} a(n)/n^2 = 1/4, lim inf_{n -> oo} a(n)/n^2 = 2/9.
For n >= 2, a(2^n-1) = 2^(2n-2) - 1; a(2^n) = 2^(2n-2) + 1;
   a(2^n+1) = 2^(2n-2) + 2^(n-1) + 1; a(2^n + 2^(n-1)) = 2^(2n-1) + 1.
Recursion for n > 2: a(n) = 2^(2k-q) + 1 + 2^p*a(m), where k = floor(log_2(n-1)), and p, q and m are determined as follows:
Case 1: If n = 2^(k+1), then p = 0, q = 0, m = 1;
Case 2: If 2^k < n < 2^k+2^(k-1), then p = k-floor(log_2(i))-1 with i = n-2^k, q = 2, m = 2^floor(log_2(i)) + i;
Case 3: If n = 2^k + 2^(k-1), then p = 0, q = 1, m = 1;
Case 4: If 2^k + 2^(k-1) < n < 2^(k+1), then p = k-floor(log_2(j))-1 with j = n-2^k-2^(k-1), q = 1, m = 2*2^floor(log_2(j))+j.
Non-recursive formula:
Let n >= 3, m = floor(log_2(n)), p = floor((3*2^(m-1)-1)/n), then
a(n) = 2^(2*m-1-p) + 1 + p*(1-(-1)^n)*2^(m-1-p) + sum_{k=1 .. m-1-p} (floor((n-(3-p)*2^(m-1))/2^(m-1-k)) mod 2)*(2^k+2^(2*m-1-p-k)). [Typo at the last exponent of the third sum term eliminated by the author, Sep 05 2018]
a(n) = 2^(2*m-2) + 1 + 2*floor((n-2^m)/2^(m-1)) + 2^(m-1)*floor((1/2)*min(n+1-2^m,2^(m-1)+1)) + 3*2^(m-1)*floor((1/2)*max(n+1-3*2^(m-1),0)) + 3*sum_{j=2 .. m-1} floor((n+2^(j-1)-2^m)/2^j)*2^(m-j). [Seems correct for n > 3. - The Editors]
Inversion formula: The index of any binary palindrome b = a(n) > 0 is n = palindromicIndex(b) = ((5-(-1)^m)/2 + Sum_{k=1..[m/2]} ([b/2^k] mod 2)/2^k)*2^[m/2], where [.] = floor(.) and m = [log_2(b)].
(End)
G.f.: g(x) = x^2 + 3x^3 + sum_{j=1..oo}( 3*2^j*(1-x^floor((j+1)/2))/(1-x)*x^((1/2)-floor((j+1)/2)) + f_j(x) - f_j(1/x))*x^(2*2^floor(j/2)+3*2^floor((j-1)/2)-(1/2)), where the f_j(x) are defined as follows:
  f_1(x) = x^(1/2), and for j > 1,
  f_j(x) = x^(1/2)*sum_{i=0..2^floor((j-1)/2)-1}((3+(1/2)*sum_{k=1..floor((j-1)/2)}(1-(-1)^floor(2i/2^k))*b(j,k))*x^i), where b(j,k) = 2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k > 1, and b(j,1) = (2+(-1)^j)*2^(floor((j-1)/2)+1).  - Hieronymus Fischer, Apr 04 2012
A044051(n) = (a(n)+1)/2 for n > 0. - Reinhard Zumkeller, Apr 20 2015
A145799(a(n)) = a(n). - Reinhard Zumkeller, Sep 24 2015
Sum_{n>=2} 1/a(n) = A244162. - Amiram Eldar, Oct 17 2020

Edited and extended by Hieronymus Fischer, Feb 21 2012

Edited by M. F. Hasler, Feb 23 2018

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

A262038 Least palindrome >= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 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, 77, 77, 77, 77, 77, 77, 77, 77

Offset: 0

M. F. Hasler, Sep 08 2015

Could be called nextpalindrome() in analogy to the nextprime() function A007918. As for the latter (A151800), there is the variant "next strictly larger palindrome" which equals a(n+1), and thus differs from a(n) iff n is a palindrome; see PARI code.

Might also be called palindromic ceiling function in analogy to the name "palindromic floor" proposed for A261423.

M. F. Hasler, 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, A261423, A262037.

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

a262038 n = a262038_list !! n
a262038_list = f 0 a002113_list where
   f n ps'@(p:ps) = p : f (n + 1) (if p > n then ps' else ps)
-- Reinhard Zumkeller, Sep 16 2015

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Table[k = n; While[! palQ@ k, k++]; k, {n, 0, 80}] (* Michael De Vlieger, Sep 09 2015 *)

{A262038(n, d=digits(n), p(d)=sum(i=1, #d\2, (10^(i-1)+10^(#d-i))*d[i],if(bittest(#d,0),10^(#d\2)*d[#d\2+1])))= for(i=(#d+3)\2,#d,d[i]>d[#d+1-i]&&break;(d[i]9||return(p(d));d[i]=0);10^#d+1} \\ For a function "next strictly larger palindrome", delete the i==#d and n<10... part. - M. F. Hasler, Sep 09 2015

def A262038(n):
    sl = len(str(n))
    l = sl>>1
    if sl&1:
        w = 10**l
        n2 = w*10
        for y in range(n//(10**l),n2):
            k, m = y//10, 0
            while k >= 10:
                k, r = divmod(k,10)
                m = 10*m + r
            z = y*w + 10*m + k
            if z >= n:
                return z
    else:
        w = 10**(l-1)
        n2 = w*10
        for y in range(n//(10**l),n2):
            k, m = y, 0
            while k >= 10:
                k, r = divmod(k,10)
                m = 10*m + r
            z = y*n2 + 10*m + k
            if z >= n:
                return z # Chai Wah Wu, Sep 14 2022

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 *)

A178225 Characteristic function of A006995 (binary palindromes).

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0

Offset: 0

Jeremy Gardiner, May 23 2010

a(n)=1 if n is in A006995, a(n)=0 otherwise.
For n<43, identical to parity of A175096.
Comment by Franklin T. Adams-Watters: (Start)
Any permutation of the runs of n gives another such permutation when reversed. This pairs up all non-palindromic permutations of the runs of n. Thus the parity of A175096(n) is the parity of the number of palindromic run-permutations of n. For small n, this is 1 when n is a binary palindrome, and 0 otherwise.
The first exception is 43, binary 101011, which has a nontrivial palindromic run-permutation 45, binary 101101. Another kind of exception occurs first for n = 365, binary 101101101, which is a palindrome, but has another palindromic run-permutation 427, binary 110101011. (End)
Given an index n such that a(n)=1, then the following A164126(A206915(n))-1 terms will be 0. n'=A164126(A206915(n)) is the next term with a(n')=1. Therefore, if we subtract 1 from each term of A164126, we get the sequence of run lengths of 0's. - Hieronymus Fischer, Feb 19 2012.
Given an index n such that a(n)=0, then p=A206913(n) is the greatest index pA206914(n) is the least index q>n such that a(q)=1, which implies a(k)=0 for all k with n<=kHieronymus Fischer, Feb 19 2012.
Binary palindromes are distributed symmetrically with respect to threefold multiples of powers of 2. This becomes obvious by the generating function g(x) below. Example for the resulting factors of x^(3*2^5)=x^96: the factors are x^q and x^(-q) for q=3,11,23,31. Thus, the palindromes are 96+3, 96-3, 96+11, 96-11, 96+23, 96-23, 96+31, 96-31. The respective number of palindromes with this property is 2^(floor(m/2)), where m is the exponent of the corresponding power of 2. - Hieronymus Fischer, Apr 04 2012

			a(3)=1, since 3 is binary palindromic;
a(4)=0, since 4 is not palindromic.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for characteristic functions

Cf. A006995, A175096, A178226

Cf. A136522. See A206915 for the partial sums.

a178225 n = fromEnum $ n == a030101 n  -- Reinhard Zumkeller, Oct 21 2011

A178225[n_]:=Boole[PalindromeQ[IntegerDigits[n,2]]];
Array[A178225,100,0] (* Paolo Xausa, Oct 15 2023 *)

a(n) = my(b=binary(n)); b == Vecrev(b); \\ Michel Marcus, Feb 13 2019

a187225 = lambda n: int(bin(n)[2:] == bin(n)[:1:-1]) # David Radcliffe, May 05 2023

a(A006995(n)) = 1; a(A154809(n)) = 0. - Reinhard Zumkeller, Oct 21 2011
a(n) = if A030101(n) = n then 1, otherwise 0. - Reinhard Zumkeller, Jan 17 2012
a(n) = 1 - (A206916(n) - A206915(n)). - Hieronymus Fischer, Feb 18 2012
G.f.: g(x) = 1 + x + x^3 + Sum{j>=1} x^(3*2^j)*(f_j(x)+f_j(1/x)), where the f_j(x) are defined as follows:
  f_1(x)=x, and for j > 1,
  f_j(x) = x^3*Product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k) = 2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k > 1, and b(j,1) = (2+(-1)^j)*2^(floor((j-1)/2)+1). The first explicit terms of this g.f. are
  g(x) = 1 + x + x^3 + (f_1(x) + f_1(1/x))*x^6 + (f_2(x) + f_2(1/x))*x^12 + (f_3(x)+f_3(1/x))*x^24 + (f_4(x) + f_4(1/x))*x^48 + (f_5(x) + f_5(1/x))*x^96 + ... = 1 + x + x^3 + (x+1/x)*x^6 + (x^3+1/x^3)*x^12 + (x^3*(1+x^4) + (1+1/x^4)/x^3)*x^24 + (x^3*(1+x^12) + (1+1/x^12)/x^3)*x^48 + (x^3*(1+x^8)(1+x^20) + (1+1/x^20)(1+1/x^8)/x^3)*x^96 + ...  - Hieronymus Fischer, Apr 02 2012

A154809 Numbers whose binary expansion is not palindromic.

Original entry on oeis.org

2, 4, 6, 8, 10, 11, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 29, 30, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88

Offset: 1

Omar E. Pol, Jan 24 2009

Complement of A006995.
The (a(n)-n+1)-th binary palindrome equals the greatest binary palindrome <= a(n). The corresponding formula identity is: A006995(a(n)-n+1)=A206913(a(n)). - Hieronymus Fischer, Mar 18 2012
A145799(a(n)) < a(n). - Reinhard Zumkeller, Sep 24 2015

			a(1)=2, since 2 = 10_2 is not binary palindromic.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006995, A154810, A164861, A206913, A206915.

Cf. A145799.

a154809 n = a154809_list !! (n-1)
a154809_list = filter ((== 0) . a178225) [0..]

[n: n in [0..600] | not (Intseq(n, 2) eq Reverse(Intseq(n, 2)))]; // Vincenzo Librandi, Jul 05 2015

ispali:= proc(n) local L;
L:= convert(n,base,2);
ListTools:-Reverse(L)=L
end proc:
remove(ispali, [$1..1000]); # Robert Israel, Jul 05 2015

palQ[n_Integer, base_Integer]:=Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], ! palQ[#, 2] &] (* Vincenzo Librandi, Jul 05 2015 *)

isok(n) = binary(n) != Vecrev(binary(n)); \\ Michel Marcus, Jul 05 2015

def A154809(n):
    def f(x): return n+(x>>(l:=x.bit_length())-(k:=l+1>>1))-(int(bin(x)[k+1:1:-1],2)>(x&(1<Chai Wah Wu, Jul 24 2024

A030101(n) != n. - David W. Wilson, Jun 09 2009
A178225(a(n)) = 0. - Reinhard Zumkeller, Oct 21 2011
From Hieronymus Fischer, Feb 19 2012 and Mar 18 2012: (Start)
Inversion formula: If d is any number from this sequence, then the index number n for which a(n)=d can be calculated by n=d+1-A206915(A206913(d)).
Explicitly: Let p=A206913(d), m=floor(log_2(p)) and p>2, then: a(n)=d+1+(((5-(-1)^m)/2) + sum(k=1...floor(m/2)|(floor(p/2^k) mod 2)/2^k))*2^floor(m/2).
Example 1: d=1000, A206913(d)=975, A206915(975)=62, hence n=1001-62=939.
Example 2: d=10^6, A206913(d)=999471, A206915(999471)=2000, hence n=1000001-2000=998001.
Recursion formulas:
a(n+1)=a(n)+1+A178225(a(n)+1)
Also:
  Case 1: a(n+1)=a(n)+2, if A206914(a(n))=a(n)+1;
  Case 2: a(n+1)=a(n)+1, else.
Also:
  Case 1: a(n+1)=a(n)+1, if A206914(a(n))>a(n)+1;
  Case 2: a(n+1)=a(n)+2, else.
Iterative calculation formula:
Let f(0):=n+1, f(j):=n-1+A206915(A206913(f(j-1)) for j>0; then a(n)=f(j), if f(j)=f(j-1). The number of necessary steps is typically <4 and is limited by O(log_2(n)).
Example 3: n=1000, f(0)=1001, f(1)=1061, f(2)=1064=f(3), hence a(1000)=1064.
Example 4: n=10^6, f(0)=10^6+1, f(1)=1001999, f(2)=1002001=f(3), hence a(10^6)=1002001.
Formula identity:
a(n) = n-1 + A206915(A206913(a(n))).
(End)

Extended by Ray Chandler, Mar 14 2010

A206915 The index (in A006995) of the greatest binary palindrome <= n; also the 'lower inverse' of A006995.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Hieronymus Fischer, Feb 15 2012

The greatest m such that A006995(m)<= n;
The number of binary palindromes <= n;
n is palindromic iff a(n)=A206916(n);
a(n) is the number of the binary palindrome A206913(n);
if n is a binary palindrome, then A006995(a(n))=n, so a(n) is 'inverse' with respect to A006995.
Partial sums of the binary palindromic characteristic function A178225.

			a(1)=2 since 2 is the index number of the greatest binary palindrome <= 1;
a(5)=4 since there are only 4 binary palindromes (namely 0,1,3 and 5) which are less than or equal to 5;
a(10)=6 since A006995(6)=9<=10, but A006995(7)=15>10, and so that, 6 is the index number of greatest binary palindrome <= 10;

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A006995, A206914, A206916, A206920.

A178225[n_]:=Boole[PalindromeQ[IntegerDigits[n,2]]];
Accumulate[Array[A178225,100,0]] (* Paolo Xausa, Oct 15 2023 *)

def A206915(n):
    l = n.bit_length()
    k = l+1>>1
    return (n>>l-k)-(int(bin(n)[k+1:1:-1] or '0',2)>(n&(1<Chai Wah Wu, Jul 24 2024

a(n) = max(m | A006995(m) <= n);
a(A006995(n)) = n;
A006995(a(n)) <= n, equality holds true iff n is a binary palindrome;
Let p = A206913(n), m = floor(log_2(p)) and p>2, then:
a(n) = (((5-(-1)^m)/2) + sum_{k=1..floor(m/2)} (floor(p/2^k) mod 2)/2^k)) * 2^floor(m/2).
a(n) = (1/2)*((6-(-1)^m)*2^floor(m/2) - 1 - sum_{k=1..floor(m/2)} (-1)^floor(p/2^k) * 2^(floor(m/2)-k))).
a(n) = (5-(-1)^m) * 2^floor(m/2)/2 - 3*sum_{k=2..floor(m/2)} (floor(p/2^k) * 2^floor(m/2)/2^k) + (floor(p/2) * 2^floor(m/2)/2 - 2*floor((p/2) * 2^floor(m/2)) * floor((m-1)/m+1/2).
Partial sums S(n) = sum_{k=0..n} a(k):
S(n) = (n+1)*a(n) - A206920(a(n)).
G.f.: g(x) = (1+x+x^3+sum_{j>=1} x^(3*2^j)*(f_j(x)+f_j(1/x)))/(1-x), where the f_j(x) are defined as follows:
  f_1(x) = x, and for j>1,
  f_j(x) = x^3*product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k)=2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k>1, and b(j,1)=(2+(-1)^j)*2^(floor((j-1)/2)+1).

cp's OEIS Frontend

A006995 Binary palindromes: numbers whose binary expansion is palindromic.

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A261423 Largest palindrome <= n.

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A262038 Least palindrome >= n.

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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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