A154809 -id:A154809 - 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

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

A206923 Number of bisections of the n-th binary palindrome bit pattern until the result is not palindromic.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Hieronymus Fischer, Mar 12 2012

Let k=1, p(1)=A006995(n) and m(1)=number of bits in p(1); if p(k) is a binary palindrome > 1 then iterate k=k+1, m(k)=floor((m(k-1)+1)/2), p(k)=leftmost m(k) bits of p(k-1); else set a(n)=k endif.

			a(1)=a(2)=1, since A006995(1)=0 and A006995(2)=1;
a(5)=3, since A006995(5)=7=111_2 and so the iteration is 11==>11==>1;
a(9)=2, since A006995(9)=21=10101_2 and so the iteration is 10101==>101;
a(13)=2, since A006995(13)=45=101101_2 and so the iteration is 101101==>101;
a(15)=4, since A006995(15)=63=111111_2 and so the iteration is 111111==>111==>11==>1;
a(37)=3, since A006995(37)=341=101010101_2 and so the iteration is 101010101==>10101==>101;

A006995, A206915, A178225, A154809, A206924, A206925.

/* quasi-C program fragment, omitting formal details, n>1 */
p=n;
p1=n+1;
k=0;
while (A178225(p)==1) && (p != p1)
{
  p1=p;
  k++;
  m=int(log(p)/log(2));
  p=int(p/2^int((m+1)/2));
}
return k;

Recursion: define f(x)=floor(A006995(x)/2^floor(floor(log_2(A006995(x))+1)/2)), for x=1,2,3,...
  Case 1: a(n)=1+a(A206915(f(n))), if f(n) is a binary palindrome;
  Case 2: a(n)=1, else.
Formally: a(n)=if (A178225(f(n))==1) then a(A206915(f(n)))+1 else 1.

A145799 a(n) = the largest integer that is an (odd) palindrome when represented in binary and that occurs in the binary representation of n.

Original entry on oeis.org

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

Offset: 1

Leroy Quet, Oct 19 2008

The binary expansion of a(n) is the largest (odd) palindrome that appears as a substring of the binary expansion of n. Nonzero binary palindromes are necessarily odd (see A006995).
For n = 2^k, a(n) = 1 is the largest binary palindrome in the binary representation of n.
a(2^k*A006995(n)) = A006995(n). - Ray Chandler, Oct 26 2008
a(m) = m iff m is a palindrome: a(A006995(n)) = A006995(n), a(A154809(n)) < A154809(n). - Reinhard Zumkeller, Sep 24 2015

			20 in binary is 10100. The largest binary palindrome included in this binary representation is 101, which is 5 in decimal. So a(20) = 5.

Reinhard Zumkeller, Table of n, a(n) for n = 1..16384

Cf. A006995, A145800.

Cf. A154809.

a145799 = maximum . map (foldr (\b v -> 2 * v + b) 0) .
                    filter (\bs -> bs == reverse bs && head bs == 1) .
                    substr . bin where
   substr [] = []
   substr us'@(_:us) = sub us' ++ substr us where
      sub [] = []; sub (v:vs) = [v] : [v : ws | ws <- sub vs ]
   bin 0 = []; bin n = b : bin n' where (n', b) = divMod n 2
-- Reinhard Zumkeller, Sep 24 2015

Block[{nn = 87, s}, s = Reverse@ Select[IntegerDigits[#, 2] & /@ Range[2^Log2@ nn], PalindromeQ]; Table[With[{d = IntegerDigits[n, 2]}, FromDigits[#, 2] &@ SelectFirst[s, SequenceCount[d, #] > 0 &]], {n, nn}]] (* Michael De Vlieger, Sep 23 2017 *)

Extended by Ray Chandler, Oct 26 2008

A164861 Odd positive integers that are not palindromes when written in binary.

Original entry on oeis.org

11, 13, 19, 23, 25, 29, 35, 37, 39, 41, 43, 47, 49, 53, 55, 57, 59, 61, 67, 69, 71, 75, 77, 79, 81, 83, 87, 89, 91, 95, 97, 101, 103, 105, 109, 111, 113, 115, 117, 121, 123, 125, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151, 155, 157, 159, 161, 163, 167

Offset: 1

Leroy Quet, Aug 28 2009

These are the odd members of A154809.

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

Cf. A154809.

a164861 n = a164861_list !! (n-1)
a164861_list = filter ((== 0) . a178225) a005408_list
-- Reinhard Zumkeller, Oct 21 2011

npbQ[n_]:=Module[{idn2=IntegerDigits[n,2]},idn2!=Reverse[idn2]]; Select[ Range[1,201,2],npbQ] (* Harvey P. Dale, May 17 2012 *)

A178225(a(n)) * (1 - A000035(a(n))) = 0. [Reinhard Zumkeller, Oct 21 2011]

Extended by Ray Chandler, Mar 14 2010

A272670 Numbers whose binary expansion is not palindromic but which when reversed and leading zeros omitted, does form a palindrome.

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 30, 32, 34, 36, 40, 42, 48, 54, 56, 60, 62, 64, 66, 68, 72, 80, 84, 90, 96, 102, 108, 112, 120, 124, 126, 128, 130, 132, 136, 144, 146, 160, 168, 170, 180, 186, 192, 198, 204, 214, 216, 224, 238, 240, 248, 252, 254

Offset: 1

N. J. A. Sloane, May 19 2016

Decimal interpretation of A273245. Twice A057890.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A006995, A057890, A273245, A273329, subsequence of A154809.

isPal := proc(L::list)
    local i;
    for i from 1 to nops(L)/2 do
        if op(i,L) <> op(-i,L) then
            return false;
        end if;
    end do:
    true ;
end proc:
isA272670 := proc(n)
    local bdgs ;
    bdgs := convert(n,base,2) ;
    if isPal(bdgs) then
        return false;
    else
        A000265(n) ;
        bdgs := convert(%,base,2) ;
        isPal(bdgs) ;
    end if;
end proc:
for n from 1 to 500 do
    if isA272670(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, May 20 2016

A272670_list = [n for n in range(1,10**4) if bin(n)[2:] != bin(n)[:1:-1] and bin(n)[2:].rstrip('0') == bin(n)[:1:-1].lstrip('0')] # Chai Wah Wu, May 21 2016

A154810 Nonpalindromic numbers with binary digits only.

Original entry on oeis.org

10, 100, 110, 1000, 1010, 1011, 1100, 1101, 1110, 10000, 10010, 10011, 10100, 10110, 10111, 11000, 11001, 11010, 11100, 11101, 11110, 100000, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101110, 101111

Offset: 1

Omar E. Pol, Jan 24 2009

A154809 written in base 2.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A007088, A029742, A057148, A154809.

Map[FromDigits, Select[IntegerDigits[Range[50], 2], !PalindromeQ[#] &]] (* Paolo Xausa, Jul 24 2024 *)

def A154810(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

a(n) = A007088(A154809(n)). - Michel Marcus, Jul 24 2024

Extended by Ray Chandler, Mar 14 2010

A206921 Rank of the n-th binary palindrome. The minimal number of iterations A206915(A206915(...A206915(A006995(n))...)) such that the result is not a binary palindrome, a(3)=1.

Original entry on oeis.org

2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1

Offset: 1

Hieronymus Fischer, Mar 12 2012

The number of iterations such that A006995(n) = A006995(A006995(A006995(...(A206922(n))...))) [For n<>3].

			a(1)=2, since A006995(1)=0=A006995(A006995(2)) [==> 2 iterations; 2 is not a binary palindrome];
a(3)=1 by definition;
a(4)=1, since A006995(4)=5=A006995(4) [==> 1 iteration; 4 is not a binary palindrome];
a(7)=3, since A006995(7)=15=A006995(A006995(A006995(4))) [==> 3 iterations; 4 is not a binary palindrome];

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A006995, A206922, A178225, A206915, A154809.

/* C program fragment, omitting formal details, n!=3 */
k=0;
p=A006995(n);
while A178225(p)==1
{
  k++;
  p=A206915(p);
}
return k;

up_to = 65537;
A178225(n) = (Vecrev(n=binary(n))==n);
A206915list(up_to) = { my(v=vector(up_to+1), s=0); for(n=1,up_to+1,s += A178225(n-1); v[n] = s); (v); };
v206915 = A206915list(up_to);
A206915(n) = v206915[1+n];
A206921(n) = if((3==n)||!A178225(n),1,1+A206921(A206915(n))); \\ Antti Karttunen, Nov 14 2018

a(n)=k, where k can be determined by the following iteration: set k=0, p(0)=A006995(n). Repeat while A178225(p(k))==1, set k=k+1, p(k)=A206915(p(k-1)) end repeat [for n<>3].
Recursion for n<>3:
  Case 1: a(n)=1, if n is not a binary palindrome;
  Case 2: a(n)=a(A206915(n))+1, else.
Formally: a(n)=if (A178225(n)==0) then 1 else a(A206915(n))+1

cp's OEIS Frontend

A178226 Characteristic function of A154809 (numbers that are not binary palindromes).

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A175392 a(n) is the smallest positive integer that, when written in binary, occurs in binary A154809(n) but not in binary A030101(A154809(n)).

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A006995 Binary palindromes: numbers whose binary expansion is palindromic.

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A178225 Characteristic function of A006995 (binary palindromes).

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A206923 Number of bisections of the n-th binary palindrome bit pattern until the result is not palindromic.

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A145799 a(n) = the largest integer that is an (odd) palindrome when represented in binary and that occurs in the binary representation of n.

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