A178225 -id:A178225 - cp's OEIS frontend

A048700 Binary palindromes of odd length (written in base 10).

1, 5, 7, 17, 21, 27, 31, 65, 73, 85, 93, 99, 107, 119, 127, 257, 273, 297, 313, 325, 341, 365, 381, 387, 403, 427, 443, 455, 471, 495, 511, 1025, 1057, 1105, 1137, 1161, 1193, 1241, 1273, 1285, 1317, 1365, 1397, 1421, 1453, 1501, 1533, 1539, 1571, 1619, 1651

Offset: 1

Antti Karttunen, Mar 07 1999

Note: you get A006995 (all binary palindromes) if you take (after zero) alternatively 2^n (starting from 2^0 = 1) terms from A048700 and as many from A048701 and then each time, twice as many from both.
A178225(a(n)) = 1. - Reinhard Zumkeller, Oct 21 2011
Comment from Altug Alkan, Dec 03 2015: (Start)
a(6*k) is divisible by 9 for k > 0.
a(3*k+(-1)^k-2) is divisible by 3 for k > 1.
The minimum value of a(n+1) - a(n) occurs when n = 2.
A014551(n) appears in this sequence for n > 0. (End)

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

Cf. A048701 (binary palindromes of even length), A002113 (decimal palindromes), A006995 (all binary palindromes).

Cf. also A178225.

import Data.Set (singleton, deleteFindMin, insert)
import Data.List (unfoldr)
a048700 n = a048700_list !! (n-1)
a048700_list = f 1 $ singleton 1 where
   f z s = m : f (z+1) (insert (c 0) (insert (c 1) s')) where
     c d = foldl (\v d -> 2 * v + d) 0 $ (reverse b) ++ [d] ++ b
     b = unfoldr
         (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2) z
     (m,s') = deleteFindMin s
-- Reinhard Zumkeller, Oct 21 2011

bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);
floor_log_2 := proc(n) local nn,i: nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi: nn := floor(nn/2); od: end:

Select[Range@ 1651, # == Reverse@ # && OddQ@ Length@ # &@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Dec 03 2015 *)

{a(n) = local(f); if( n<1, 0, f = length(binary(n)) - 1; 2^f*n + sum(i=1, f, bittest(n,i) * 2^(f-i)))}; /* Michael Somos, Nov 27 2002 */

def A048700(n):
    s = bin(n)[2:]
    return int(s+s[-2::-1],2) # Chai Wah Wu, Feb 26 2021

a(n) = (2^(floor_log_2(n)))*n + sum('(bit_i(n, i)*(2^(floor_log_2(n)-i)))', 'i'=1..floor_log_2(n));

a(A047264(n)) mod 3 = 0, for n > 1. - Altug Alkan, Dec 03 2015

A141707 Least k>0 such that (2n-1)k is palindromic in base 2.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 5, 1, 1, 27, 1, 89, 13, 1, 49, 1, 1, 13, 69, 5, 25, 3, 1, 103, 29, 1, 63, 3, 9, 103, 7, 1, 1, 19, 37, 147, 1, 13, 3, 19, 11, 45, 1, 37, 23, 3, 1, 27, 61, 1, 233, 47, 13, 1, 21, 23, 59, 525, 5, 1, 93, 23, 41, 1, 1, 49, 27, 13, 187, 87, 269, 15, 111, 13, 29, 7, 1, 13, 3

Offset: 1

M. F. Hasler, Jul 17 2008

Even numbers cannot be palindromic in base 2 (unless leading zeros are considered), that's why we search only for odd numbers 2n-1 the k-values such that k(2n-1) is palindromic in base 2. Obviously they are necessarily also odd.

a(A044051(n)) = 1. - Reinhard Zumkeller, Apr 20 2015

			a(1..5)=1 since 1,3,5,7,9 are already palindromic in base 2.
a(6)=3 since 2*6-1=11 and 2*11=22 are not palindromic in base 2, but 3*11=33 is.

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

Cf. A050782, A141708, A178225, A044051.

a141707 n = head [k | k <- [1, 3 ..], a178225 (k * (2 * n - 1)) == 1]
-- Reinhard Zumkeller, Apr 20 2015

lkp[n_]:=Module[{k=1,n2=2n-1},While[IntegerDigits[k*n2,2]!= Reverse[ IntegerDigits[ k*n2,2]],k++];k]; Array[lkp,80] (* Harvey P. Dale, Mar 19 2016 *)

A141707(n,L=10^9)={ n=2*n-1; forstep(k=1,L,2, binary(k*n)-vecextract(binary(k*n),"-1..1") || return(k))}

def binpal(n): b = bin(n)[2:]; return b == b[::-1]
def a(n):
    m = 2*n - 1
    km = m
    while not binpal(km): km += m
    return km//m
print([a(n) for n in range(1, 80)]) # Michael S. Branicky, Mar 20 2022

A141709 Least positive multiple of n which is palindromic in base 2, allowing for leading zeros (or: ignoring trailing zeros).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 33, 12, 65, 14, 15, 16, 17, 18, 513, 20, 21, 66, 2047, 24, 325, 130, 27, 28, 1421, 30, 31, 32, 33, 34, 455, 36, 2553, 1026, 195, 40, 1025, 42, 129, 132, 45, 4094, 4841, 48, 1421, 650, 51, 260, 3339, 54, 165, 56, 513, 2842, 6077, 60, 427, 62

Offset: 1

M. F. Hasler, Jul 17 2008

Even numbers cannot be palindromic in base 2, unless leading zeros are considered (or, equivalently, resp. more precisely, trailing zeros are discarded). This is done in this version of A141708, which therefore does not need to be restricted to odd n as it has been done for A141707 and A141708.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A050782, A141707-A141708, A062279, A203070, A000265, A178225.

a141709 n = until ((== 1) . a178225 . a000265) (+ n) n
-- Reinhard Zumkeller, Nov 06 2012

notpalbinQ[i_]:=Module[{id=IntegerDigits[i,2]},While[Last[id]==0,id=Most[id]];id!= Reverse[id]]; lm[n_]:=Module[{k=1},While[notpalbinQ[k n],k++];k n]; Array[lm,70] (* Harvey P. Dale, Dec 28 2011 *)

A141709(n)=forstep(k=n,10^9,n,vecextract(t=binary(k>>valuation(k,2)),"-1..1")-t || return(k))

A178225(A000265(a(n))) = 1. - Reinhard Zumkeller, Nov 06 2012

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

A164126 First differences of A006995.

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 4, 6, 4, 2, 12, 6, 12, 2, 8, 12, 8, 6, 8, 12, 8, 2, 24, 12, 24, 6, 24, 12, 24, 2, 16, 24, 16, 12, 16, 24, 16, 6, 16, 24, 16, 12, 16, 24, 16, 2, 48, 24, 48, 12, 48, 24, 48, 6, 48, 24, 48, 12, 48, 24, 48, 2, 32, 48, 32, 24, 32, 48, 32, 12, 32, 48, 32, 24, 32, 48, 32, 6

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 10 2009

Contribution from Hieronymus Fischer, Feb 18 2012: (Start)
From the formula section it follows that a(2^m - 1 + 2^(m-1) - k) = a(2^m - 1 + k) for 0 <= k <= 2^(m-1), as well as a(2^m - 1 + 2^(m-1) - k) = 2 for k=0, 2^(m-1) and a(2^m - 1 + 2^(m-1) - k) = 6 for k=2^(m-2), hence, starting from positions n=2^m-1, the following 2^(m-1) terms form symmetric tuples limited on the left and on the right by a '2' and always having a '6' as the center element.
Example: for n = 15 = 2^4 - 1, we have the (2^3+1)-tuple (2,8,12,8,6,8,12,8,2).
Further on, since a(2^m - 1 + 2^(m-1) + k) = a(2^(m+1) - 1 - k) for 0 <= k <= 2^(m-1) an analogous statement holds true for starting positions n = 2^m + 2^(m-1) - 1.
Example: for n = 23 = 2^4 + 2^3 - 1, we find the (2^3+1)-tuple (2,24,12,24,6,24,12,24,2).
If we group the sequence terms according to the value of m=floor(log_2(n)), writing those terms together in separate lines and opening each new line for n >= 2^m + 2^(m-1), then a kind of a 'logarithmic shaped' cone end will be formed, where both the symmetry and the calculation rules become obvious. The first 63 terms are depicted below:
                          1
                          2
                          2
                        2  2
                        6  2
                      4 6  4  2
                     12 6 12  2
                8 12  8 6  8 12  8 2
               24 12 24 6 24 12 24 2
   16 24 16 12 16 24 16 6 16 24 16 12 16 24 16 2
   48 24 48 12 48 24 48 6 48 24 48 12 48 24 48 2
.
(End)
Decremented by 1, also the sequence of run lengths of 0's in A178225. - Hieronymus Fischer, Feb 19 2012

			a(1) = A006995(2) - A006995(1) = 1 - 0 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A006995, A164124, A029971, A016041, A164125.

See A178225.

f[n_]:=FromDigits[RealDigits[n,2][[1]]]==FromDigits[Reverse[RealDigits[n, 2][[1]]]]; a=1;lst={};Do[If[f[n],AppendTo[lst,n-a];a=n],{n,1,8!,1}]; lst

def A164126(n):
    if n == 1: return 1
    m = (a:=1<<(l:=n.bit_length()-2))|(n&a-1)
    k = (m<Chai Wah Wu, Jun 11 2024

a(n) = A006995(n+1) - A006995(n).
Contribution from Hieronymus Fischer, Feb 17 2012: (Start)
a(4*2^m - 1) = a(6*2^m - 1) = 2;
a(5*2^m - 1) = a(7*2^m - 1) = 6 (for m > 0);
Let m = floor(log_2(n)), then
  Case 1: a(n) = 2, if n+1 = 2^(m+1) or n+1 = 3*2^(m-1);
  Case 2: a(n) = 2^(m-1), if n = 0(mod 2) and n < 3*2^(m-1);
  Case 3: a(n) = 3*2^(m-1), if n = 0(mod 2) and n >= 3*2^(m-1);
  Case 4: a(n) = 3*2^(m-1)/gcd(n+1-2^m, 2^m), otherwise.
Cases 2-4 above can be combined as
  Case 2': a(n) = (2 - (-1)^(n-(n-1)*floor(2*n/(3*2^m))))*2^(m-1)/gcd(n+1-2^m, 2^m).
Recursion formula:
Let m = floor(log_2(n)); then
  Case 1: a(n) = 2*a(n-2^(m-1)), if 2^m <= n < 2^m + 2^(m-2) - 1;
  Case 2: a(n) = 6, if n = 2^m + 2^(m-2) - 1;
  Case 3: a(n) = a(n-2^(m-2)), if 2^m + 2^(m-2) <= n < 2^m + 2^(m-1) - 1;
  Case 4: a(n) = 2, if n = 2^m + 2^(m-1) - 1;
  Case 5: a(n) = (2 + (-1)^n)*a(n-2^(m-1)), otherwise (which means 2^m + 2^(m-1) <= n < 2^(m+1)).
(End)

a(1) changed to 1 and keyword:base added by R. J. Mathar, Aug 26 2009

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

Original entry on oeis.org

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

Offset: 0

Jeremy Gardiner, May 23 2010

a(n)=1 if n is in A154809, a(n)=0 otherwise.

Identical to parity of A086757 (Smallest prime p such that n is a palindrome in base p representation)

Cf. A030101, A057891, A086757, A154809, A178225, A280505, A280506.

Table[If[IntegerDigits[n,2]==Reverse[IntegerDigits[n,2]],0,1],{n,0,120}] (* Harvey P. Dale, Aug 07 2023 *)

A178226(n) = (n!=subst(Polrev(binary(n)),x,2)); \\ Antti Karttunen, Dec 15 2017

a(n) = 1 - A178225(n). - Antti Karttunen, Dec 15 2017

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

A206922 Root of the n-th binary palindrome. Least number r > 1 such that A006995(n) can be represented by a finite or infinite number of iterations A006995(A006995(A006995(...(...(r))...).

Original entry on oeis.org

2, 2, 3, 4, 4, 6, 4, 8, 6, 10, 11, 12, 13, 14, 4, 16, 8, 18, 19, 20, 6, 22, 23, 24, 25, 26, 10, 28, 29, 30, 11, 32, 12, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 13, 46, 47, 48, 49, 50, 14, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 4, 64, 16, 66, 67, 68

Offset: 1

Hieronymus Fischer, Mar 12 2012

If n is not a binary palindrome, then a(n)=n.

For n>3: a(n)

For n<>3: The number of iterations such that A006995(n)= A006995(A006995(A006995(...(...(r))...) is given by A206921(n).

			a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).
a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).
a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).
a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).

Cf. A006995, A206921, 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 p;

a(n) <= n for n > 1.
a(n)=p(k), where p(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)=n, if n is not a binary palindrome;
  Case 2: a(n)=a(A206915(n)), else.
Formally: a(n)=if (A178225(n)==0) then n else a(A206915(n)).

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cp's OEIS Frontend

A048700 Binary palindromes of odd length (written in base 10).

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A141707 Least k>0 such that (2n-1)k is palindromic in base 2.

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A141709 Least positive multiple of n which is palindromic in base 2, allowing for leading zeros (or: ignoring trailing zeros).

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

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A164126 First differences of A006995.

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

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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.

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