A195064 -id:A195064 - cp's OEIS frontend

A035928 Numbers n such that BCR(n) = n, where BCR = binary-complement-and-reverse = take one's complement then reverse bit order.

2, 10, 12, 38, 42, 52, 56, 142, 150, 170, 178, 204, 212, 232, 240, 542, 558, 598, 614, 666, 682, 722, 738, 796, 812, 852, 868, 920, 936, 976, 992, 2110, 2142, 2222, 2254, 2358, 2390, 2470, 2502, 2618, 2650, 2730, 2762, 2866, 2898, 2978, 3010, 3132, 3164, 3244

Offset: 1

Mike Keith

Numbers n such that A036044(n) = n.
Also: numbers such that n+BR(n) is in A000225={2^k-1} (with BR = binary reversed). - M. F. Hasler, Dec 17 2007
Also called "antipalindromes". - Jeffrey Shallit, Feb 04 2022

			38 is such a number because 38=100110; complement to get 011001, then reverse bit order to get 100110.

Reinhard Zumkeller, 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.
Aayush Rajasekaran, Jeffrey Shallit, and Tim Smith, Sums of Palindromes: an Approach via Nested-Word Automata, preprint arXiv:1706.10206 [cs.FL], June 30 2017.

Cf. A061855.
Cf. A000225, A036044, A062383.
Intersection of A195064 and A195066; cf. A195063, A195065.

a035928 n = a035928_list !! (n-1)
a035928_list = filter (\x -> a036044 x == x) [0,2..]
-- Reinhard Zumkeller, Sep 16 2011

[seq(ReflectBinSeq(j,(floor_log_2(j)+1)),j=1..256)];
ReflectBinSeq := (x,n) -> (((2^n)*x)+binrevcompl(x));
binrevcompl := proc(nn) local n,z; n := nn; z := 0; while(n <> 0) do z := 2*z + ((n+1) mod 2); n := floor(n/2); od; RETURN(z); end;
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; # Computes essentially the same as floor(log[2](n))
# alternative Maple program:
q:= n-> (l-> is(n=add((1-l[-i])*2^(i-1), i=1..nops(l))))(Bits[Split](n)):
select(q, [$1..3333])[];  # Alois P. Heinz, Feb 10 2021

bcrQ[n_]:=Module[{idn2=IntegerDigits[n,2]},Reverse[idn2/.{1->0,0->1}] == idn2]; Select[Range[3200],bcrQ] (* Harvey P. Dale, May 24 2012 *)

for(n=1,1000,l=length(binary(n)); b=binary(n); if(sum(i=1,l,abs(component(b,i)-component(b,l+1-i)))==l,print1(n,",")))

for(i=1,999,if(Set(vecextract(t=binary(i),"-1..1")+t)==[1],print1(i","))) \\ M. F. Hasler, Dec 17 2007

a(n) = my (b=binary(n)); (n+1)*2^#b-fromdigits(Vecrev(b),2)-1 \\ Rémy Sigrist, Mar 15 2021

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
def aupto(limit): return [m for m in range(limit+1) if BCR(m) == m]
print(aupto(3244)) # Michael S. Branicky, Feb 10 2021

from itertools import count, islice
def A035928_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:n==int(format(~n&(1<<(m:=n.bit_length()))-1,'0'+str(m)+'b')[::-1],2),count(max(startvalue,1)))
A035928_list = list(islice(A035928_gen(),30)) # Chai Wah Wu, Jun 30 2022

If offset were 0, a(2n+1) - a(2n) = 2^floor(log_2(n)+1).

a(n) = n * A062383(n) + A036044(n). - Rémy Sigrist, Jun 11 2022

More terms from Erich Friedman

A036044 BCR(n): write in binary, complement, reverse.

Original entry on oeis.org

1, 0, 2, 0, 6, 2, 4, 0, 14, 6, 10, 2, 12, 4, 8, 0, 30, 14, 22, 6, 26, 10, 18, 2, 28, 12, 20, 4, 24, 8, 16, 0, 62, 30, 46, 14, 54, 22, 38, 6, 58, 26, 42, 10, 50, 18, 34, 2, 60, 28, 44, 12, 52, 20, 36, 4, 56, 24, 40, 8, 48, 16, 32, 0, 126, 62, 94, 30, 110, 46, 78, 14, 118, 54, 86

Offset: 0

N. J. A. Sloane

a(0) could be considered to be 0 if the binary representation of zero were chosen to be the empty string. - Jason Kimberley, Sep 19 2011
From Bernard Schott, Jun 15 2021: (Start)
Except for a(0) = 1, every term is even.
For each q >= 0, there is one and only one odd number h such that a(n) = 2*q iff n = h*2^m-1 for m >= 1 when q = 0, and for m >= 0 when q >= 1 (see A345401 and some examples below).
a(n) =  0 iff n =    2^m-1 for m >= 1 (Mersenne numbers) (A000225).
a(n) =  2 iff n =  3*2^m-1 for m >= 0 (A153893).
a(n) =  4 iff n =  7*2^m-1 for m >= 0 (A086224).
a(n) =  6 iff n =  5*2^m-1 for m >= 0 (A153894).
a(n) =  8 iff n = 15*2^m-1 for m >= 0 (A196305).
a(n) = 10 iff n = 11*2^m-1 for m >= 0 (A086225).
a(n) = 12 iff n = 13*2^m-1 for m >= 0 (A198274).
For k >= 1, a(n) = 2^k iff n = (2^(k+1)-1)*2^m - 1 for m >= 0.
Explanation for a(n) = 2:
   For m >= 0, A153893(m) = 3*2^m-1 -> 1011...11 -> 0100...00 -> 10 -> 2 where 1011...11_2 is 10 followed by m 1's. (End)

			4 -> 100 -> 011 -> 110 -> 6.

Indranil Ghosh, Table of n, a(n) for n = 0..10000 (first 1024 terms from T. D. Noe)

Cf. A030101, A056539, A329369, A345401.
Cf. A035928 (fixed points), A195063, A195064, A195065, A195066.
Indices of terms 0, 2, 4, 6, 8, 10, 12, 14, 18, 22, 26, 30: A000225 \ {0}, A153893, A086224, A153894, A196305, A086225, A198274, A052996\{1,3}, A291557, A198276, A171389, A198275.

import Data.List (unfoldr)
a036044 0 = 1
a036044 n = foldl (\v d -> 2 * v + d) 0 (unfoldr bc n) where
   bc 0 = Nothing
   bc x = Just (1 - m, x') where (x',m) = divMod x 2
-- Reinhard Zumkeller, Sep 16 2011

A036044:=func; // Jason Kimberley, Sep 19 2011

A036044 := proc(n)
    local bcr ;
    if n = 0 then
        return 1;
    end if;
    convert(n,base,2) ;
    bcr := [seq(1-i,i=%)] ;
    add(op(-k,bcr)*2^(k-1),k=1..nops(bcr)) ;
end proc:
seq(A036044(n),n=0..200) ; # R. J. Mathar, Nov 06 2017

dtn[ L_ ] := Fold[ 2#1+#2&, 0, L ]; f[ n_ ] := dtn[ Reverse[ 1-IntegerDigits[ n, 2 ] ] ]; Table[ f[ n ], {n, 0, 100} ]
Table[FromDigits[Reverse[IntegerDigits[n,2]/.{1->0,0->1}],2],{n,0,80}] (* Harvey P. Dale, Mar 08 2015 *)

a(n)=fromdigits(Vecrev(apply(n->1-n,binary(n))),2) \\ Charles R Greathouse IV, Apr 22 2015

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
print([BCR(n) for n in range(75)]) # Michael S. Branicky, Jun 14 2021

def A036044(n): return -int((s:=bin(n)[-1:1:-1]),2)-1+2**len(s) # Chai Wah Wu, Feb 04 2022

a(2n) = 2*A059894(n), a(2n+1) = a(2n) - 2^floor(log_2(n)+1). - Ralf Stephan, Aug 21 2003

Conjecture: a(n) = (-1)^A023416(n)*b(n) for n > 0 with a(0) = 1 where b(2^m) = (-1)^m*(2^(m+1) - 2) for m >= 0, b(2n+1) = b(n) for n > 0, b(2n) = b(n) + b(n - 2^f(n)) + b(2n - 2^f(n)) for n > 0 and where f(n) = A007814(n) (see A329369). - Mikhail Kurkov, Dec 13 2024

A195063 Numbers n such that BCR(n) is less than n, where BCR = binary-complement-and-reverse = A036044.

Original entry on oeis.org

6, 14, 22, 26, 28, 30, 46, 50, 54, 58, 60, 62, 78, 86, 90, 94, 98, 102, 106, 108, 110, 114, 116, 118, 120, 122, 124, 126, 158, 166, 174, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 220, 222, 226, 228, 230, 234, 236, 238, 242, 244, 246, 248, 250, 252

Offset: 1

Reinhard Zumkeller, Sep 16 2011

A035928(a(n)) < n.

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

Complement of A195066; subsequence of A195064; cf. A035928, A195065.

a195063 n = a195063_list !! (n-1)
a195063_list = filter (\x -> a036044 x < x) [0,2..]

A195065 Numbers n such that BCR(n) is greater than n, where BCR = binary-complement-and-reverse = A036044.

Original entry on oeis.org

0, 4, 8, 16, 18, 20, 24, 32, 34, 36, 40, 44, 48, 64, 66, 68, 70, 72, 74, 76, 80, 82, 84, 88, 92, 96, 100, 104, 112, 128, 130, 132, 134, 136, 138, 140, 144, 146, 148, 152, 154, 156, 160, 162, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 208, 216, 224

Offset: 1

Reinhard Zumkeller, Sep 16 2011

A035928(a(n)) > n.

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

Complement of A195064; subsequence of A195066.

Cf. A035928, A036044, A195063.

a195065 n = a195065_list !! (n-1)
a195065_list = filter (\x -> a036044 x > x) [0,2..]

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
def aupto(limit): return [m for m in range(limit+1) if BCR(m) > m]
print(aupto(224)) # Michael S. Branicky, Jun 14 2021

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A035928 Numbers n such that BCR(n) = n, where BCR = binary-complement-and-reverse = take one's complement then reverse bit order.

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A036044 BCR(n): write in binary, complement, reverse.

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A195063 Numbers n such that BCR(n) is less than n, where BCR = binary-complement-and-reverse = A036044.

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A195065 Numbers n such that BCR(n) is greater than n, where BCR = binary-complement-and-reverse = A036044.

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