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

1,1

Comments

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

Links

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.

Formula

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

Example

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

Maple

[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

Mathematica

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

Prog

(PARI) 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, ", ")))
(PARI) for(i=1, 999, if(Set(vecextract(t=binary(i), "-1..1")+t)==[1], print1(i", "))) \\ M. F. Hasler, Dec 17 2007
(PARI) a(n) = my (b=binary(n)); (n+1)*2^#b-fromdigits(Vecrev(b), 2)-1 \\ Rémy Sigrist, Mar 15 2021
(Haskell)
a035928 n = a035928_list !! (n-1)
a035928_list = filter (\x -> a036044 x == x) [0, 2..]
-- Reinhard Zumkeller, Sep 16 2011
(Python)
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
(Python)
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

Crossrefs

Cf. A061855.

Cf. A000225, A036044, A062383.

Intersection of A195064 and A195066; cf. A195063, A195065.

Sequence in context: A176978 A186630 A154391 * A014486 A166751 A216649

Adjacent sequences: A035925 A035926 A035927 * A035929 A035930 A035931

Keyword

nonn,nice,easy,base

Author

Mike Keith (domnei(AT)aol.com)

Extensions

More terms from Erich Friedman

Status

approved