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

Author

Mike Keith

Keywords

• nonn
• nice
• easy
• base

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

Examples

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

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.

Crossrefs

Cf. A061855.

Cf. A000225, A036044, A062383.

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

Programs

Haskell

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

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 *)

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

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

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

Extensions

More terms from Erich Friedman