A209492 -id:A209492 - cp's OEIS frontend

A030101 a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 13, 3, 11, 7, 15, 1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 1, 33, 17, 49, 9, 41, 25, 57, 5, 37, 21, 53, 13, 45, 29, 61, 3, 35, 19, 51, 11, 43, 27, 59, 7, 39, 23, 55, 15, 47, 31, 63, 1, 65, 33, 97, 17, 81, 49, 113, 9, 73, 41, 105, 25, 89, 57

Offset: 0

David W. Wilson

As with decimal reversal, initial zeros are ignored; otherwise, the reverse of 1 would be 1000000... ad infinitum.
Numerators of the binary van der Corput sequence. - Eric Rowland, Feb 12 2008
It seems that in most cases A030101(x) = A000265(x) and that if A030101(x) <> A000265(x), the next time A030101(y) = A000265(x), A030101(x) = A000265(y). Also, it seems that if a pair of values exist at one index, they will exist at any index where one of them exist. It also seems like the greater of the pair always shows up on A000265 first. - Dylan Hamilton, Aug 04 2010
The number of occasions A030101(n) = A000265(n) before n = 2^k is A053599(k) + 1. For n = 0..2^19, the sequences match less than 1% of the time. - Andrew Woods, May 19 2012
For n > 0: a(a(n)) = n if and only if n is odd; a(A006995(n)) = A006995(n). - Juli Mallett, Nov 11 2010, corrected: Reinhard Zumkeller, Oct 21 2011
n is binary palindromic if and only if a(n) = n. - Reinhard Zumkeller, corrected: Jan 17 2012, thanks to Hieronymus Fischer, who pointed this out; Oct 21 2011
Given any n > 1, the set of numbers A030109(i) = (A030101(i) - 1)/2 for indexes i ranging from 2^n to 2^(n + 1) - 1 is a permutation of the set of consecutive integers {0, 1, 2, ..., 2^n - 1}. This is important in the standard FFT algorithms (starting or ending bit-reversal permutation). - Stanislav Sykora, Mar 15 2012
Row n of A030308 gives the binary digits of a(n), prepended with zero at even positions. - Reinhard Zumkeller, Jun 17 2012
The binary van der Corput sequence is the infinite sequence of fractions { A030101(n)/A062383(n), n = 0, 1, 2, 3, ... }, and begins 0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16, 5/16, 13/16, 3/16, 11/16, 7/16, 15/16, 1/32, 17/32, 9/32, 25/32, 5/32, 21/32, 13/32, 29/32, 3/32, 19/32, 11/32, 27/32, 7/32, 23/32, 15/32, 31/32, 1/64, 33/64, 17/64, 49/64, ... - N. J. A. Sloane, Dec 01 2019
Record highs occur at n = A209492(m) (for n>=1) with values a(n) = A224195(m) (for n>=3). - Bill McEachen, Aug 02 2023

			a(100) = 19 because 100 (base 10) = 1100100 (base 2) and R(1100100 (base 2)) = 10011 (base 2) = 19 (base 10).

Hlawka E. The theory of uniform distribution. Academic Publishers, Berkhamsted, 1984. See pp. 93, 94 for the van der Corput sequence. - N. J. A. Sloane, Dec 01 2019

T. D. Noe, Table of n, a(n) for n = 0..10000
Sean Anderson, Bit Twiddling Hacks, for fixed-width reversals.
Joerg Arndt, Matters Computational (The Fxtbook), section 1.14 Reversing the bits of a word, page 33.
Harold L. Dorwart, Seventeenth annual USA Mathematical Olympiads, Math. Mag., 62 (1989), 210-214 (#3).
Bernd Fischer and Lothar Reichel, Newton interpolation in Fejér and Chebyshev points, Mathematics of Computation 53.187 (1989): 265-278. See pp. 266, 267 for the van der Corput sequence. - _N. J. A. Sloane_, Dec 01 2019
Michael Gilleland, Some Self-Similar Integer Sequences
E. Hlawka, The theory of uniform distribution. Academic Publishers, Berkhamsted, 1984. See pp. 93, 94 for the van der Corput sequence. - _N. J. A. Sloane_, Dec 01 2019
Project Euler, Problem 463: A weird recurrence relation
Wikipedia, van der Corput sequence.
Index entries for sequences related to binary expansion of n

Cf. A030102 - A030109, A036044, A004086, A005408.
Cf. A055944 (reverse and add), A178225, A273258.
Cf. A056539, A057889 (bijective variants), A224195, A209492.

a030101 = f 0 where
   f y 0 = y
   f y x = f (2 * y + b) x'  where (x', b) = divMod x 2
-- Reinhard Zumkeller, Mar 18 2014, Oct 21 2011

([: #. [: |. #:)"0 NB. Stephen Makdisi, May 07 2018

A030101:=func; // Jason Kimberley, Sep 19 2011

A030101 := proc(n)
    convert(n,base,2) ;
    ListTools[Reverse](%) ;
    add(op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, Mar 10 2015
# second Maple program:
a:= proc(n) local m, r; m:=n; r:=0;
      while m>0 do r:=r*2+irem(m, 2, 'm') od; r
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, Nov 17 2015

Table[FromDigits[Reverse[IntegerDigits[i, 2]], 2], {i, 0, 80}]
bitRev[n_] := Switch[Mod[n, 4], 0, bitRev[n/2], 1, 2 bitRev[(n + 1)/2] - bitRev[(n - 1)/4], 2, bitRev[n/2], 3, 3 bitRev[(n - 1)/2] - 2 bitRev[(n - 3)/4]]; bitRev[0] = 0; bitRev[1] = 1; bitRev[3] = 3; Array[bitRev, 80, 0] (* Robert G. Wilson v, Mar 18 2014 *)

a(n)=if(n<1,0,subst(Polrev(binary(n)),x,2))

a(n) = fromdigits(Vecrev(binary(n)), 2); \\ Michel Marcus, Nov 10 2017

def a(n): return int(bin(n)[2:][::-1], 2) # Indranil Ghosh, Apr 24 2017

def A030101(n): return Integer(bin(n).lstrip("0b")[::-1],2) if n!=0 else 0
[A030101(n) for n in (0..78)]  # Peter Luschny, Aug 09 2012

(0 to 127).map(n => Integer.parseInt(Integer.toString(n, 2).reverse, 2)) // Alonso del Arte, Feb 11 2020

a(n) = 0, a(2n) = a(n), a(2n+1) = a(n) + 2^(floor(log_2(n)) + 1). For n > 0, a(n) = 2*A030109(n) - 1. - Ralf Stephan, Sep 15 2003
a(n) = b(n, 0) with b(n, r) = r if n = 0, otherwise b(floor(n/2), 2*r + n mod 2). - Reinhard Zumkeller, Mar 03 2010
a(1) = 1, a(3) = 3, a(2n) = a(n), a(4n+1) = 2a(2n+1) - a(n), a(4n+3) = 3a(2n+1) - 2a(n) (as in the Project Euler problem). To prove this, expand the recurrence into binary strings and reversals. - David Applegate, Mar 16 2014, following a posting to the Sequence Fans Mailing List by Martin Møller Skarbiniks Pedersen.
Conjecture: a(n) = 2*w(n) - 2*w(A053645(n)) - 1 for n > 0, where w = A264596. - Velin Yanev, Sep 12 2017

Edits (including correction of an erroneous date pointed out by J. M. Bergot) by Jon E. Schoenfield, Mar 16 2014

Name clarified by Antti Karttunen, Nov 09 2017

A255264 Total number of ON cells in the "Ulam-Warburton" two-dimensional cellular automaton of A147562 after A048645(n) generations.

Original entry on oeis.org

1, 5, 9, 21, 25, 37, 85, 89, 101, 149, 341, 345, 357, 405, 597, 1365, 1369, 1381, 1429, 1621, 2389, 5461, 5465, 5477, 5525, 5717, 6485, 9557, 21845, 21849, 21861, 21909, 22101, 22869, 25941, 38229, 87381, 87385, 87397, 87445, 87637

Offset: 1

Omar E. Pol, Feb 19 2015

It appears that these are the terms of A147562, A162795, A169707, A255366, A256250, A256260, whose indices have binary weight 1 or 2.

			Also, written as an irregular triangle in which row lengths are the terms of A028310 the sequence begins:
      1;
      5;
      9,    21;
     25,    37,    85;
     89,   101,   149,   341;
    345,   357,   405,   597,  1365;
   1369,  1381,  1429,  1621,  2389,  5461;
   5465,  5477,  5525,  5717,  6485,  9557, 21845;
  21849, 21861, 21909, 22101, 22869, 25941, 38229, 87381;
  ...
Right border gives the positive terms of A002450.
It appears that the second leading diagonal gives the odd terms of A206374.

Cf. A002450, A028310, A032925, A048645, A075897, A139250, A147562, A162795, A169707, A206374, A209492, A255263, A255366, A256250, A256260.

a(n) = A147562(A048645(n)).
Conjecture 1: a(n) = A162795(A048645(n)).
Conjecture 2: a(n) = A169707(A048645(n)).
Conjecture 3: a(n) = A255366(A048645(n)).
Conjecture 4: a(n) = A256250(A048645(n)).
Conjecture 5: a(n) = A256260(A048645(n)).
a(n) = A032925(A209492(n-1)) (conjectured). - Jon Maiga, Dec 17 2021

A340836 a(n) is the least k such that the binary reversal of k is greater than or equal to n.

Original entry on oeis.org

0, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 11, 11, 15, 15, 17, 17, 19, 19, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 31, 31, 33, 33, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 39, 39, 39, 39, 39, 39, 39, 39, 47, 47, 47, 47, 63, 63, 65, 65, 67, 67

Offset: 0

Rémy Sigrist, Mar 13 2021

A030101 gives the binary reversal of a number.
All positive terms belong to A209492.
This sequence is nondecreasing.

			For n = 8:
- A030101(k) < 8 for any k <= 8,
- A030101(9) = 9 >= 8,
- so a(8) = 9.

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Cf. A030101, A209492, A340835 (decimal analog).

{ base = 2; k = 0; r = 0; for (n=0, 67, while (r

def A340836(n):
    if n == 0:
        return 0
    s = bin(n)[2:]
    i = s.find('0')
    if i == -1:
        return n
    s1, s2 = s[:i+1], s[i+1:]
    if s2 == '':
        return n+1
    if int(s2) <= 1:
        return int('1'+s2[-2::-1]+s1[::-1],2)
    else:
        return int('1'+'0'*(len(s2)-1)+bin(int(s1,2)+1)[:1:-1],2) # Chai Wah Wu, Mar 14 2021

a(n) <= n + 1.

cp's OEIS Frontend

A030101 a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

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A255264 Total number of ON cells in the "Ulam-Warburton" two-dimensional cellular automaton of A147562 after A048645(n) generations.

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A340836 a(n) is the least k such that the binary reversal of k is greater than or equal to n.

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PARI

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