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 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. 157
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS 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 101/A062383(n), n >= 0}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 REFERENCES 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 LINKS T. D. Noe, Table of n, a(n) for n = 0..10000 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 Project Euler, Problem 463: A weird recurrence relation Wikipedia, van der Corput sequence. FORMULA 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 EXAMPLE a(100) = 19 because 100 (base 10) = 1100100 (base 2) and R(1100100 (base 2)) = 10011 (base 2) = 19 (base 10). MAPLE 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 MATHEMATICA 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 *) PROG (PARI) a(n)=if(n<1, 0, subst(Polrev(binary(n)), x, 2)) (PARI) a(n) = fromdigits(Vecrev(binary(n)), 2); \\ Michel Marcus, Nov 10 2017 (Magma) A030101:=func; // Jason Kimberley, Sep 19 2011 (Haskell) 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 (Sage) 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 (Python) def a(n): return int(bin(n)[2:][::-1], 2) # Indranil Ghosh, Apr 24 2017 (J) ([: #. [: |. #:)"0 NB. Stephen Makdisi, May 07 2018 (Scala) (0 to 127).map(n => Integer.parseInt(Integer.toString(n, 2).reverse, 2)) // Alonso del Arte, Feb 11 2020 CROSSREFS Cf. A030102 - A030109, A036044, A004086, A005408. Cf. A055944 (reverse and add), A178225, A273258. Cf. A056539, A057889 (bijective variants). Sequence in context: A106609 A171968 A093474 * A162742 A348363 A081432 Adjacent sequences:  A030098 A030099 A030100 * A030102 A030103 A030104 KEYWORD nonn,base,nice,hear,look AUTHOR EXTENSIONS 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 STATUS approved

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Last modified August 18 02:34 EDT 2022. Contains 356204 sequences. (Running on oeis4.)