A030102 -id:A030102 - cp's OEIS frontend

A263273 Bijective base-3 reverse: a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).

0, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 19, 12, 13, 22, 21, 16, 25, 18, 11, 20, 15, 14, 23, 24, 17, 26, 27, 28, 55, 30, 37, 64, 57, 46, 73, 36, 31, 58, 39, 40, 67, 66, 49, 76, 63, 34, 61, 48, 43, 70, 75, 52, 79, 54, 29, 56, 33, 38, 65, 60, 47, 74, 45, 32, 59, 42, 41, 68, 69, 50, 77, 72, 35, 62, 51, 44, 71, 78, 53, 80, 81

Offset: 0

Antti Karttunen, Dec 05 2015

Here the base-3 reverse has been adjusted so that the maximal suffix of trailing zeros (in base-3 representation A007089) stays where it is at the right side, and only the section from the most significant digit to the least significant nonzero digit is reversed, thus making this sequence a self-inverse permutation of nonnegative integers.
Because successive powers of 3 and 9 modulo 2, 4 and 8 are always either constant 1, 1, 1, ... or alternating 1, -1, 1, -1, ... it implies similar simple divisibility rules for 2, 4 and 8 in base 3 as e.g. 3, 9 and 11 have in decimal base (see the Wikipedia-link). As these rules do not depend on which direction they are applied from, it means that this bijection preserves the fact whether a number is divisible by 2, 4 or 8, or whether it is not. Thus natural numbers are divided to several subsets, each of which is closed with respect to this bijection. See the Crossrefs section for permutations obtained from these sections.
When polynomials over GF(3) are encoded as natural numbers (coefficients presented with the digits of the base-3 expansion of n), this bijection works as a multiplicative automorphism of the ring GF(3)[X]. This follows from the fact that as there are no carries involved, the multiplication (and thus also the division) of such polynomials could be as well performed by temporarily reversing all factors (like they were seen through mirror). This implies also that the sequences A207669 and A207670 are closed with respect to this bijection.

			For n = 15, A007089(15) = 120. Reversing this so that the trailing zero stays at the right yields 210 = A007089(21), thus a(15) = 21 and vice versa, a(21) = 15.

Antti Karttunen, Table of n, a(n) for n = 0..6561
Wikipedia, Divisibility rule
Index entries for sequences that are permutations of the natural numbers

Bisections: A264983, A264984.
Cf. A000035, A007089, A010873, A030102, A038500, A038502, A207669, A207670.
Cf. also A057889, A264965, A264966, A264980.
Permutations induced by various sections: A263272 (a(2n)/2), A264974 (a(4n)/4), A264978 (a(8n)/8), A264985, A264989.
Cf. also A004488, A140263, A140264, A246207, A246208 (other base-3 related permutations).

r[n_] := FromDigits[Reverse[IntegerDigits[n, 3]], 3]; b[n_] := n/ 3^IntegerExponent[n, 3]; c[n_] := n/b[n]; a[0]=0; a[n_] := r[b[n]]*c[n]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Dec 29 2015 *)

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a(n): return 0 if n==0 else a030102(a038502(n))*a038500(n) # Indranil Ghosh, May 22 2017

(define (A263273 n) (if (zero? n) n (* (A030102 (A038502 n)) (A038500 n))))

a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
A010873(a(n)) = 0 if and only if A010873(n) = 0. [See the comments section.]

A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

Original entry on oeis.org

1, 2, 4, 8, 16, 64, 32, 184, 352, 704, 1408, 1880, 2824, 14032, 10328, 56128, 100576, 145784, 189472, 370304, 731752, 4388248, 2924096, 11175712, 15965704, 31930448, 63861880, 383165344, 255439712, 1021772344, 510875648, 2550188248, 5619691648, 9689861048, 17830350904, 79068724264, 34109913224, 192259976368, 133338241880

Offset: 0

Antti Karttunen, Dec 05 2015

			2^5 = 32 in base 3 = "1012" (= A007089(32)) as 1*27 + 1*3 + 2*1 = 32. 2^6 = 64 in base 3 = "2101" as 2*27 + 1*9 + 1*1 = 64. "1012" reversed is "2101" and vice versa, thus a(5) = 64 and a(6) = 32.

Antti Karttunen, Table of n, a(n) for n = 0..512

Leftmost column of A265345.
Cf. A000079, A007089, A030102, A263273, A265210, A265342.
Cf. also A036215.

base(n) = {my(a=[n%3]); while(0Altug Alkan, Dec 29 2015

a(n) = A030102(A000079(n)) = A263273(A000079(n)).

a(0) = 1, for n >= 1, a(n) = A265342(a(n-1)).

A004086 Read n backwards (referred to as R(n) in many sequences).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95, 6, 16, 26, 36, 46, 56, 66, 76, 86, 96, 7, 17, 27, 37, 47

Offset: 0

N. J. A. Sloane

Also called digit reversal of n.

Leading zeros (after the reversal has taken place) are omitted. - N. J. A. Sloane, Jan 23 2017

Indranil Ghosh, Table of n, a(n) for n = 0..50000 (first 1001 terms from Franklin T. Adams-Watters)
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625 [math.CO], Dec 08 2020.
Michael Penn, A digit moving number puzzle., YouTube video, 2022.

Cf. A004185, A004186, A009996, A073138, A188649, A221714.

Cf. A030101, A030102, A030103, A030104, A030105, A030107, A030108.

a004086 = read . reverse . show  -- Reinhard Zumkeller, Apr 11 2011

|.&.": i.@- 1e5 NB. Stephen Makdisi, May 14 2018

read transforms; A004086 := digrev; #cf "Transforms" link at bottom of page
A004086:=proc(n) local s,t; if n<10 then n else s:=irem(n,10,'t'); while t>9 do s:=s*10+irem(t,10,'t') od: s*10+t fi end; # M. F. Hasler, Jan 29 2012

Table[FromDigits[Reverse[IntegerDigits[n]]], {n, 0, 75}]
IntegerReverse[Range[0,80]](* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 13 2018 *)

dig(n) = {local(m=n,r=[]); while(m>0,r=concat(m%10,r);m=floor(m/10));r}
A004086(n) = {local(b,m,r);r=0;b=1;m=dig(n);for(i=1,matsize(m)[2],r=r+b*m[i];b=b*10);r} \\ Michael B. Porter, Oct 16 2009

A004086(n)=fromdigits(Vecrev(digits(n))) \\ M. F. Hasler, Nov 11 2010, updated May 11 2015, Sep 13 2019

def A004086(n):
    return int(str(n)[::-1]) # Chai Wah Wu, Aug 30 2014

For n > 0, a(a(n)) = n iff n mod 10 != 0. - Reinhard Zumkeller, Mar 10 2002
a(n) = d(n,0) with d(n,r) = r if n=0, otherwise d(floor(n/10), r*10+(n mod 10)). - Reinhard Zumkeller, Mar 04 2010
a(10*n+x) = x*10^m + a(n) if 10^(m-1) <= n < 10^m and 0 <= x <= 9. - Robert Israel, Jun 11 2015

Extended by Ray Chandler, Dec 30 2004

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.

Original entry on oeis.org

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

A263272 Self-inverse permutation of nonnegative integers: a(n) = A263273(2*n) / 2.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 11, 8, 9, 10, 7, 12, 13, 14, 15, 32, 23, 18, 29, 20, 33, 38, 17, 24, 35, 26, 27, 28, 19, 30, 37, 16, 21, 34, 25, 36, 31, 22, 39, 40, 41, 42, 95, 68, 45, 86, 59, 96, 113, 50, 69, 104, 77, 54, 83, 56, 87, 110, 47, 60, 101, 74, 99, 92, 65, 114, 119, 44, 51, 98, 71, 72, 89, 62, 105, 116, 53, 78, 107, 80, 81

Offset: 0

Antti Karttunen, Dec 05 2015

Antti Karttunen, Table of n, a(n) for n = 0..9841
Index entries for sequences that are permutations of the natural numbers

Cf. A000035, A263273.
Bisections: A264986, A264987.
Cf. also A246200, A264967, A264968, A264974, A264978, A264975, A264976, A264984.

f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@ g[n] h[n]]]; Table[f[2 n]/2, {n, 0, 81}] (* Michael De Vlieger, Jan 04 2016,after Jean-François Alcover at A263273 *)

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a(n): return a263273(2*n)/2 # Indranil Ghosh, May 23 2017

(define (A263272 n) (/ (A263273 (+ n n)) 2))

a(n) = A263273(2*n) / 2 = A264984(n) / 2.
As a composition of related permutations:
a(n) = A264974(A264975(n)) = A264976(A264974(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
A264974(n) = a(2n)/2. [Thus the restriction onto even numbers induces yet another permutation.]

A264974 Self-inverse permutation of natural numbers: a(n) = A263273(4*n) / 4.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 16, 9, 10, 19, 12, 13, 14, 15, 8, 17, 18, 11, 20, 21, 34, 43, 48, 25, 52, 27, 28, 55, 30, 37, 46, 57, 22, 49, 36, 31, 58, 39, 40, 41, 42, 23, 50, 45, 32, 59, 24, 35, 44, 51, 26, 53, 54, 29, 56, 33, 38, 47, 60, 61, 142, 63, 88, 169, 102, 115, 124, 129, 70, 151, 144, 97, 178, 75, 106, 133, 156, 79, 160, 81

Offset: 0

Antti Karttunen, Dec 05 2015

Antti Karttunen, Table of n, a(n) for n = 0..4921
Index entries for sequences that are permutations of the natural numbers

Terms of A264986 halved.
Cf. A263272, A263273, A264978.
Cf. also A264975, A264976.

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a(n): return a263273(4*n)/4 # Indranil Ghosh, May 25 2017

a(n) = A263273(4*n) / 4.
a(n) = A264986(n) / 2 = A263272(2*n) / 2.
As a composition of related permutations:
a(n) = A264975(A263272(n)) = A263272(A264976(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
A264978(n) = a(2n)/2. [Thus the restriction onto even numbers induces yet another permutation.]

A264985 Self-inverse permutation of nonnegative integers: a(n) = (A264983(n)-1) / 2.

Original entry on oeis.org

0, 1, 3, 2, 4, 9, 6, 10, 12, 5, 7, 11, 8, 13, 27, 18, 28, 36, 15, 19, 33, 24, 31, 30, 21, 37, 39, 14, 16, 32, 23, 22, 29, 20, 34, 38, 17, 25, 35, 26, 40, 81, 54, 82, 108, 45, 55, 99, 72, 85, 90, 63, 109, 117, 42, 46, 96, 69, 58, 87, 60, 100, 114, 51, 73, 105, 78, 94, 84, 57, 91, 111, 48, 64, 102, 75, 112, 93, 66, 118, 120, 41

Offset: 0

Antti Karttunen, Dec 05 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..9841
Index entries for sequences that are permutations of the natural numbers

Cf. A263273, A264983.

Cf. also A264989, A264991, A264992.

f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@ g[n] h[n]]]; t = Select[f /@ Range@ 1000, OddQ]; Table[(t[[n + 1]] - 1)/2, {n, 0, 81}] (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A263273 *)

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a(n): return (a263273(2*n + 1) - 1)/2 # Indranil Ghosh, May 22 2017

(define (A264985 n) (/ (- (A264983 n) 1) 2))

a(n) = (A264983(n)-1) / 2 = (1/2) * (A263273(2n + 1) - 1).

A134028 Reversal of n in balanced ternary representation.

Original entry on oeis.org

0, 1, -2, 1, 4, -11, -2, 7, -8, 1, 10, -5, 4, 13, -38, -11, 16, -29, -2, 25, -20, 7, 34, -35, -8, 19, -26, 1, 28, -17, 10, 37, -32, -5, 22, -23, 4, 31, -14, 13, 40, -119, -38, 43, -92, -11, 70, -65, 16, 97, -110, -29, 52, -83, -2, 79, -56, 25, 106, -101, -20, 61, -74, 7, 88, -47, 34, 115, -116, -35, 46, -89, -8, 73, -62, 19, 100

Offset: 0

Reinhard Zumkeller, Oct 19 2007

As the graph demonstrates, the sequence makes large negative steps at terms (3^i+1)/2. These steps divide the graph into conspicuous blocks. - N. J. A. Sloane, Jul 03 2016

			20 = 1*3^3 - 1*3^2 + 1*3^1 - 1*3^0 == '+-+-'
=> a(20) = -1*3^3 + 1*3^2 - 1*3^1 + 1*3^0 = -20;
21 = 1*3^3 - 1*3^2 + 1*3^1 + 0*3^0 == '+-+0'
=> a(21) = 0*3^3 + 1*3^2 - 1*3^1 + 1*3^0 = 7;
22 = 1*3^3 - 1*3^2 + 1*3^1 + 1*3^0 == '+-++'
=> a(22) = 1*3^3 + 1*3^2 - 1*3^1 + 1*3^0 = 34;
23 = 1*3^3 + 0*3^2 - 1*3^1 - 1*3^0 == '+0--'
=> a(23) = -1*3^3 - 1*3^2 + 0*3^1 + 1*3^0 = -35;
24 = 1*3^3 + 0*3^2 - 1*3^1 + 0*3^0 == '+0-0'
=> a(24) = 0*3^3 - 1*3^2 + 0*3^1 + 1*3^0 = -8;
25 = 1*3^3 + 0*3^2 - 1*3^1 + 1*3^0 == '+0-+'
=> a(25) = 1*3^3 - 1*3^2 + 0*3^1 + 1*3^0 = 19.

D. E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, Vol 2, pp 173-175.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Reversal
Wikipedia, Balanced Ternary

Cf. A117966 (balanced ternary representation), A030102, A134021, A274107.

A134027 gives the numbers whose balanced ternary representation is palindromic.

def a(n):
    if n==0: return 0
    s=[]
    x=0
    while n>0:
        x=n%3
        n=n//3
        if x==2:
            x=-1
            n+=1
        s.append(x)
    l=s[::-1]
    return sum(l[i]*3**i for i in range(len(l)))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 10 2017

a(A134027(n)) = A134027(n);

A134021(ABS(a(n))) <= A134021(n).

A255588 Convert n to base 3, move the least significant digit to the most significant one and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 1, 4, 7, 2, 5, 8, 3, 12, 21, 4, 13, 22, 5, 14, 23, 6, 15, 24, 7, 16, 25, 8, 17, 26, 9, 36, 63, 10, 37, 64, 11, 38, 65, 12, 39, 66, 13, 40, 67, 14, 41, 68, 15, 42, 69, 16, 43, 70, 17, 44, 71, 18, 45, 72, 19, 46, 73, 20, 47, 74, 21, 48, 75, 22, 49, 76, 23

Offset: 0

Paolo P. Lava, Feb 27 2015

a(3*n) = n.

Fixed points of the transform are listed in A048328.

			10 in base 3 is 101: moving the least significant digit to the most significant one we have 110 that is 12 in base 10.

Indranil Ghosh, Table of n, a(n) for n = 0..19683 (terms 0..1000 Paolo P. Lava)

Cf. A030102, A038572, A048328, A255589-A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,3);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 3; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 69]) (* Michael De Vlieger, Mar 04 2015 *)
Table[FromDigits[RotateRight[IntegerDigits[n,3]],3],{n,0,70}] (* Harvey P. Dale, Feb 20 2022 *)

def A255588(n):
    x=A007089(n)
    return int(x[-1]+x[:-1], 3) # Indranil Ghosh, Feb 03 2017

A265352 Permutation of nonnegative integers: a(n) = A263273(A263272(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 6, 19, 8, 9, 10, 5, 12, 13, 22, 21, 64, 23, 18, 55, 20, 57, 58, 25, 24, 73, 26, 27, 28, 11, 30, 31, 16, 15, 46, 17, 36, 37, 14, 39, 40, 67, 66, 199, 68, 63, 190, 65, 192, 193, 70, 69, 208, 71, 54, 163, 56, 165, 166, 61, 60, 181, 62, 171, 172, 59, 174, 175, 76, 75, 226, 77, 72, 217, 74, 219, 220, 79, 78, 235, 80, 81

Offset: 0

Antti Karttunen, Dec 07 2015

Composition of A263273 with the permutation obtained from its even bisection.

Antti Karttunen, Table of n, a(n) for n = 0..9841
Index entries for sequences that are permutations of the natural numbers

Inverse: A265351.
Cf. A263272, A263273, A264974, A265368.
Cf. also A265354, A265355, A265356

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a(n): return a263273(a263273(2*n)/2) # Indranil Ghosh, Jun 08 2017

(define (A265352 n) (A263273 (A263272 n)))

a(n) = A263273(A263272(n)).
As a composition of other related permutations:
a(n) = A265368(A264974(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).

cp's OEIS Frontend

A263273 Bijective base-3 reverse: a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).

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A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

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A004086 Read n backwards (referred to as R(n) in many sequences).

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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.

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A263272 Self-inverse permutation of nonnegative integers: a(n) = A263273(2*n) / 2.

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A264974 Self-inverse permutation of natural numbers: a(n) = A263273(4*n) / 4.

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