A048724 -id:A048724 - cp's OEIS frontend

A235042 Factorization-preserving bijection from GF(2)[X]-polynomials to nonnegative integers, version which fixes the elements that are irreducible in both semirings.

0, 1, 2, 3, 4, 9, 6, 7, 8, 21, 18, 11, 12, 13, 14, 27, 16, 81, 42, 19, 36, 49, 22, 39, 24, 5, 26, 63, 28, 33, 54, 31, 32, 93, 162, 91, 84, 37, 38, 99, 72, 41, 98, 15, 44, 189, 78, 47, 48, 77, 10, 243, 52, 57, 126, 17, 56, 117, 66, 59, 108, 61, 62, 147, 64, 441

Offset: 0

Antti Karttunen, Jan 02 2014

nonn

Like A091203 this is a factorization-preserving isomorphism from GF(2)[X]-polynomials to integers. The former are encoded in the binary representation of n like this: n=11, '1011' in binary, stands for polynomial x^3+x+1, n=25, '11001' in binary, stands for polynomial x^4+x^3+1. However, this version does not map the irreducible GF(2)[X] polynomials (A014580) straight to the primes (A000040), but instead fixes the intersection of those two sets (A091206), and maps the elements in their set-wise difference A014580 \ A000040 (= A091214) in numerical order to the set-wise difference A000040 \ A014580 (= A091209).
The composite values are defined by the multiplicativity. E.g., we have a(A048724(n)) = 3*a(n) and a(A001317(n)) = A000244(n) = 3^n for all n.
This map satisfies many of the same identities as A091203, e.g., we have A091219(n) = A008683(a(n)), A091220(n) = A000005(a(n)), A091221(n) = A001221(a(n)), A091222(n) = A001222(a(n)), A091225(n) = A010051(a(n)) and A091247(n) = A066247(a(n)) for all n >= 1.

			Here (t X u) = A048720(t,u):
a(2)=2, a(3)=3 and a(7)=7, as 2, 3 and 7 are all in A091206.
a(4) = a(2 X 2) = a(2)*a(2) = 2*2 = 4.
a(5) = a(3 X 3) = a(3)*a(3) = 3*3 = 9.
a(9) = a(3 X 7) = a(3)*a(7) = 3*7 = 21.
a(10) = a(2 X 3 X 3) = a(2)*a(3)*a(3) = 2*3*3 = 18.
a(15) = a(3 X 3 X 3) = a(3)^3 = 3^3 = 27.
a(17) = a(3 X 3 X 3 X 3) = a(3)^4 = 3^4 = 81.
a(21) = a(7 X 7) = a(7)*a(7) = 7*7 = 49.
a(25) = 5, as 25 is the first term of A091214 and 5 is the first term of A091209.
a(50) = a(2 X 25) = a(2)*a(25) = 2*5 = 10.

Inverse: A235041. Fixed points: A235045.
Cf. A010051, A004247, A091225, A091209, A235044.
Similar cross-multiplicative permutations: A091203, A091205, A106443, A106445, A106447.

a(0)=0, a(1)=1, a(p) = p for those irreducible GF(2)[X]-polynomials whose binary encoding is a prime (i.e., p is in A091206), and for the rest of irreducible GF(2)[X]-polynomials (those which are encoded by a composite natural number, i.e., q is in A091214), a(q) = A091209(A235044(q)), and for reducible polynomials, a(i X j X k X ...) = a(i) * a(j) * a(k) * ..., where each i, j, k, ... is in A014580, X stands for carryless multiplication of GF(2)[X] polynomials (A048720) and * for the ordinary multiplication of integers (A004247).

A268389 a(n) = greatest k such that polynomial (X+1)^k divides the polynomial (in polynomial ring GF(2)[X]) that is encoded in the binary expansion of n. (See the comments for details).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 4, 1, 0, 2, 0, 0, 1, 1, 0, 0, 2, 0, 1, 3, 0, 0, 1, 4, 0, 1, 0, 0, 2, 2, 0, 0, 1, 0, 3, 1, 0, 1, 0, 0, 5, 0, 1, 2, 0, 0, 2, 1, 0, 3, 0, 0, 1, 0, 2, 1, 0, 4, 0, 0, 1, 1, 0, 0, 3, 0, 1, 2, 0, 2, 0, 0, 1, 0, 6, 1, 0, 0, 1, 3, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 1, 5, 0, 0, 3, 1, 0, 2, 0, 0, 1, 0, 1, 2, 0, 1, 0, 0, 4, 3

Offset: 1

Antti Karttunen, Feb 10 2016

a(n) gives the number of iterations of map k -> A006068(k)/2 that are required (when starting from k = n) until k is an odd number.
A001317 gives the record positions and particularly, A001317(n) gives the first occurrence of n in this sequence.
When polynomials over GF(2) are encoded in the binary representation of n in a natural way where each polynomial b(n)*X^n+...+b(0)*X^0 over GF(2) is represented by the binary number b(n)*2^n+...+b(0)*2^0 in N (each coefficient b(k) is either 0 or 1), then a(n) = the number of times polynomial X+1 (encoded by 3, "11" in binary) divides the polynomial encoded by n.

			For n = 5 ("101" in binary) which encodes polynomial x^2 + 1, we see that it can be factored over GF(2) as (X+1)(X+1), and thus a(5) = 2.
For n = 8 ("1000" in binary) which encodes polynomial x^3, we see that it is not divisible in ring GF(2)[X] by polynomial X+1, thus a(8) = 0.
For n = 9 ("1001" in binary) which encodes polynomial x^3 + 1, we see that it can be factored over GF(2) as (X+1)(X^2 + X + 1), and thus a(9) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences operating on GF(2)[X]-polynomials

Cf. A001317, A006068, A007949, A048720, A048723, A048724, A057889, A268384, A235042, A268670.
Cf. A268669 (quotient left after (X+1)^a(n) has been divided out).
Cf. A268395 (partial sums).
Cf. A000069 (positions of zeros), A268679 (this sequence without zeros).
Cf. also A037096, A037097, A136386.

f[n_] := Which[n == 1, 0, OddQ@ #, 0, EvenQ@ #, 1 + f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Array[f, {120}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

a(n) = {my(b = binary(n), p = Pol(binary(n))*Mod(1,2), k = poldegree(p)); while (type(p/(x+1)^k*Mod(1,2)) != "t_POL", k--); k;} \\ Michel Marcus, Feb 12 2016

;; This employs the given recurrence and uses memoization-macro definec:
(definec (A268389 n) (if (odd? (A006068 n)) 0 (+ 1 (A268389 (/ (A006068 n) 2)))))
(define (A268389 n) (let loop ((n n) (s 0)) (let ((k (A006068 n))) (if (odd? k) s (loop (/ k 2) (+ 1 s)))))) ;; Computed in a loop, no memoization.

If A006068(n) is odd, then a(n) = 0, otherwise a(n) = 1 + a(A006068(n)/2).
Other identities. For all n >= 0:
a(A001317(n)) = n. [The sequence works as a left inverse of A001317.]
A048720(A268669(n),A048723(3,a(n))) = A048720(A268669(n),A001317(a(n))) = n.
A048724^a(n) (A268669(n)) = n. [The a(n)-th fold application (power) of A048724 when applied to A268669(n) gives n back.]
a(n) = A007949(A235042(n)).
a(A057889(n)) = a(n).

A163617 a(2n) = 2a(n), a(2n + 1) = 2a(n) + 2 + (-1)^n, for all n in Z.

Original entry on oeis.org

0, 3, 6, 7, 12, 15, 14, 15, 24, 27, 30, 31, 28, 31, 30, 31, 48, 51, 54, 55, 60, 63, 62, 63, 56, 59, 62, 63, 60, 63, 62, 63, 96, 99, 102, 103, 108, 111, 110, 111, 120, 123, 126, 127, 124, 127, 126, 127, 112, 115, 118, 119, 124, 127, 126, 127, 120, 123, 126, 127, 124, 127, 126

Offset: 0

Michael Somos, Aug 01 2009

nonn

Fibbinary numbers (A003714) give all integers n >= 0 for which a(n) = 3*n.
From Antti Karttunen, Feb 21 2016: (Start)
Fibbinary numbers also give all integers n >= 0 for which a(n) = A048724(n).
Note that there are also other multiples of three in the sequence, for example, A163617(99) = 231 ("11100111" in binary) = 3*77, while 77 ("1001101" in binary) is not included in A003714. Note that 99 is "1100011" in binary.
(End)

			G.f. = 3*x + 6*x^2 + 7*x^3 + 12*x^4 + 15*x^5 + 14*x^6 + 15*x^7 + 24*x^8 + 27*x^9 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A003986, A048724, A213370, A163618.

Cf. also A269161.

import Data.Bits ((.|.), shiftL)
a163617 n = n .|. shiftL n 1 :: Integer
-- Reinhard Zumkeller, Mar 06 2013

using IntegerSequences
A163617List(len) = [Bits("OR", n, n<<1) for n in 0:len]
println(A163617List(62))  # Peter Luschny, Sep 26 2021

A163617 := n -> Bits:-Or(2*n, n):
seq(A163617(n), n=0..62); # Peter Luschny, Sep 23 2019

Table[BitOr[n, 2*n], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 19 2011 *)

PARI
```
{a(n) = bitor(n, n<<1)};
```

{a(n) = if( n==0 || n==-1, n, 2 * a(n \ 2) + (n%2) * (2 + (-1)^(n \ 2)))};

(define (A163617 n) (A003986bi n (+ n n))) ;; Here A003986bi implements dyadic bitwise-OR operation (see A003986) - Antti Karttunen, Feb 21 2016

a(n) = -A163618(-n) for all n in ZZ.

Conjecture: a(n) = A003188(n) + (6*n + 1 - (-1)^n)/4. - Velin Yanev, Dec 17 2016

Comment about Fibbinary numbers rephrased by Antti Karttunen, Feb 21 2016

A235041 Factorization-preserving bijection from nonnegative integers to GF(2)[X]-polynomials, version which fixes the elements that are irreducible in both semirings.

Original entry on oeis.org

0, 1, 2, 3, 4, 25, 6, 7, 8, 5, 50, 11, 12, 13, 14, 43, 16, 55, 10, 19, 100, 9, 22, 87, 24, 321, 26, 15, 28, 91, 86, 31, 32, 29, 110, 79, 20, 37, 38, 23, 200, 41, 18, 115, 44, 125, 174, 47, 48, 21, 642, 89, 52, 117, 30, 227, 56, 53, 182, 59, 172, 61, 62, 27, 64

Offset: 0

Antti Karttunen, Jan 02 2014

nonn

Like A091202 this is a factorization-preserving isomorphism from integers to GF(2)[X]-polynomials. The latter are encoded in the binary representation of n like this: n=11, '1011' in binary, stands for polynomial x^3+x+1, n=25, '11001' in binary, stands for polynomial x^4+x^3+1. However, this version does not map the primes (A000040) straight to the irreducible GF(2)[X] polynomials (A014580), but instead fixes the intersection of those two sets (A091206), and maps the elements in their set-wise difference A000040 \ A014580 (= A091209) in numerical order to the set-wise difference A014580 \ A000040 (= A091214).
The composite values are defined by the multiplicativity. E.g., we have a(3n) = A048724(a(n)) and a(3^n) = A001317(n) for all n.
This map satisfies many of the same identities as A091202, e.g., we have A000005(n) = A091220(a(n)), A001221(n) = A091221(a(n)), A001222(n) = A091222(a(n)) and A008683(n) = A091219(a(n)) for all n >= 1.

			Here (t X u) = A048720(t,u):
a(2)=2, a(3)=3 and a(7)=7, as 2, 3 and 7 are all in A091206.
a(4) = a(2*2) = a(2) X a(2) = 2 X 2 = 4.
a(9) = a(3*3) = a(3) X a(3) = 3 X 3 = 5.
a(5) = 25, as 5 is the first term of A091209 and 25 is the first term of A091214.
a(10) = a(2*5) = a(2) X a(5) = 2 X 25 = 50.
Similarly, a(17) = 55, as 17 is the second term of A091209 and 55 is the second term of A091214.
a(21) = a(3*7) = a(3) X a(7) = 3 X 7 = 9.

Inverse: A235042. Fixed points: A235045.
Cf. A010051, A048720, A091225, A091214, A235043, A048724, A115857, A115872.
Similar cross-multiplicative permutations: A091202, A091204, A106442, A106444, A106446.

a(0)=0, a(1)=1, a(p) = p for those primes p whose binary representations encode also irreducible GF(2)[X]-polynomials (i.e., p is in A091206), and for the rest of the primes q (those whose binary representation encode composite GF(2)[X]-polynomials, i.e., q is in A091209), a(q) = A091214(A235043(q)), and for composite natural numbers, a(p * q * r * ...) = a(p) X a(q) X a(r) X ..., where p, q, r, ... are primes and X stands for the carryless multiplication (A048720) of GF(2)[X] polynomials encoded as explained in the Comments section.

A269174 Formula for Wolfram's Rule 124 cellular automaton: a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).

Original entry on oeis.org

0, 3, 6, 7, 12, 15, 14, 11, 24, 27, 30, 31, 28, 31, 22, 19, 48, 51, 54, 55, 60, 63, 62, 59, 56, 59, 62, 63, 44, 47, 38, 35, 96, 99, 102, 103, 108, 111, 110, 107, 120, 123, 126, 127, 124, 127, 118, 115, 112, 115, 118, 119, 124, 127, 126, 123, 88, 91, 94, 95, 76, 79, 70, 67, 192, 195, 198, 199, 204, 207, 206, 203, 216

Offset: 0

Antti Karttunen, Feb 22 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..32767
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
Stephen Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata

Cf. A003986, A003987, A004198, A048724, A048725, A057889, A269173, A161903.
Cf. A269175.
Cf. A269176 (numbers not present in this sequence).
Cf. A269177 (same sequence sorted into ascending order, duplicates removed).
Cf. A269178 (numbers that occur only once).
Cf. A267357 (iterates from 1 onward).

package main
import "fmt"
func main() {
    for n:=0; n<=100; n++{
    fmt.Println((n|2*n)&((n^(2*n))|(n^(4*n))))}
} // Indranil Ghosh, Apr 19 2017

a[n_] := BitAnd[BitOr[n, 2n], BitOr[BitXor[n, 2n], BitXor[n, 4n]]];
a /@ Range[0, 100] (* Jean-François Alcover, Feb 23 2020 *)

def a(n): return (n|2*n)&((n^(2*n))|(n^(4*n))) # Indranil Ghosh, Apr 19 2017

(define (A269174 n) (A004198bi (A163617 n) (A003986bi (A048724 n) (A048725 n))))

a(n) = A163617(n) AND A269173(n).
a(n) = A163617(n) AND (A048724(n) OR A048725(n)).
a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).
Other identities. For all n >= 0:
a(2*n) = 2*a(n).
a(n) = A057889(A161903(A057889(n))). [Rule 124 is the mirror image of rule 110.]
G.f.: (-3*x^3 - 2*x^2 - 3*x)/(x^4 - 1) + Sum_{k>=1}((2^(k + 1)*x^(2^k) - 2^(k + 1)*x^(14*2^(k - 2)))/((x^(2^(k + 2)) - 1)*(x - 1))). - Miles Wilson, Jan 25 2025

A246200 Self-inverse permutation of natural numbers: a(n) = A057889(3*n) / 3.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 19, 14, 15, 16, 17, 18, 13, 20, 21, 22, 27, 24, 35, 38, 23, 28, 39, 30, 31, 32, 33, 34, 25, 36, 41, 26, 29, 40, 37, 42, 43, 44, 75, 54, 59, 48, 67, 70, 51, 76, 83, 46, 55, 56, 71, 78, 47, 60, 79, 62, 63, 64, 65, 66, 49, 68, 81, 50, 57, 72, 73, 82, 45, 52, 77, 58, 61, 80, 69

Offset: 0

Antti Karttunen, Aug 27 2014

In binary system, 3 ("11" in binary), has a similar shortcut rule for divisibility as eleven has in decimal system. This rule doesn't depend on which end of the number representation it is applied from, thus, if we reverse the number 3*n with "balanced bit-reverse" (A057889), the result should still be divisible by 3. Moreover, because the reversing operation is itself a self-inverse involution, and the prime factorization of any natural number is unique, we get a self-inverse permutation of nonnegative integers when we divide the bit-reversed result with 3.

Antti Karttunen, Table of n, a(n) for n = 0..10921
Index entries for sequences that are permutations of the natural numbers

Cf. A036215, A057889, A003714, A048724, A083822, A083824.

def a057889(n):
    x=bin(n)[2:]
    y=x[::-1]
    return int(str(int(y))+(len(x) - len(str(int(y))))*'0', 2)
def a(n): return a057889(3*n)//3
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 11 2017

(define (A246200 n) (/ (A057889 (* 3 n)) 3))

a(n) = A057889(3*n) / 3.

A106409 n XOR (greatest proper divisor of n).

Original entry on oeis.org

0, 3, 2, 6, 4, 5, 6, 12, 10, 15, 10, 10, 12, 9, 10, 24, 16, 27, 18, 30, 18, 29, 22, 20, 28, 23, 18, 18, 28, 17, 30, 48, 42, 51, 36, 54, 36, 53, 42, 60, 40, 63, 42, 58, 34, 57, 46, 40, 54, 43, 34, 46, 52, 45, 60, 36, 42, 39, 58, 34, 60, 33, 42, 96, 76, 99, 66, 102, 82, 101, 70

Offset: 1

Reinhard Zumkeller, May 02 2005

a(n) = n XOR A032742(n);
a(A000040(n)) = A000040(n) - 1 for n>1;
a(2*n) = A048724(n);
a(A022340(n)) = A022340(n) + A032742(A022340(n)).

Alois P. Heinz, Table of n, a(n) for n = 1..16383

a:= n-> Bits[Xor](n, max(1, (numtheory[divisors](n) minus {n})[])):
seq(a(n), n=1..100);  # Alois P. Heinz, May 15 2016

a[1] = 0; a[n_] := BitXor[n, Divisors[n][[-2]]]; Array[a, 100] (* Jean-François Alcover, Mar 12 2019 *)

gpd(n) = if(n==1, 1, n/factor(n)[1, 1]); \\ A032742
a(n) = bitxor(n, gpd(n)); \\ Michel Marcus, Mar 12 2019

A268669 a(n) = polynomial quotient (computed over GF(2), result is its binary encoding) that is left after all instances of polynomial (X+1) have been factored out of the polynomial that is encoded by the binary expansion of n. (See comments for details).

Original entry on oeis.org

1, 2, 1, 4, 1, 2, 7, 8, 7, 2, 11, 4, 13, 14, 1, 16, 1, 14, 19, 4, 21, 22, 13, 8, 25, 26, 7, 28, 11, 2, 31, 32, 31, 2, 35, 28, 37, 38, 11, 8, 41, 42, 25, 44, 7, 26, 47, 16, 49, 50, 1, 52, 19, 14, 55, 56, 13, 22, 59, 4, 61, 62, 21, 64, 21, 62, 67, 4, 69, 70, 61, 56, 73, 74, 13, 76, 59, 22, 79, 16, 81, 82, 49, 84, 1

Offset: 1

Antti Karttunen, Feb 10 2016

When polynomials over GF(2) are encoded in the binary representation of n in a natural way where each polynomial b(n)*X^n+...+b(0)*X^0 over GF(2) is represented by the binary number b(n)*2^n+...+b(0)*2^0 in N (each coefficient b(k) is either 0 or 1), then a(n) = representation of the polynomial that is left as a quotient when all X+1 polynomials (encoded by 3, "11" in binary) have been divided out.
Each a(n) is one of the "Garden of Eden" patterns of Rule-60 one-dimensional cellular automaton, a seed pattern which after A268389(n) generations yields the configuration encoded in binary expansion of n.
No terms of A001969 occur so all terms are odious (in A000069). Each odious number occurs an infinitely many times.

			For n = 5 ("101" in binary) which encodes polynomial x^2 + 1, we observe that it can be factored in ring GF(2)[X] as (X+1)(X+1), and thus a(5) = 1, because after dividing both instances of (X+1) off, we are left with the quotient polynomial 1 which is encoded by 1.
For n = 8 ("1000" in binary) which encodes polynomial x^3, we observe that it is not divisible in ring GF(2)[X] by polynomial X+1, thus a(8) = 8.
For n = 9 ("1001" in binary) which encodes polynomial x^3 + 1, we observe that it can be factored over GF(2) as (X+1)(X^2 + X + 1), and thus a(9) = 7, because the quotient polynomial X^2 + X + 1 is encoded by 7 ("111" in binary).

Antti Karttunen, Table of n, a(n) for n = 1..65537
Eric Weisstein's World of Mathematics, Rule 60
Index entries for sequences related to cellular automata
Index entries for sequences operating on GF(2)[X]-polynomials

Cf. A001317 (positions of ones).
Cf. A000069, A001969, A003188, A003987, A010060, A048720, A048723, A048724, A268384, A268670.
Cf. A268389 (the highest exponent for (X+1)).
Cf. also A136386.

f[n_] := If[OddQ@ #, n, f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Array[f, {85}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

a(n) = {p = Pol(binary(n))*Mod(1,2); q = (x+1)*Mod(1,2); while (type(r = p/q) == "t_POL", p = r); np = Polrev(vector(poldegree(p)+1, k, k--; lift(polcoeff(p, k)))); subst(np, x, 2);} \\ Michel Marcus, Feb 12 2016

;; This employs the given recurrence and uses memoization-macro definec:
(definec (A268669 n) (if (odd? (A006068 n)) n (A268669 (/ (A006068 n) 2))))
(define (A268669 n) (let loop ((n n)) (let ((k (A006068 n))) (if (odd? k) n (loop (/ k 2)))))) ;; Computed in a loop, no memoization.

If A006068(n) is odd, then a(n) = n, otherwise a(n) = a(A006068(n)/2).
Other identities and observations. For all n >= 1:
a(n) = A003188(A268670(n)).
A010060(a(n)) = 1. [All terms are odious.]
a(n) <= n.
More precisely, a(A000069(n)) = A000069(n) and a(A001969(n)) < A001969(n).
The equivalence of the following two formulas stems from the additive nature of Rule-60 cellular automaton. Or more plainly, because carryless binary multiplication A048720 distributes over carryless binary sum, XOR A003987:
A048724^A268389(n) (a(n)) = n. [Starting from k = a(n), and iterating map k -> A048724(k) exactly A268389(n) times yields n back.]
A048720(a(n),A048723(3,A268389(n))) = A048720(a(n),A001317(A268389(n))) = n.

A178729 a(n) = n XOR 3n, where XOR is bitwise XOR.

Original entry on oeis.org

0, 2, 4, 10, 8, 10, 20, 18, 16, 18, 20, 42, 40, 42, 36, 34, 32, 34, 36, 42, 40, 42, 84, 82, 80, 82, 84, 74, 72, 74, 68, 66, 64, 66, 68, 74, 72, 74, 84, 82, 80, 82, 84, 170, 168, 170, 164, 162, 160, 162, 164, 170, 168, 170, 148, 146, 144, 146, 148, 138, 136, 138, 132, 130

Offset: 0

Dmitry Kamenetsky, Jun 08 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Helena A. Verrill, Note on the negative base xor sequence n⊕-b(-n), arXiv:2506.09509 [math.NT], 2025.

Cf. A048724, A048725, A178731, A178732, A178733, A178734, A178735, A178736. - Robert G. Wilson v, Jun 09 2010

Cf. A184617.

[BitwiseXor(n, 3*n): n in [0..80]]; // Vincenzo Librandi, Jul 11 2017

f[n_] := BitXor[n, 3 n]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 09 2010 *)

a(n) = bitxor(n, 3*n); \\ Michel Marcus, Jul 11 2017

def A178729(n): return n^ 3*n # Chai Wah Wu, Jun 29 2022

a(n) = A005351(n) XOR A005352(n) (conjectured). Proved by Verrill link.

a(n) = 2 * A184617(n). - Alois P. Heinz, Jul 21 2017

a(30) onwards from Robert G. Wilson v, Jun 09 2010

A234027 Self-inverse permutation of nonnegative integers, A054429-conjugate of blue code: a(n) = A054429(A193231(A054429(n))).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Dec 28 2013

nonn

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences that are permutations of the natural numbers

Cf. A054429, A059894, A193231, A234024-A234026.

def a065621(n): return n^(2*(n - (n&-n)))
def a048724(n): return n^(2*n)
def a054429(n): return 1 if n==1 else 2*a054429(int(n/2)) + 1 - n%2
def a193231(n):
    if n<2: return n
    if n%2==0: return a048724(a193231(n/2))
    else: return a065621(1 + a193231((n - 1)/2))
def a(n): return n if n<2 else a054429(a193231(a054429(n))) # Indranil Ghosh, Jun 05 2017

(define (A234027 n) (A054429 (A193231 (A054429 n))))

a(n) = A054429(A193231(A054429(n))).
a(n) = A234025(A054429(n)).
a(n) = A054429(A234026(n)).
a(n) = A059894(A234024(A059894(n))).

cp's OEIS Frontend

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A235041 Factorization-preserving bijection from nonnegative integers to GF(2)[X]-polynomials, version which fixes the elements that are irreducible in both semirings.

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