A268727 -id:A268727 - cp's OEIS frontend

A268717 Permutation of natural numbers: a(0) = 0, a(n) = A003188(1+A006068(n-1)), where A003188 is binary Gray code and A006068 is its inverse.

0, 1, 3, 6, 2, 12, 4, 7, 5, 24, 8, 11, 9, 13, 15, 10, 14, 48, 16, 19, 17, 21, 23, 18, 22, 25, 27, 30, 26, 20, 28, 31, 29, 96, 32, 35, 33, 37, 39, 34, 38, 41, 43, 46, 42, 36, 44, 47, 45, 49, 51, 54, 50, 60, 52, 55, 53, 40, 56, 59, 57, 61, 63, 58, 62, 192, 64, 67, 65, 69, 71, 66, 70, 73, 75, 78, 74, 68, 76, 79, 77, 81

Offset: 0

Antti Karttunen, Feb 12 2016

nonn

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

Inverse: A268718.
Cf. A000523, A003188, A003987, A006068, A010060, A066194, A101080, A171977, A268718, A268726, A268727.
Row 1 and column 1 of array A268715 (without the initial zero).
Row 1 of array A268820.
Cf. A092246 (fixed points).
Cf. A268817 ("square" of this permutation).
Cf. A268821 ("shifted square"), A268823 ("shifted cube") and also A268825, A268827 and A268831 ("shifted higher powers").

A003188[n_] := BitXor[n, Floor[n/2]]; A006068[n_] := If[n == 0, 0, BitXor @@ Table[Floor[n/2^m], {m, 0, Floor[Log[2, n]]}]]; a[n_] := If[n == 0, 0, A003188[1 + A006068[n-1]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 23 2016 *)

A003188(n) = bitxor(n, floor(n/2));
A006068(n) = if(n<2, n, {my(m = A006068(floor(n/2))); 2*m + (n%2 + m%2)%2});
for(n=0, 100, print1(if(n<1, 0, A003188(1 + A006068(n - 1)))", ")) \\ Indranil Ghosh, Mar 31 2017

def A003188(n): return n^(n//2)
def A006068(n):
    if n<2: return n
    m = A006068(n//2)
    return 2*m + (n%2 + m%2)%2
def a(n): return 0 if n<1 else A003188(1 + A006068(n - 1))
print([a(n) for n in range(0, 101)]) # Indranil Ghosh, Mar 31 2017

def A268717(n):
    k, m = n-1, n-1>>1
    while m > 0:
        k ^= m
        m >>= 1
    return k+1^ k+1>>1 # Chai Wah Wu, Jun 29 2022

(define (A268717 n) (if (zero? n) n (A003188 (A066194 n))))

a(n) = A003188(A066194(n)) = A003188(1+A006068(n-1)).
Other identities. For all n >= 0:
A101080(n,a(n+1)) = 1. [The Hamming distance between n and a(n+1) is always one.]
A268726(n) = A000523(A003987(n, a(n+1))). [A268726 gives the index of the toggled bit.]
From Alan Michael Gómez Calderón, May 29 2025: (Start)
a(2*n) = (2*n-1) XOR (2-A010060(n-1)) for n >= 1;
a(n) = (A268718(n-1)-1) XOR (A171977(n-1)+1) for n >= 2. (End)

A268833 Square array A(n, k) = A101080(k, A003188(n+A006068(k))), read by descending antidiagonals, where A003188 is the binary Gray code, A006068 is its inverse, and A101080(x,y) gives the Hamming distance between binary expansions of x and y.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 2, 3, 2, 0, 1, 2, 3, 2, 3, 0, 1, 2, 1, 2, 1, 2, 0, 1, 2, 3, 2, 3, 2, 1, 0, 1, 2, 1, 2, 3, 4, 3, 2, 0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 0, 1, 2, 3, 2, 3, 4, 3, 2, 1, 4, 3, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 1, 2, 1, 2

Offset: 0

Antti Karttunen, Feb 15 2016

The entry at row n, column k, gives the Hamming distance between binary expansions of k and A003188(n+A006068(k)). When Gray code is viewed as a traversal of vertices of an infinite dimensional hypercube by bit-flipping (see the illustration "Visualized as a traversal of vertices of a tesseract" in the Wikipedia's "Gray code" article) the argument k is the "address" (the binary code given inside each vertex) of the starting vertex, and argument n tells how many edges forward along the Gray code path we should hop from it (to the direction that leads away from the vertex with code 0000...). A(n, k) gives then the Hamming distance between the starting and the ending vertex. For how this works with case n=3, see comments in A268676. - Antti Karttunen, Mar 11 2024

			The top left [0 .. 24] X [0 .. 24] section of the array:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 1, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 3, 3, 1, 3, 1, 3, 3, 3
2, 2, 4, 4, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 2
1, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3
4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 2, 2, 4
3, 3, 3, 1, 5, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 1, 5, 3, 3, 5, 3, 3, 3, 1, 3
2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 2
3, 1, 3, 3, 3, 5, 5, 3, 3, 5, 5, 3, 3, 1, 3, 3, 3, 5, 5, 3, 3, 1, 3, 3, 3
2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 2
1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
3, 5, 5, 3, 3, 1, 3, 3, 5, 3, 3, 5, 3, 5, 5, 3, 5, 3, 3, 5, 3, 5, 5, 3, 3
4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
5, 3, 3, 5, 3, 3, 3, 1, 5, 5, 5, 3, 5, 3, 5, 5, 5, 5, 5, 3, 5, 3, 5, 5, 5
4, 4, 4, 4, 4, 4, 2, 2, 6, 6, 4, 4, 4, 4, 6, 6, 6, 6, 4, 4, 4, 4, 6, 6, 4
3, 3, 3, 3, 3, 3, 1, 3, 5, 5, 3, 5, 3, 5, 5, 5, 5, 5, 3, 5, 3, 5, 5, 5, 3
2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2

Antti Karttunen, Table of n, a(n) for n = 0..32895; the first 256 antidiagonals of array
Wikipedia, Gray code.

Transpose A268834.
Main diagonal: A268835.
Column 0: A005811.
Row 0: A000004, Row 1: A000012, Row 2: A007395, Row 3: A268676.
Cf. A003188, A006068.
Cf. also A268726, A268727.

A101080[n_, k_]:= DigitCount[BitXor[n, k], 2, 1];A003188[n_]:=BitXor[n, Floor[n/2]]; A006068[n_]:=If[n<2, n, Block[{m=A006068[Floor[n/2]]}, 2m + Mod[Mod[n,2] + Mod[m, 2], 2]]]; a[r_, 0]:= 0; a[0, c_]:=c; a[r_, c_]:= A003188[1 + A006068[a[r - 1, c - 1]]]; A[r_, c_]:=A101080[c, a[r, r + c]]; Table[A[c, r - c], {r, 0, 20}, {c, 0, r}] // Flatten (* Indranil Ghosh, Apr 02 2017 *)

b(n) = if(n<1, 0, b(n\2) + n%2);
A101080(n, k) = b(bitxor(n, k));
A003188(n) = bitxor(n, n\2);
A006068(n) = if(n<2, n, {my(m = A006068(n\2)); 2*m + (n%2 + m%2)%2});
A268820(r, c) = if(r==0, c, if(c==0, 0, A003188(1 + A006068(A268820(r - 1, c - 1)))));
A(r, c) = A101080(c, A268820(r, r + c));
for(r=0, 20, for(c=0, r, print1(A(c, r - c),", ");); print();) \\ Indranil Ghosh, Apr 02 2017

up_to = 32895; \\ = binomial(1+256,2)-1.\\ A003188 and A006068 as above.
A268833sq(n, k) = hammingweight(bitxor(n,A003188(k+A006068(n))));
A268833list(up_to) = { my(v = vector(up_to), i=0); for(a=0,oo, for(col=0,a, i++; if(i > up_to, return(v)); v[i] = A268833sq(a-col,col))); (v); };
v268833 = A268833list(1+up_to);
A268833(n) = v268833[1+n]; \\ Antti Karttunen, Mar 11 2024

def A101080(n, k): return bin(n^k)[2:].count("1")
def A003188(n): return n^(n//2)
def A006068(n):
    if n<2: return n
    else:
        m=A006068(n//2)
        return 2*m + (n%2 + m%2)%2
def A268820(r, c): return c if r<1 else 0 if c<1 else A003188(1 + A006068(A268820(r - 1, c - 1)))
def a(r, c): return A101080(c, A268820(r, r + c))
for r in range(21):
    print([a(c, r - c) for c in range(r + 1)]) # Indranil Ghosh, Apr 02 2017

(define (A268833 n) (A268833bi (A002262 n) (A025581 n)))
(define (A268833bi row col) (A101080bi col (A268820bi row (+ row col))))

A(row,col) = A101080(col, A268820(row, row+col)).

A(n, k) = A101080(k, A003188(n+A006068(k))). - Antti Karttunen, Mar 11 2024

Definition simplified by Antti Karttunen, Mar 11 2024

A268726 Index of the toggled bit between n and A268717(n+1): a(n) = A000523(A003987(n, A268717(1+n))).

Original entry on oeis.org

0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 6, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 7, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0

Offset: 0

Antti Karttunen, Feb 13 2016

A fractal sequence, because a permutation of A007814. Removing zeros yields A268727(n) = a(n)+1.

Antti Karttunen, Table of n, a(n) for n = 0..16383
Indranil Ghosh, C program to generate the sequence

One less than A268727.
Cf. A003188, A006068.
Cf. A000523, A003987, A007814, A101080, A268717.
Cf. also array A268833.

A003188[n_]:=BitXor[n, Floor[n/2]]; A006068[n_]:=If[n<2, n, Block[{m=A006068[Floor[n/2]]}, 2m + Mod[Mod[n,2] + Mod[m, 2], 2]]]; A268717[n_]:=If[n<1, 0, A003188[1 + A006068[n - 1]]]; a[n_]:= Floor[Log[2, BitXor[n, A268717[n + 1]]]]; Table[a[n], {n, 0, 200}] (* Indranil Ghosh, Apr 02 2017 *)

A003188(n) = bitxor(n, n\2);
A006068(n) = if(n<2, n, {my(m = A006068(n\2)); 2*m + (n%2 + m%2)%2});
A268717(n) = if(n<1, 0, A003188(1 + A006068(n - 1)));
for(n=0, 200, print1(logint(bitxor(n, A268717(n + 1)), 2),", "))  \\ Indranil Ghosh, Apr 02 2017

def A003188(n): return n^(n//2)
def A006068(n):
    if n<2: return n
    else:
        m=A006068(n//2)
        return 2*m + (n%2 + m%2)%2
def A268717(n): return 0 if n<1 else A003188(1 + A006068(n - 1))
print([int(math.floor(math.log(n^A268717(n + 1), 2))) for n in range(201)]) # Indranil Ghosh, Apr 02 2017

(define (A268726 n) (A000523 (A003987bi n (A268717 (+ 1 n))))) ;; A003987bi implements the bitwise-XOR, defined in A003987.

a(n) = A007814(1 + A006068(n)).
a(n) = A000523(A003987(n, A268717(1+n))).
a(n) = floor(log_2(n XOR A003188(1 + A006068(n)))).
Other identities:
For all n >= 1, a(A003188(n-1)) = A007814(n).

cp's OEIS Frontend

A268717 Permutation of natural numbers: a(0) = 0, a(n) = A003188(1+A006068(n-1)), where A003188 is binary Gray code and A006068 is its inverse.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Python

Python

Scheme

Formula

A268833 Square array A(n, k) = A101080(k, A003188(n+A006068(k))), read by descending antidiagonals, where A003188 is the binary Gray code, A006068 is its inverse, and A101080(x,y) gives the Hamming distance between binary expansions of x and y.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Scheme

Formula

Extensions

A268726 Index of the toggled bit between n and A268717(n+1): a(n) = A000523(A003987(n, A268717(1+n))).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Python

Scheme

Formula