A268833 -id:A268833 - cp's OEIS frontend

A268834 Transpose of array A268833.

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

Offset: 0

Antti Karttunen, Feb 15 2016

See comments in A268833.

			The top left [0 .. 16] x [0 .. 16] section of the array:
  0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2
  0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2
  0, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2
  0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 2
  0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2
  0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2
  0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 2
  0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2
  0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2
  0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2
  0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 2
  0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2
  0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2
  0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2
  0, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2
  0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 2
  0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2

Antti Karttunen, Table of n, a(n) for n = 0..8255; the first 128 antidiagonals of array

Transpose of A268833.

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[r - c, 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(r - c, c),", ");); print();) \\ Indranil Ghosh, Apr 02 2017

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(0, 21):
    print([a(r - c, c) for c in range(0, r + 1)]) # Indranil Ghosh, Apr 02 2017

(define (A268834 n) (A268833bi (A025581 n) (A002262 n))) ;; Code for A268833bi given in A268833.

A268835 Main diagonal of arrays A268833 & A268834: a(n) = A101080(n, A268820(n, 2*n)).

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 2, 3, 2, 3, 4, 5, 2, 1, 4, 3, 2, 3, 4, 5, 4, 3, 6, 5, 2, 3, 2, 1, 4, 5, 4, 3, 2, 3, 4, 5, 4, 3, 6, 5, 4, 5, 4, 3, 6, 7, 6, 5, 2, 3, 4, 3, 2, 3, 2, 3, 4, 5, 6, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 3, 6, 5, 4, 5, 4, 3, 6, 7, 6, 5, 4, 5, 6, 5, 4, 5, 4, 5, 6, 7, 8, 7, 6, 5, 6, 5, 2, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 5, 2, 1, 4, 3, 4, 5, 6, 5, 6, 5, 6, 5, 4

Offset: 0

Antti Karttunen, Feb 15 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..1024
Indranil Ghosh, C program to generate the sequence

Cf. A101080, A268820, A268833, A268834.

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]]]; Flatten@ Table[A101080[n, a[n, 2n]], {n, 0, 300}] (* 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)))));
for(n=0, 300, print1(A101080(n, A268820(n, 2*n)),", ")) \\ Indranil Ghosh, Apr 02 2017

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 A268717(n): return 0 if n<1 else A003188(1 + A006068(n - 1))
def A268820(r, c): return c if r<1 else 0 if c<1 else A003188(1 + A006068(A268820(r - 1, c - 1)))
print([A101080(n, A268820(n, 2*n)) for n in range(301)]) # Indranil Ghosh, Apr 02 2017

(define (A268835 n) (A101080bi n (A268820bi n (* 2 n))))
(define (A268835 n) (A268833bi n n)) ;; Code for A268833bi given in A268833.

a(n) = A101080(n, A268820(n, 2*n)).

A268820 Square array A(r,c): A(0,c) = c, A(r,0) = 0, A(r>=1,c>=1) = A003188(1+A006068(A(r-1,c-1))) = A268717(1+A(r-1,c-1)), read by descending antidiagonals as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ...

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 6, 3, 1, 0, 5, 2, 2, 3, 1, 0, 6, 12, 7, 2, 3, 1, 0, 7, 4, 6, 6, 2, 3, 1, 0, 8, 7, 13, 5, 6, 2, 3, 1, 0, 9, 5, 12, 7, 7, 6, 2, 3, 1, 0, 10, 24, 5, 15, 4, 7, 6, 2, 3, 1, 0, 11, 8, 4, 13, 5, 5, 7, 6, 2, 3, 1, 0, 12, 11, 25, 4, 14, 12, 5, 7, 6, 2, 3, 1, 0, 13, 9, 24, 12, 15, 4, 4, 5, 7, 6, 2, 3, 1, 0, 14, 13, 9, 27, 12, 10, 13, 4, 5, 7, 6, 2, 3, 1, 0

Offset: 0

Antti Karttunen, Feb 14 2016

			The top left [0 .. 16] x [0 .. 19] section of the array:
0, 1, 2, 3, 4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
0, 1, 3, 6, 2, 12,  4,  7,  5, 24,  8, 11,  9, 13, 15, 10, 14, 48, 16, 19
0, 1, 3, 2, 7,  6, 13, 12,  5,  4, 25, 24,  9,  8, 15, 14, 11, 10, 49, 48
0, 1, 3, 2, 6,  5,  7, 15, 13,  4, 12, 27, 25,  8, 24, 14, 10,  9, 11, 51
0, 1, 3, 2, 6,  7,  4,  5, 14, 15, 12, 13, 26, 27, 24, 25, 10, 11,  8,  9
0, 1, 3, 2, 6,  7,  5, 12,  4, 10, 14, 13, 15, 30, 26, 25, 27, 11,  9, 24
0, 1, 3, 2, 6,  7,  5,  4, 13, 12, 11, 10, 15, 14, 31, 30, 27, 26,  9,  8
0, 1, 3, 2, 6,  7,  5,  4, 12, 15, 13,  9, 11, 14, 10, 29, 31, 26, 30,  8
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 14, 15,  8,  9, 10, 11, 28, 29, 30, 31
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 10, 14, 24,  8, 11,  9, 20, 28, 31
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 11, 10, 25, 24,  9,  8, 21, 20
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10,  9, 11, 27, 25,  8, 24, 23
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10, 11,  8,  9, 26, 27, 24, 25
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10, 11,  9, 24,  8, 30, 26, 25
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10, 11,  9,  8, 25, 24, 31, 30
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10, 11,  9,  8, 24, 27, 25, 29
0, 1, 3, 2, 6,  7,  5,  4, 12, 13, 15, 14, 10, 11,  9,  8, 24, 25, 26, 27

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

Cf. A003188, A006068.
Inverses of these permutations can be found in table A268830.
Row 0: A001477, Row 1: A268717, Row 2: A268821, Row 3: A268823, Row 4: A268825, Row 5: A268827, Row 6: A268831, Row 7: A268933.
Rows converge towards A003188, which is also the main diagonal.
Cf. array A268715 (can be extracted from this one).
Cf. array A268833 (shows related Hamming distances with regular patterns).

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]]]; Table[a[c, r - c], {r, 0, 15}, {c, 0, r}] //Flatten (* 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});
a(r, c) = if(r==0, c, if(c==0, 0, A003188(1 + A006068(a(r - 1, c - 1)))));
for(r=0, 15, for(c=0, r, print1(a(c, r - c),", "); ); print(); ); \\ 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 a(r, c): return c if r<1 else 0 if c<1 else A003188(1 + A006068(a(r - 1, c - 1)))
for r in range(16):
    print([a(c, r - c) for c in range(r + 1)]) # Indranil Ghosh, Apr 02 2017

(define (A268820 n) (A268820bi (A002262 n) (A025581 n)))
(define (A268820bi row col) (cond ((zero? row) col) ((zero? col) 0) (else (A268717 (+ 1 (A268820bi (- row 1) (- col 1)))))))
(define (A268820bi row col) (cond ((zero? row) col) ((zero? col) 0) (else (A003188 (+ 1 (A006068 (A268820bi (- row 1) (- col 1))))))))

For row zero: A(0,k) = k, for column zero: A(n,0) = 0, and in other cases: A(n,k) = A003188(1+A006068(A(n-1,k-1)))
Other identities. For all n >= 0:
A(n,n) = A003188(n).
A(A006068(n),A006068(n)) = n.

A268821 Permutation of nonnegative integers: a(0) = 0, a(n) = A268717(1 + A268717(n-1)).

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 13, 12, 5, 4, 25, 24, 9, 8, 15, 14, 11, 10, 49, 48, 17, 16, 23, 22, 19, 18, 27, 26, 31, 30, 21, 20, 29, 28, 97, 96, 33, 32, 39, 38, 35, 34, 43, 42, 47, 46, 37, 36, 45, 44, 51, 50, 55, 54, 61, 60, 53, 52, 41, 40, 57, 56, 63, 62, 59, 58, 193, 192, 65, 64, 71, 70, 67, 66, 75, 74, 79, 78, 69, 68, 77, 76, 83, 82

Offset: 0

Antti Karttunen, Feb 14 2016

nonn

The "shifted square" of permutation A268717.

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

Inverse: A268822.
Row 2 of array A268820.
From term a(2) onward (3, 2, 7, 6, ...) also row 3 of A268715.
Cf. A268717, A268823.
Cf. also A101080, 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]]]; Table[If[n<2, n, A268717[1 + A268717[n - 1]]], {n, 0, 100}] (* Indranil Ghosh, Apr 01 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, 100, print1(if(n<2, n, A268717(1 + A268717(n - 1))), ", ")) \\ Indranil Ghosh, Apr 01 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))
def a(n): return A268717(1 + A268717(n-1)) if n>0 else 0
print([a(n) for n in range(101)]) # Indranil Ghosh, Apr 01 2017

(define (A268821 n) (if (zero? n) n (A268717 (+ 1 (A268717 (- n 1))))))

a(0) = 0, for n >= 1, a(n) = A268717(1 + A268717(n-1)).
Other identities. For all n >= 0:
A101080(n, a(n+2)) = 2.

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).

A268727 One-based index of the toggled bit between n and A268717(n+1): a(n) = A070939(A003987(n,A268717(1+n))).

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 1, 2, 5, 1, 1, 2, 1, 2, 3, 1, 6, 1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 4, 1, 1, 2, 7, 1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 2, 3, 1, 4, 1, 1, 2, 5, 1, 1, 2, 1, 2, 3, 1, 8, 1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 2, 3, 1, 4, 1, 1, 2, 5, 1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 4, 1, 1, 2, 5, 1, 1, 2, 1, 2, 3, 1, 6, 1, 1, 2, 1, 2, 3, 1, 1

Offset: 0

Antti Karttunen, Feb 13 2016

A fractal sequence like A268726.

Antti Karttunen, Table of n, a(n) for n = 0..8191

One more than A268726.
Cf. A003188, A006068.
Cf. A001511, A003987, A010059, A070939, A101080, A268717.
Cf. also array A268833.

(define (A268727 n) (A070939 (A003987bi n (A268717 (+ 1 n)))))  ;; A003987bi implements the bitwise-XOR, defined in A003987.

a(n) = A001511(1+A006068(n)).
a(n) = A070939(A003987(n,A268717(1+n))).
a(n) = 1 + floor(log_2(n XOR A003188(1+A006068(n)))).
a(n) = A001511(n)*(1-A010059(n)) + 1. - Alan Michael Gómez Calderón, Jun 15 2025

A268676 a(n) = A101080(n,A268823(3+n)), where A101080(x,y) gives the Hamming distance between binary expansions of x and y.

Original entry on oeis.org

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, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 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, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 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

Offset: 0

Antti Karttunen, Feb 19 2016

nonn

It seems that A001969 gives the positions of 1's, while A000069 gives the positions of 3's.

The above observation follows because by definition, a(n) gives the Hamming distance between binary expansions of n and A003188(3+A006068(n)). To see how this leads to the stated claim, consider the illustration "Visualized as a traversal of vertices of a tesseract" in the Wikipedia article "Gray code". Starting from any vertex with either (A) an even, or (B) an odd number of 1-bits, traverse three edges along the red path, to the direction indicated by the arrow, and then note the Hamming distance between the starting and the ending vertex. It is always 1 in case (A), and 3 in case (B), because the position of the flipped bit is given by sequence A007814, with its every other term zero, so in case (A) the third flip cancels the first flip (both toggling the rightmost bit), which leaves only the second bit-flip effective. Note that the properties of a tesseract generalize to those of an infinite dimensional hypercube. - Antti Karttunen, Mar 11 2024

Antti Karttunen, Table of n, a(n) for n = 0..8191
Wikipedia, Gray code.

Row 3 of A268833.

Cf. A000069, A000120, A001969, A003188, A006068, A007814, A010060, A101080, A106400, A268823.

a := n -> 2-(-1)^add(convert(n, base, 2)):
seq(a(n), n = 0 .. 120); # Lorenzo Sauras Altuzarra, Mar 10 2024

A003188(n) = bitxor(n, n>>1);
A006068(n) = { my(s=1, ns); while(1, ns = n >> s; if(0==ns, return(n)); n = bitxor(n, ns); s <<= 1); };
A268676(n) = hammingweight(bitxor(n,A003188(3+A006068(n)))); \\ Antti Karttunen, Mar 11 2024

A268676(n) = 2-((-1)^hammingweight(n)); \\ Antti Karttunen, Mar 11 2024, after Lorenzo Sauras Altuzarra's Maple-code.

(define (A268676 n) (A101080bi n (A268823 (+ 3 n))))
;; Where A101080bi implements the dyadic function A101080(x,y) which gives the Hamming distance between binary expansions of x and y.

a(n) = A101080(n,A268823(3+n)), where A101080(x,y) gives the Hamming distance between binary expansions of x and y.
a(n) = 2 - (-1)^A000120(n) = 2 - A106400(n). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 + 2 * A010060(n). - Joerg Arndt, Mar 11 2024

cp's OEIS Frontend

A268834 Transpose of array A268833.

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A268835 Main diagonal of arrays A268833 & A268834: a(n) = A101080(n, A268820(n, 2*n)).

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A268820 Square array A(r,c): A(0,c) = c, A(r,0) = 0, A(r>=1,c>=1) = A003188(1+A006068(A(r-1,c-1))) = A268717(1+A(r-1,c-1)), read by descending antidiagonals as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ...

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A268821 Permutation of nonnegative integers: a(0) = 0, a(n) = A268717(1 + A268717(n-1)).

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

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A268727 One-based index of the toggled bit between n and A268717(n+1): a(n) = A070939(A003987(n,A268717(1+n))).

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A268676 a(n) = A101080(n,A268823(3+n)), where A101080(x,y) gives the Hamming distance between binary expansions of x and y.

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