A101080 -id:A101080 - cp's OEIS frontend

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.

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

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

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)).

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.

Original entry on oeis.org

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)

A163252 a(0) = 0, and a(n) is the least positive integer not occurring earlier in the sequence such that a(n-1) and a(n) differ in only one bit when written in binary.

Original entry on oeis.org

0, 1, 3, 2, 6, 4, 5, 7, 15, 11, 9, 8, 10, 14, 12, 13, 29, 21, 17, 16, 18, 19, 23, 22, 20, 28, 24, 25, 27, 26, 30, 31, 63, 47, 39, 35, 33, 32, 34, 38, 36, 37, 45, 41, 40, 42, 43, 59, 51, 49, 48, 50, 54, 52, 53, 55, 119, 87, 71, 67, 65, 64, 66, 70, 68, 69, 77, 73, 72, 74, 75, 79, 78

Offset: 0

Keenan Pepper, Jul 23 2009

Step from a(488) = 237 = 11101101_2 to a(489) = 749 = 1011101101_2 is the first case when one term is two binary digits longer than the previous. Considering the leading zeros, though, they still differ in only one bit. - Ivan Neretin, Jun 25 2015

Paul Tek, Table of n, a(n) for n = 0..10000

Cf. A003188, A101080, A360307 (inverse).

N:= 10: # to get all terms before the first where a(n) >= 2^N
B:= Array(0..2^N-1):
B[0]:= 1:
a[0]:= 0:
L:= Vector([0$N]):
for n from 1 do
  cands:= select(t -> B[t[1]]=0, [seq(`if`(L[i]=0,[a[n-1]+2^(i-1),i],[a[n-1]-2^(i-1),i]),i=1..N)]);
  if nops(cands)=0 then break fi;
  j:= min[index](map(t->t[1],cands));
  a[n]:= cands[j][1];
  i:= cands[j][2];
  B[a[n]]:= 1;
  L[i]:= 1 - L[i];
od:
seq(a[i],i=0..n-1); # Robert Israel, Jun 25 2015

Nest[Append[#, Min[Complement[BitXor[#[[-1]], 2^Range[0, Floor[Log2[#[[-1]]]] + 2]], #]]] &, {0, 1}, 71] (* Ivan Neretin, Jun 25 2015 *)

From Alois P. Heinz, Feb 02 2023: (Start)
A101080(a(n),a(n+1)) = 1.
a(A360307(n)) = n = A360307(a(n)). (End)

A230415 Square array T(i,j) giving the number of differing digits in the factorial base representations of i and j, for i >= 0, j >= 0, read by antidiagonals.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Nov 10 2013

This table relates to the factorial base representation (A007623) in a somewhat similar way as A101080 relates to the binary system. See A231713 for another analog.

			The top left corner of this square array begins as:
0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, ...
1, 0, 2, 1, 2, 1, 2, 1, 3, 2, 3, ...
1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, ...
2, 1, 1, 0, 2, 1, 3, 2, 2, 1, 3, ...
1, 2, 1, 2, 0, 1, 2, 3, 2, 3, 1, ...
2, 1, 2, 1, 1, 0, 3, 2, 3, 2, 2, ...
1, 2, 2, 3, 2, 3, 0, 1, 1, 2, 1, ...
2, 1, 3, 2, 3, 2, 1, 0, 2, 1, 2, ...
2, 3, 1, 2, 2, 3, 1, 2, 0, 1, 1, ...
3, 2, 2, 1, 3, 2, 2, 1, 1, 0, 2, ...
2, 3, 2, 3, 1, 2, 1, 2, 1, 2, 0, ...
...
For example, T(1,2) = T(2,1) = 2 as 1 has factorial base representation '...0001' and 2 has factorial base representation '...0010', and they differ by their two least significant digits.
On the other hand, T(3,5) = T(5,3) = 1, as 3 has factorial base representation '...0011' and 5 has factorial base representation '...0021', and they differ only by their second rightmost digit.
Note that as A007623(6)='100' and A007623(10)='120', we have T(6,10) = T(10,6) = 1 (instead of 2 as in A231713, cf. also its Example section), as here we count only the number of differing digit positions, but ignore the magnitudes of their differences.

Antti Karttunen, The first 121 antidiagonals of the table, flattened

The topmost row and the leftmost column: A060130.

Only the lower triangular region: A230417. Related arrays: A230419, A231713. Cf. also A101080, A084558, A230410.

nn = 14; m = 1; While[m! < nn, m++]; m; Table[Function[w, Count[Subtract @@ Map[PadLeft[#, Max@ Map[Length, w]] &, w], k_ /; k != 0]]@ Map[IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, {i - j, j}], {i, 0, nn}, {j, 0, i}] // Flatten (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

(define (A230415 n) (A230415bi (A025581 n) (A002262 n)))
(define (A230415bi x y) (let loop ((x x) (y y) (i 2) (d 0)) (cond ((and (zero? x) (zero? y)) d) (else (loop (floor->exact (/ x i)) (floor->exact (/ y i)) (+ i 1) (+ d (if (= (modulo x i) (modulo y i)) 0 1)))))))

T(n,0) = T(0,n) = A060130(n).

Each entry T(i,j) <= A231713(i,j).

A261073 Semiprimes whose prime factors are of equal binary length and which differ from each other in one bit position only.

Original entry on oeis.org

6, 35, 323, 437, 713, 899, 1763, 1961, 2021, 2537, 3233, 4757, 5561, 5609, 6497, 7313, 9797, 10403, 10961, 11009, 18209, 19043, 21353, 22499, 23393, 26969, 27221, 29177, 37001, 38021, 39203, 45113, 71273, 72899, 79523, 87953, 95477, 98201, 99221, 106793, 114857, 114929, 123353

Offset: 1

Antti Karttunen, Sep 22 2015

			6 = 2*3 is present, as 2 in binary is "10" and 3 in binary is "11", so both have two (significant) bits and they differ only in one bit-position from each other.
35 = 5*7 is present, as 5 in binary is "101" and 7 in binary is "111", which both have three bits, differing only in the middle position from each other.

Antti Karttunen, Table of n, a(n) for n = 1..5000

Cf. A000523, A001222, A006530, A020639, A101080.
Cf. also A261074, A261075.
Cf. A071697 (a subsequence).
Intersection of A085721 and A261077.

Select[Range[10^6], And[Length@ # == 2, IntegerLength[#1, 2] == IntegerLength[#2, 2] & @@ #, Total@ BitXor[IntegerDigits[#1, 2], IntegerDigits[#2, 2]] == 1 & @@ #] &@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ #] &] (* Michael De Vlieger, Oct 08 2016 *)

A000523 = n -> logint(n, 2);
A020639(n) = if(1==n,n,vecmin(factor(n)[, 1]));
isA261073(n) = { my(a,b); if(bigomega(n)!=2, 0, a=A020639(n); b = (n/a); ((A000523(a) == A000523(b)) && (1 == norml2(binary(bitxor(a,b)))))); };
i=0; n=0; while(i < 5000, n++; if(isA261073(n), i++; write("b261073.txt", i, " ", n)));

;; With Antti Karttunen's IntSeq-library.
(define A261073 (MATCHING-POS 1 1 (lambda (n) (and (= 2 (A001222 n)) (= (A000523 (A020639 n)) (A000523 (A006530 n))) (= 1 (A101080bi (A020639 n) (A006530 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.

A231713 Square array A(i,j) = the sum of absolute values of digit differences in the matching positions of the factorial base representations of i and j, for i >= 0, j >= 0, read by antidiagonals.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Nov 12 2013

This table relates to the factorial base representation (A007623) in a similar way as A101080 relates to the binary system. See A230415 for another analog.

			The top left corner of this square array begins as:
0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, ...
1, 0, 2, 1, 3, 2, 2, 1, 3, 2, 4, ...
1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, ...
2, 1, 1, 0, 2, 1, 3, 2, 2, 1, 3, ...
2, 3, 1, 2, 0, 1, 3, 4, 2, 3, 1, ...
3, 2, 2, 1, 1, 0, 4, 3, 3, 2, 2, ...
1, 2, 2, 3, 3, 4, 0, 1, 1, 2, 2, ...
2, 1, 3, 2, 4, 3, 1, 0, 2, 1, 3, ...
2, 3, 1, 2, 2, 3, 1, 2, 0, 1, 1, ...
3, 2, 2, 1, 3, 2, 2, 1, 1, 0, 2, ...
3, 4, 2, 3, 1, 2, 2, 3, 1, 2, 0, ...
...
For example, A(1,2) = A(2,1) = 2 as 1 has factorial base representation '...0001' and 2 has factorial base representation '...0010', and adding the absolute values of the digit differences, we get 1+1 = 2.
On the other hand, A(3,5) = A(5,3) = 1, as 3 has factorial base representation '...0011' and 5 has factorial base representation '...0021', and they differ only by their second rightmost digit, the absolute value of difference being 1.
Note that as A007623(6)='100' and A007623(10)='120', we have A(6,10) = A(10,6) = 2.

Antti Karttunen, The first 121 antidiagonals of the table, flattened

The topmost row and the leftmost column: A034968.

Only the lower triangular region: A231714. Related tables: A230415, A230419. Cf. also A101080, A231717.

(define (A231713 n) (A231713bi (A025581 n) (A002262 n)))
(define (A231713bi x y) (let loop ((x x) (y y) (i 2) (d 0)) (cond ((and (zero? x) (zero? y)) d) (else (loop (floor->exact (/ x i)) (floor->exact (/ y i)) (+ i 1) (+ d (abs (- (modulo x i) (modulo y i)))))))))

Each entry A(i,j) >= A230415(i,j) and also each entry A(i,j) >= abs(A230419(i,j)).

A261074 Semiprimes whose prime factors are of equal binary length and which differ from each other in exactly two bit positions.

Original entry on oeis.org

143, 391, 493, 589, 667, 1517, 1739, 1927, 2257, 2419, 2501, 2773, 2867, 3599, 4891, 5293, 5767, 5893, 6499, 6901, 7081, 7169, 7171, 7387, 7811, 7957, 8137, 8453, 8611, 9379, 9991, 10033, 10057, 10379, 10573, 11021, 11227, 11413, 11663, 13081, 13589, 13843, 17947, 19781, 21509, 21877, 22657, 23449, 23701, 23707, 25217, 25283, 26069, 26441, 27029

Offset: 1

Antti Karttunen, Sep 22 2015

			143 = 11*13 is included because 11 ("1011" in binary) and 13 ("1101" in binary) differ from each other in exactly two bit-positions.
56153 = 233 * 241 is included (as term a(119)) because 233 ("11101001" in binary) and 241 ("11110001" in binary) differ from each other in exactly two bit-positions.

Antti Karttunen, Table of n, a(n) for n = 1..10000
N. S. Dattani & N. Bryans, Quantum factorization of 56153 with only 4 qubits, arXiv:1411.6758 [quant-ph], 2014.

Cf. A000523, A001222, A006530, A020639, A101080.
Cf. also A261073, A261075.
Subsequence of A085721.

Select[Range[10^5], And[Length@ # == 2, IntegerLength[#1, 2] == IntegerLength[#2, 2] & @@ #, Total@ BitXor[IntegerDigits[#1, 2], IntegerDigits[#2, 2]] == 2 & @@ #] &@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ #] &] (* Michael De Vlieger, Oct 08 2016 *)

A000523 = n -> logint(n, 2);
A020639(n) = if(1==n,n,vecmin(factor(n)[, 1]));
isA261074(n) = { my(a,b); if(bigomega(n)!=2, 0, a = A020639(n); b = (n/a); ((A000523(a) == A000523(b)) && (2 == norml2(binary(bitxor(a,b)))))); };
i=0; n=0; while(i < 10000, n++; if(isA261074(n), i++; write("b261074.txt", i, " ", n)));

;; With Antti Karttunen's IntSeq-library.
(define A261074 (MATCHING-POS 1 1 (lambda (n) (and (= 2 (A001222 n)) (= (A000523 (A020639 n)) (A000523 (A006530 n))) (= 2 (A101080bi (A020639 n) (A006530 n)))))))

cp's OEIS Frontend

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.

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

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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.

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A163252 a(0) = 0, and a(n) is the least positive integer not occurring earlier in the sequence such that a(n-1) and a(n) differ in only one bit when written in binary.

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A230415 Square array T(i,j) giving the number of differing digits in the factorial base representations of i and j, for i >= 0, j >= 0, read by antidiagonals.

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A261073 Semiprimes whose prime factors are of equal binary length and which differ from each other in one bit position only.

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