A268835 -id:A268835 - 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

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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