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

Author

Antti Karttunen, Feb 15 2016

Keywords

• nonn
• tabl

Comments

See comments in A268833.

Examples

			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
		

Links

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

Crossrefs

Transpose of A268833.

Programs

Mathematica

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 *)

PARI

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

Python

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

Scheme

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