A268719 Triangular table T(n>=0,k=0..n) = A003188(A006068(n) + A006068(k)), read by rows as A(0,0), A(1,0), A(1,1), A(2,0), A(2,1), A(2,2), ...
0, 1, 3, 2, 6, 5, 3, 2, 7, 6, 4, 12, 15, 13, 9, 5, 4, 13, 12, 11, 10, 6, 7, 4, 5, 14, 15, 12, 7, 5, 12, 4, 10, 14, 13, 15, 8, 24, 27, 25, 29, 31, 26, 30, 17, 9, 8, 25, 24, 31, 30, 27, 26, 19, 18, 10, 11, 8, 9, 26, 27, 24, 25, 22, 23, 20, 11, 9, 24, 8, 30, 26, 25, 27, 18, 22, 21, 23, 12, 13, 14, 15, 8, 9, 10, 11, 28, 29, 30, 31, 24
Offset: 0
Examples
The first fifteen rows of the triangle: 0 1 3 2 6 5 3 2 7 6 4 12 15 13 9 5 4 13 12 11 10 6 7 4 5 14 15 12 7 5 12 4 10 14 13 15 8 24 27 25 29 31 26 30 17 9 8 25 24 31 30 27 26 19 18 10 11 8 9 26 27 24 25 22 23 20 11 9 24 8 30 26 25 27 18 22 21 23 12 13 14 15 8 9 10 11 28 29 30 31 24 13 15 10 14 24 8 11 9 20 28 31 29 25 27 14 10 9 11 27 25 8 24 23 21 28 20 26 30 29
Links
Crossrefs
Programs
-
Mathematica
a88[n_] := BitXor[n, Floor[n/2]]; a68[n_] := BitXor @@ Table[Floor[n/2^m], {m, 0, Floor[Log[2, n]]}]; a68[0] = 0; T[n_, k_] := a88[a68[n] + a68[k]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 19 2019 *)
-
Python
def a003188(n): return n^(n>>1) def a006068(n): s=1 while True: ns=n>>s if ns==0: break n=n^ns s<<=1 return n def T(n, k): return a003188(a006068(n) + a006068(k)) for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Jun 07 2017
-
Scheme
(define (A268719 n) (A268715bi (A003056 n) (A002262 n)))