A268823 Permutation of nonnegative integers: a(0) = 0, a(n) = A268717(1 + A268821(n-1)).

0, 1, 3, 2, 6, 5, 7, 15, 13, 4, 12, 27, 25, 8, 24, 14, 10, 9, 11, 51, 49, 16, 48, 22, 18, 17, 19, 26, 30, 29, 31, 23, 21, 28, 20, 99, 97, 32, 96, 38, 34, 33, 35, 42, 46, 45, 47, 39, 37, 44, 36, 50, 54, 53, 55, 63, 61, 52, 60, 43, 41, 56, 40, 62, 58, 57, 59, 195, 193, 64, 192, 70, 66, 65, 67, 74, 78, 77, 79, 71, 69, 76, 68, 82, 86, 85

Offset: 0

Author

Antti Karttunen, Feb 14 2016

Keywords

• nonn

Comments

The "third shifted power" of permutation A268717.

Links

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

Crossrefs

Inverse: A268824.

Cf. A003188, A006068, A268717, A268821, A268825, A268676.

Row 3 of array A268820.

Programs

PARI

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});
A268717(n) = if(n<1, 0, A003188(1 + A006068(n - 1)));
for(n=0, 100, print1(if(n<2, n, A268717(1 + A268717(1 + A268717(n - 2)))),", ")) \\ Indranil Ghosh, Mar 31 2017

Python

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(1 + A268717(n - 2))) if n>1 else n
print([a(n) for n in range(101)]) # Indranil Ghosh, Mar 31 2017

Scheme

(define (A268823 n) (if (<= n 1) n (A268717 (+ 1 (A268717 (+ 1 (A268717 (- n 2))))))))

Formula

a(0), for n >= 1, a(n) = A268717(1 + A268821(n-1)).

a(0) = 0, a(1) = 1, and for n > 1, a(n) = A268717(1 + A268717(1 + A268717(n-2))).

For n >= 3, a(n) = A003188(3+A006068(n-3)). - Antti Karttunen, Mar 11 2024