A163234 Inverse permutation to A163233.

0, 1, 2, 4, 6, 3, 11, 7, 9, 13, 5, 8, 24, 18, 17, 12, 28, 21, 37, 29, 10, 15, 16, 22, 58, 48, 47, 38, 31, 39, 23, 30, 35, 43, 27, 34, 62, 52, 51, 42, 14, 19, 20, 26, 32, 25, 41, 33, 112, 98, 97, 84, 73, 85, 61, 72, 70, 59, 83, 71, 40, 49, 50, 60, 120, 105, 137, 121, 78

Offset

0,3

Links

Antti Karttunen, Table of n, a(n) for n = 0..4095

Index entries for sequences that are permutations of the natural numbers

Formula

a(n) = A001477bi(A006068(A059905(n)),A006068(A059906(n))), where A001477bi(x,y) = (((x+y)^2)+x+(3y))/2.

Prog

(Scheme:) (define (A163234 n) (A001477bi (A006068 (A059905 n)) (A006068 (A059906 n))))
(define (A001477bi x y) (/ (+ (expt (+ x y) 2) x (* 3 y)) 2))
(Python)
def A(x, y): return (((x + y)**2) + x + 3*y)//2
def a006068(n):
    s=1
    while True:
        ns=n>>s
        if ns==0: break
        n=n^ns
        s<<=1
    return n
def a059905(n): return sum([(n>>2*i&1)<<i for i in range(int(len(bin(n)[2:])//2) + 1)])
def a059906(n):
    x=[int(k) for k in list(bin(n)[2:])][::-1]
    return sum([x[2*i + 1]*2**i for i in range(len(x)//2)])
def a(n): return A(a006068(a059905(n)), a006068(a059906(n)))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 25 2017

Crossrefs

Inverse: A163233. a(n) = A163236(A057300(n)). Cf. A163236.

Sequence in context: A108236 A002849 A329492 * A366111 A348022 A072984

Adjacent sequences: A163231 A163232 A163233 * A163235 A163236 A163237

Keyword

nonn

Author

Antti Karttunen, Jul 29 2009

Status

approved