A163236 Inverse permutation to A163235.

0, 2, 1, 4, 9, 5, 13, 8, 6, 11, 3, 7, 24, 17, 18, 12, 35, 27, 43, 34, 14, 20, 19, 26, 62, 51, 52, 42, 32, 41, 25, 33, 28, 37, 21, 29, 58, 47, 48, 38, 10, 16, 15, 22, 31, 23, 39, 30, 112, 97, 98, 84, 70, 83, 59, 71, 73, 61, 85, 72, 40, 50, 49, 60, 135, 119, 151, 134, 90

Offset

0,2

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

Prog

(Scheme:) (define (A163236 n) (A001477bi (A006068 (A059906 n)) (A006068 (A059905 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(a059906(n)), a006068(a059905(n)))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 25 2017

Crossrefs

Inverse: A163235. a(n) = A163234(A057300(n)). Cf. A163234.

Sequence in context: A086933 A077878 A128058 * A325008 A325016 A362984

Adjacent sequences: A163233 A163234 A163235 * A163237 A163238 A163239

Keyword

nonn

Author

Antti Karttunen, Jul 29 2009

Status

approved