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A163236
Inverse permutation to A163235.
4
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
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
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 29 2009
STATUS
approved