A265904 Self-inverse permutation of nonnegative integers: a(n) = A263272(A263273(A263272(n))).
0, 1, 2, 3, 4, 11, 6, 29, 8, 9, 10, 5, 12, 13, 38, 33, 92, 17, 18, 83, 20, 87, 110, 35, 24, 89, 26, 27, 28, 7, 30, 37, 32, 15, 86, 23, 36, 31, 14, 39, 40, 119, 114, 281, 44, 99, 254, 65, 276, 335, 98, 51, 260, 71, 54, 245, 56, 249, 326, 101, 60, 263, 74, 261, 272, 47, 330, 353, 116, 105, 278, 53, 72, 251, 62, 267, 332, 107, 78, 269, 80, 81, 82, 19
Offset: 0
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Programs
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Mathematica
f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@g[n] h[n]]]; t = Table[f[2 n]/2, {n, 0, 1000}]; Table[t[[f[t[[n + 1]]] + 1]], {n, 0, 83}] (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A263273 *)
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Python
from sympy import factorint from sympy.ntheory.factor_ import digits from operator import mul def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3) def a038502(n): f=factorint(n) return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f]) def a038500(n): return n/a038502(n) def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n) def a263272(n): return a263273(2*n)/2 def a(n): return a263272(a263273(a263272(n))) # Indranil Ghosh, May 25 2017
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Scheme
(define (A265904 n) (A263272 (A263273 (A263272 n))))
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