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A338478
Let b be an odd function such that b(0) = 0, b(1) = 1, and for any n > 1 such that 3^x < 2*n < 3^(x+1) for some x > 0, b(n) = b(3^x-n) - 3^x; a(n) = abs(b(n)) for any n >= 0.
1
0, 1, 2, 3, 4, 13, 12, 11, 8, 9, 10, 7, 6, 5, 32, 33, 34, 37, 36, 35, 38, 39, 40, 31, 30, 29, 26, 27, 28, 25, 24, 23, 14, 15, 16, 19, 18, 17, 20, 21, 22, 103, 102, 101, 98, 99, 100, 97, 96, 95, 104, 105, 106, 109, 108, 107, 110, 111, 112, 121, 120, 119, 116
OFFSET
0,3
COMMENTS
This sequence is a self-inverse permutation of the nonnegative integers.
It is possible to build a continuous injective complex-valued function of a real-variable, say f, such that Im(f(r)) = 0 iff r is an integer and for any n in Z, f(n) = b(n) (see illustration in Links section).
LINKS
Rémy Sigrist, Representation in the complex plane of a function f as described in Comments section (the corresponding curve divides the complex plane into two simply connected regions rendered with different colors)
FORMULA
a(n) = n iff abs(n - 3^x) <= 1 for some x >= 0.
EXAMPLE
For n = 3:
- we have 3^1 < 2*3 < 3^(1+1),
- so b(3) = b(3 - 3) - 3 = 0 - 3 = -3,
- a(3) = abs(b(3)) = 3.
PROG
(PARI) b(n) = { if (n<0, return (-b(-n)), n==0, return (0), n==1, return (1), for (x=1, oo, my (w=3^x, h=w\2); if (w<2*n && 2*n<3*w, return (b(w-n)-w)))) }
a(n) = abs(b(n))
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Oct 29 2020
STATUS
approved