|
| |
|
|
A059893
|
|
Reverse the order of all but the most significant bit in binary expansion of n: if n = 1ab..yz then a(n) = 1zy..ba.
|
|
110
|
|
|
|
1, 2, 3, 4, 6, 5, 7, 8, 12, 10, 14, 9, 13, 11, 15, 16, 24, 20, 28, 18, 26, 22, 30, 17, 25, 21, 29, 19, 27, 23, 31, 32, 48, 40, 56, 36, 52, 44, 60, 34, 50, 42, 58, 38, 54, 46, 62, 33, 49, 41, 57, 37, 53, 45, 61, 35, 51, 43, 59, 39, 55, 47, 63, 64, 96, 80, 112, 72, 104, 88, 120
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
A self-inverse permutation of the natural numbers.
a(n) is the position in B of the reversal of the n-th term of B, where B is the left-to-right binary enumeration sequence (A081242 with the empty word attached as first term). - Clark Kimberling, Mar 12 2003
When certain Stern-Brocot tree-related permutations are conjugated with this permutation, they induce a permutation on Z (folded to N), which is an infinite siteswap permutation (see, e.g., figure 7 in the Buhler and Graham paper, which is permutation A065174). We get:
Every nonnegative integer has a unique representation c(1) + c(2)*2 + c(3)*2^2 + c(4)*2^3 + ..., where every c(i) is 0 or 1. Taking tuples of coefficients in lexical order (i.e., 0, 1; 01,11; 001,011,101,111; ...) yields A059893. - Clark Kimberling, Mar 15 2015
The reduced rationals can be ordered either as the Calkin-Wilf tree A002487(n)/A002487(n+1) or the Stern-Brocot tree A007305(n+2)/A047679(n). The present sequence gives the order of matching rationals in the other sequence.
For reference, the Calkin-Wilf tree is 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8, 8/5, 5/7, 7/2, 2/7, 7/5, 5/8, 8/3, 3/7, 7/4, 4/5, ..., which is A002487(n)/A002487(n+1).
The Stern-Brocot tree is 1, 1/2, 2, 1/3, 2/3, 3/2, 3, 1/4, 2/5, 3/5, 3/4, 4/3, 5/3, 5/2, 4, 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5, 5/4, 7/5, 8/5, 7/4, 7/3, 8/3, 7/2, ..., which is A007305(n+2)/A047679(n).
There is a great little OEIS-is-useful story here. I had code for the position of fractions in the Calkin-Wilf tree. The best I had for positions of fractions in the Stern-Brocot tree was the paper "Locating terms in the Stern-Brocot tree" by Bruce Bates, Martin Bunder, Keith Tognetti. The method was opaque to me, so I used my Calkin-Wilf code on the Stern-Brocot fractions, and got A059893. And thus the problem was solved. (End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = A030109(n) + A053644(n). If 2*2^k <= n < 3*2^k then a(n) = 2*a(n-2^k); if 3*2^k <= n < 4*2^k then a(n) = 1 + a(n-2^k) starting with a(1)=1. - Henry Bottomley, Sep 13 2001
|
|
|
EXAMPLE
|
a(11) = a(1011) = 1110 = 14.
With empty word e prefixed, A081242 becomes (e,1,2,11,21,12,22,111,211,121,221,112,...); (reversal of term #9) = (term #12); i.e., a(9)=12 and a(12)=9. - Clark Kimberling, Mar 12 2003
This sequence regarded as a triangle with rows of lengths 1, 2, 4, 8, 16, ...:
1;
2, 3;
4, 6, 5, 7;
8, 12, 10, 14, 9, 13, 11, 15;
16, 24, 20, 28, 18, 26, 22, 30, 17, 25, 21, 29, 19, 27, 23, 31;
32, 48, 40, 56, 36, 52, 44, ...
|
|
|
MAPLE
|
# Implements Bottomley's formula
A059893 := proc(n) option remember; local k; if(1 = n) then RETURN(1); fi; k := floor_log_2(n)-1; if(2 = floor(n/(2^k))) then RETURN(2*A059893(n-(2^k))); else RETURN(1+A059893(n-(2^k))); fi; end;
floor_log_2 := proc(n) local nn, i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;
# second Maple program:
a:= proc(n) local i, m, r; m, r:= n, 0;
for i from 0 while m>1 do r:= 2*r +irem(m, 2, 'm') od;
r +2^i
end:
|
|
|
MATHEMATICA
|
ro[n_]:=Module[{idn=IntegerDigits[n, 2]}, FromDigits[Join[{First[idn]}, Reverse[ Rest[idn]]], 2]]; Array[ro, 80] (* Harvey P. Dale, Oct 24 2012 *)
|
|
|
PROG
|
(PARI) { for (n=1, 1023, a=1; b=n; while (b>1, a=2*a + 2*frac(b/2); b=floor(b/2); ); write("b059893.txt", n, " ", a); ) } \\ Harry J. Smith, Jun 30 2009
(PARI) a(n) = my(b=binary(n)); fromdigits(concat(b[1], Vecrev(vector(#b-1, k, b[k+1]))), 2); \\ Michel Marcus, Sep 29 2021
(Haskell)
a059893 = foldl (\v b -> v * 2 + b) 1 . init . a030308_row
(Scheme, with memoization-macro definec)
(R)
maxrow <- 6 # by choice
a <- 1
for(m in 0:maxrow) for(k in 0:(2^m-1)) {
a[2^(m+1)+ k] <- 2*a[2^m+k]
a[2^(m+1)+2^m+k] <- 2*a[2^m+k] + 1
}
a
(R)
maxblock <- 7 # by choice
a <- 1
for(n in 2:2^maxblock){
ones <- which(as.integer(intToBits(n)) == 1)
nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
anbit <- nbit
anbit[1:(length(anbit) - 1)] <- anbit[rev(1:(length(anbit)-1))]
a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
(Python)
def a(n): return int('1' + bin(n)[3:][::-1], 2)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
