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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. 114
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)=n if and only if A081242(n) is a palindrome. - Clark Kimberling, Mar 12 2003
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
From Antti Karttunen, Oct 28 2001: (Start)
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:
A065260(n) = a(A057115(a(n))),
A065266(n) = a(A065264(a(n))),
A065272(n) = a(A065270(a(n))),
A065278(n) = a(A065276(a(n))),
A065284(n) = a(A065282(a(n))),
A065290(n) = a(A065288(a(n))). (End)
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
From Ed Pegg Jr, Sep 09 2015: (Start)
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
Alois P. Heinz, Table of n, a(n) for n = 1..8191 (first 1023 terms from T. D. Noe)
Bruce Bates, Martin Bunder, and Keith Tognetti, Locating terms in the Stern-Brocot tree, European Journal of Combinatorics 31.3 (2010): 1020-1033.
Joe Buhler and R. L. Graham, Juggling Drops and Descents, Amer. Math. Monthly, 101, (no. 6) 1994, 507-519.
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, 2020-2021.
Wikipedia, Calkin-Wilf Tree.
Wikipedia, Stern-Brocot tree.
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
From Philippe Deléham, Jun 02 2015: (Start)
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, ...
Row sums = A010036. (End)
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:
seq(a(n), n=1..100); # Alois P. Heinz, Feb 28 2015
MATHEMATICA
A059893 = Reap[ For[n=1, n <= 100, n++, a=1; b=n; While[b > 1, a = 2*a + 2*FractionalPart[b/2]; b=Floor[b/2]]; Sow[a]]][[2, 1]] (* Jean-François Alcover, Jul 16 2012, after Harry J. Smith *)
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
-- Reinhard Zumkeller, May 01 2013
(Scheme, with memoization-macro definec)
(definec (A059893 n) (if (<= n 1) n (let* ((k (- (A000523 n) 1)) (r (A059893 (- n (A000079 k))))) (if (= 2 (floor->exact (/ n (A000079 k)))) (* 2 r) (+ 1 r)))))
;; Antti Karttunen, May 16 2015
(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
# Yosu Yurramendi, Mar 20 2017
(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
# Yosu Yurramendi, Apr 25 2021
(Python)
def a(n): return int('1' + bin(n)[3:][::-1], 2)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Mar 21 2017
CROSSREFS
{A000027, A054429, A059893, A059894} form a 4-group.
The set of permutations {A059893, A080541, A080542} generates an infinite dihedral group.
In other bases: A351702 (balanced ternary), A343150 (Zeckendorf), A343152 (lazy Fibonacci).
Sequence in context: A139708 A333776 A305410 * A331274 A269365 A269366
KEYWORD
easy,nonn,base,nice,look
AUTHOR
Marc LeBrun, Feb 06 2001
STATUS
approved

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Last modified May 18 08:32 EDT 2024. Contains 372618 sequences. (Running on oeis4.)