1,2

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)

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.

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

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)

# 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;

# Alternative:

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

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 *)

(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

{A000027, A054429, A059893, A059894} form a 4-group.

The set of permutations {A059893, A080541, A080542} generates an infinite dihedral group.

Cf. A000079, A000523, A007931, A030308.

In other bases: A351702 (balanced ternary), A343150 (Zeckendorf), A343152 (lazy Fibonacci).

Sequence in context: A139708 A333776 A305410 * A331274 A269365 A269366

Adjacent sequences: A059890 A059891 A059892 * A059894 A059895 A059896

easy,nonn,base,nice,look

Marc LeBrun, Feb 06 2001

approved