A332769 Permutation of the positive integers: a(n) = A258996(A054429(n)) = A054429(A258996(n)).

1, 3, 2, 5, 4, 7, 6, 13, 12, 15, 14, 9, 8, 11, 10, 21, 20, 23, 22, 17, 16, 19, 18, 29, 28, 31, 30, 25, 24, 27, 26, 53, 52, 55, 54, 49, 48, 51, 50, 61, 60, 63, 62, 57, 56, 59, 58, 37, 36, 39, 38, 33, 32, 35, 34, 45, 44

Offset: 1

Author

Yosu Yurramendi, Feb 23 2020

Keywords

• nonn

Comments

Sequence is self-inverse: a(a(n)) = n.

A002487(1+a(n)) = A162911(n) and A002487(a(n)) = A162912(n). So, a(n) generates the enumeration system of positive rationals based on Stern's sequence A002487 called 'drib'.

Given n, one can compute a(n) by taking into account the binary representation of n, and by flipping every second bit starting from the lowest until reaching the highest 1, which is not flipped.

Examples

			n = 23 =  10111_2
            x x
          10010_2 = 18 = a(n).
n = 33 = 100001_2
          x x x
         110100_2 = 52 = a(n).
		

Links

• Yosu Yurramendi, Table of n, a(n) for n = 1..8191

Crossrefs

Similar R-programs: A258996, A284447.

Programs

PARI

a(n) = bitxor(n, 2<Kevin Ryde, Mar 30 2021

R

maxrow <- 6 # by choice
a <- 1
for(m in 0:maxrow) for(k in 0:(2^m-1)){
a[2^(m+1)+2*k  ] <- 2*a[2^(m+1)-1-k] + 1
a[2^(m+1)+2*k+1] <- 2*a[2^(m+1)-1-k]
}
a

R

# Given n, compute a(n) by taking into account the binary representation of n
maxblock <- 7 # by choice
a <- c(1, 3, 2)
for(n in 4:2^maxblock){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  anbit <- nbit
  anbit[seq(1, length(anbit) - 1, 2)] <- 1 - anbit[seq(1, length(anbit) - 1, 2)]
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
# Yosu Yurramendi, Mar 30 2021

Formula

a(A054429(n)) = A054429(a(n)) = A258996(n),

a(A258996(n)) = A258996(a(n)) = A054429(n).

a(n) = A284447(A065190(n)) = A065190(A284447(n)),

a(A065190(n)) = A065190(a(n)) = A284447(n),

a(A284447(n)) = A284447(a(n)) = A065190(n).

a(A231551(n)) = A154437(n), a(A154437(n)) = A231551(n).

a(A153154(n)) = A284459(n), a(A284459(n)) = A153154(n).

a(1) = 1, a(2) = 3, a(3) = 2; for n > 3, a(2*n) = 2*a(A054429(n)) + 1, a(2*n+1) = 2*a(A054429(n)).

a(1) = 1; for m >= 0 and 0 <= k < 2^m, a(2^(m+1)+2*k) = 2*a(2^(m+1)-1-k) + 1, a(2^(m+1)+2*k+1) = 2*a(2^(m+1)-1-k).