A323710 a(n) is the symmetrical of n via transport of structure from binary trees, where the binary tree of n is built as follows: create a root with value n and recursively apply the rule {write node's value as (2^c)*(2*k+1); if c>0, create a left child with value c; if k>0, create a right child with value k}.

1, 3, 2, 7, 8, 6, 4, 5, 128, 24, 256, 14, 64, 12, 16, 15, 32, 384, 340282366920938463463374607431768211456, 56, 16777216, 768, 115792089237316195423570985008687907853269984665640564039457584007913129639936, 10, 16384, 192, 18446744073709551616, 28, 4096, 48

Offset: 1

Author

Luc Rousseau, Jan 24 2019

Keywords

• nonn

Comments

Let f denote the bijection that maps positive integers onto binary trees, defined in the name; let g be its inverse; let r denote the symmetry on binary trees (i.e., starting from the root, r recursively swaps left and right children). By definition a(n) = g(r(f(n))).

If instead of r, one uses s, the operation that swaps the left and right children of the root, without recursion, then one gets g(s(f(n))) = A117303(n).

Better leave a(39) = 2^340282366920938463463374607431768211456 not fully evaluated.

Examples

			100 = (2^2)*(2*12+1) and recursively, 2 = (2^1), 12 = (2^2)*(2*1+1). We then have the following binary tree representation of 100:
     100
     / \
    2  12
   /   / \
  1   2   1
     /
    1
Erase the numerical values, just keep the tree structure:
      o
     / \
    o   o
   /   / \
  o   o   o
     /
    o
Take its symmetrical:
      o
     / \
    o   o
   / \   \
  o   o   o
       \
        o
Compute back new numerical values from the leafs (value: 1) up:
(2*1+1) = 3; (2^1)*(2*3+1) = 14; (2^14)*(2*3+1) = 114688
   114688
     / \
   14   3
   / \   \
  1   3   1
       \
        1
So, a(100) = 114688.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..38

Crossrefs

Cf. A117303 (variant where swap left/right is not recursively applied).

Cf. A323665.

Cf. A323752 (fixed points of this sequence).

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 0, (j->
      (2*a(j)+1)*2^a((n/2^j-1)/2))(padic[ordp](n, 2)))
    end:
seq(a(n), n=1..38);  # Alois P. Heinz, Jan 24 2019

Mathematica

f[0]:=x
f[n_]:=Module[{c,k},c=IntegerExponent[n,2];k=(n/2^c-1)/2;o[f[c],f[k]]]
g[x]:=0
g[o[C_,K_]]:=(2^g[C])(2g[K]+1)
r[x]:=x
r[o[C_,K_]]:=o[r[K],r[C]]
a[n_]:=g@r@f[n]
Table[a[n], {n, 1, 30}]

Formula

a(a(n)) = n.

a(n) = n iff n is in A323752.