A323710 -id:A323710 - cp's OEIS frontend

A323752 Fixed points of A323710.

1, 6, 28, 40, 240, 544, 832, 1152, 2816, 50176, 118784, 131584, 409600, 1050624, 1056768, 1081344, 2031616, 8519680, 118489088, 201588736, 352321536, 6446645248, 15300820992, 25836912640, 104152956928, 150323855360, 1099545182208, 3315714752512, 4398583382016

Offset: 1

Luc Rousseau, Jan 26 2019

nonn

If f(n) denotes the binary tree representation of n defined in A323710, then this sequence lists the n such that f(n) is symmetrical.

			The recursive decomposition of 50176 with formula "parent = (2^left)*(2*right+1)" gives the following binary tree representation:
         o
        / \
       /   \
      /     \
     o       o
    / \     / \
   o   o   o   o
      /     \
     o       o
This tree is symmetrical, so 50176 is in the sequence.

Luc Rousseau, Table of n, a(n) for n = 1..200
Luc Rousseau, A program to compute this sequence (SWI-Prolog)

Cf. A323710.

A323665 a(n) is the number of vertices in the binary tree the root of which is assigned the value n and built recursively by the rule: write node's value as (2^c)*(2k+1); if c>0, create a left child with value c; if k>0, create a right child with value k.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 4, 5, 5, 5, 5, 5, 5, 5, 4, 5, 5, 5, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5, 5, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 4, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 6, 6, 7, 7, 7, 7, 7, 7, 6, 7, 7

Offset: 1

Luc Rousseau, Jan 23 2019

Mirroring (left <--> right) the tree corresponding to n and computing back a number gives rise to sequence A323710. Swapping the left and right subtrees of the root of the tree corresponding to n and computing back a number gives rise to sequence A117303.

			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:
     100                                         o
     / \                                        / \
    2  12                                      o   o
   /   / \              or more simply        /   / \
  1   2   1                                  o   o   o
     /                                          /
    1                                          o
7 vertices, so a(100) = 7.

Luc Rousseau, Table of n, a(n) for n = 1..10000

Cf. A117303, A323710.

a:= proc(n) option remember; `if`(n=0, 0, (j->
      1+a(j)+a((n/2^j-1)/2))(padic[ordp](n, 2)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 23 2019

nEdges[n_] :=
If[n == 0, 0,
  Module[{c, xx, k}, c = IntegerExponent[n, 2]; xx = n/2^c;
   k = (xx - 1)/2;
   Boole[c != 0]*(1 + nEdges[c]) + Boole[k != 0]*(1 + nEdges[k])]]
a[n_] := nEdges[n] + 1
Table[a[n], {n, 1, 87}]

cp's OEIS Frontend

A323752 Fixed points of A323710.

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