A286103 a(1) = 0; for n > 1, a(n) = 1 + min(a(A285734(n)), a(A285735(n))).

0, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Author

Antti Karttunen, May 02 2017

Keywords

• nonn

Comments

By invoking A285734 and A285735 recursively, any natural number n > 1 can be decomposed as a sum of successively smaller squarefree numbers, until only n instances of 1's remain. This process can be depicted as a binary tree, where 1's are leaves, and any other node n branches to the left as A285734(n) and to the right as A285735(n). This sequence gives the distance from the root of tree (n) to a leaf (1) that is nearest to the root.

Examples

			A285734(2) = A285735(2) = 1, thus a tree with root 2 has just two leaves 1 and 1, so the minimum distance to them is 1, thus a(2) = 1.
A285734(3) = 1 and A285735(3) = 2, thus a tree with root 3 has one immediate leave 1 and the subtree 2 as its other branch, so the minimum distance to a leaf (1) is one edge, thus a(3) = 1.
A285734(5) = 2 and A285735(3) = 3, thus a tree with root 5 has the subtree 2 as its other branch, and the subtree 3 as the other branch, so the minimum distance to a leaf (1) is 1 + shortest distance computed for cases 2 and 3, thus a(5) = 1 + min(1,1) = 2.
The tree with root 17 looks like this:
                                    17
                                     |
                ..................../ \..................
                7                                       10
      2......../ \........5                   5......../ \........5
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  1       1           2       3           2       3           2       3
                    1   1   1   2       1   1   1   2       1   1   1   2
                               1 1                 1 1                 1 1
We see that the shortest distance to 1 from the root is at the left border of the tree, just three edges, thus a(17) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005117, A285734, A285735, A286104, A286105, A286106.

Programs

Python

from sympy.ntheory.factor_ import core
def issquarefree(n): return core(n) == n
def a285734(n):
    if n==1: return 0
    j=n//2
    while True:
        if issquarefree(j) and issquarefree(n - j): return j
        else: j-=1
def a285735(n): return n - a285734(n)
def a286103(n): return 0 if n==1 else 1 + min(a286103(a285734(n)), a286103(a285735(n)))
print([a286103(n) for n in range(1, 121)]) # Indranil Ghosh, May 02 2017

Formula

a(1) = 0; for n > 1, a(n) = 1 + min(a(A285734(n)), a(A285735(n))).

a(1) = 0; for n > 1, a(n) = 1 + a(A286104(n)).

Other identities. For all n >= 1:

a(2*A005117(n)) = 1+a(A005117(n)).