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A286105
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a(1) = 0; for n > 1, a(n) = 1 + max(a(A285734(n)), a(A285735(n))).
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4
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0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 6, 6, 7, 6, 7, 6, 7, 7, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 7, 8, 8, 7, 7, 7, 7, 7, 8, 7, 8, 8, 8, 7, 8, 8, 7, 8, 8, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
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listen;
history;
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internal format)
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OFFSET
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1,3
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COMMENTS
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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 furthest removed from the root.
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LINKS
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FORMULA
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a(1) = 1 and for n > 1, a(n) = 1 + a(A286107(n)).
Other identities. For all n >= 1:
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EXAMPLE
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A285734(2) = A285735(2) = 1, thus a tree with root 2 has just two leaves 1 and 1, so the maximum 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 distance to a farthest leaf (1) is two edges, thus a(3) = 2.
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 maximum distance to a leaf (1) is 1 + longest distance computed for cases 2 and 3, thus a(5) = 1 + max(1,2) = 3.
The tree with root 17 looks like this:
17
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..................../ \..................
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 longest distance to 1 from the root can be found for example at the right border of the tree, five edges in total, thus a(17) = 5.
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PROG
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(Scheme, with memoization-macro definec)
(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 a286105(n): return 0 if n==1 else 1 + max(a286105(a285734(n)), a286105(a285735(n)))
print([a286105(n) for n in range(1, 121)]) # Indranil Ghosh, May 02 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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