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A285718
a(1) = 0, and for n > 1, a(n) = the least squarefree number x such that n-x is also squarefree.
4
0, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 3, 1, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 3, 5, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 3, 6, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1
OFFSET
1,5
COMMENTS
For any n > 1 there is at least one decomposition of n as a sum of two squarefree numbers (cf. A071068 and the Mathematics Stack Exchange link). Of all pairs (x,y) of positive squarefree numbers for which x <= y and x+y = n, sequences A285718 and A285719 give the unique pair for which the difference y-x is the largest possible.
Question: Are there arbitrarily large terms in this sequence?
LINKS
Mathematics Stack Exchange, Sums of square free numbers, is this conjecture equivalent to Goldbach's conjecture? (See especially the answer of Aryabhata)
K. Rogers, The Schnirelmann density of the squarefree integers, Proc. Amer. Math. Soc. 15 (1964), pp. 515-516.
FORMULA
a(n) = n - A285719(n).
EXAMPLE
For n=51 we see that 50 (2*5*5), 49 (7*7) and 48 (2^4 * 3) are all nonsquarefree (A013929). 47 (a prime) is squarefree, but 51 - 47 = 4 is not. On the other hand, both 46 (2*23) and 5 are squarefree numbers, thus a(51) = 5.
MATHEMATICA
Table[If[n == 1, 0, x = 1; While[Nand[SquareFreeQ@ x, SquareFreeQ[n - x]], x++]; x], {n, 120}] (* Michael De Vlieger, May 03 2017 *)
PROG
(Scheme)
(define (A285718 n) (if (= 1 n) 0 (let loop ((k 1)) (if (not (zero? (A008683 (- n (A005117 k))))) (A005117 k) (loop (+ 1 k))))))
(Python)
from sympy.ntheory.factor_ import core
def issquarefree(n): return core(n) == n
def a285718(n):
if n==1: return 0
x = 1
while True:
if issquarefree(x) and issquarefree(n - x):return x
else: x+=1
print([a285718(n) for n in range(1, 121)]) # Indranil Ghosh, May 02 2017
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 02 2017
STATUS
approved