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A285734
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a(1) = 0, and for n > 1, a(n) = the largest squarefree number x such that x < n-x, and n-x is also squarefree.
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9
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0, 1, 1, 2, 2, 3, 2, 3, 3, 5, 5, 6, 6, 7, 5, 6, 7, 7, 6, 10, 10, 11, 10, 11, 11, 13, 13, 14, 14, 15, 14, 15, 14, 17, 14, 17, 15, 19, 17, 19, 19, 21, 21, 22, 22, 23, 21, 22, 23, 21, 22, 26, 23, 23, 26, 26, 26, 29, 29, 30, 30, 31, 30, 31, 31, 33, 33, 34, 34, 35, 34, 35, 35, 37, 37, 38, 38, 39, 38, 39, 39, 41, 41, 42, 42, 43, 41, 42, 43, 43, 38, 46, 46, 47, 42
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OFFSET
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1,4
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COMMENTS
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For n > 1, a(n) = the largest squarefree number x <= n/2 for which n-x is also squarefree.
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 squarefree numbers for which x <= y and x+y = n, sequences A285734 and A285735 give the unique pair for which the difference y-x is the least possible.
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LINKS
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FORMULA
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PROG
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(Scheme)
(define (A285734 n) (if (= 1 n) 0 (let loop ((j 1) (k (- n 1)) (s 0)) (if (> j k) s (loop (+ 1 j) (- k 1) (max s (* j (A008966 j) (A008966 k))))))))
;; Much faster version:
(define (A285734 n) (if (= 1 n) 0 (let loop ((j (floor->exact (/ n 2)))) (if (and (= 1 (A008966 j)) (= 1 (A008966 (- n j)))) j (loop (- j 1))))))
(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
print([a285734(n) for n in range(1, 101)]) # Indranil Ghosh, May 02 2017
(PARI) a(n)=forstep(x=n\2, 1, -1, if(issquarefree(x) && issquarefree(n-x), return(x))); 0 \\ Charles R Greathouse IV, Nov 05 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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