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A285734 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. 9

%I #27 Aug 20 2022 02:23:42

%S 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,

%T 15,14,15,14,17,14,17,15,19,17,19,19,21,21,22,22,23,21,22,23,21,22,26,

%U 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

%N 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.

%C For n > 1, a(n) = the largest squarefree number x <= n/2 for which n-x is also squarefree.

%C 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.

%H Antti Karttunen, <a href="/A285734/b285734.txt">Table of n, a(n) for n = 1..10000</a>

%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec">Sums of square free numbers, is this conjecture equivalent to Goldbach's conjecture?</a> (See especially the answer of Aryabhata)

%H K. Rogers, <a href="http://dx.doi.org/10.1090/S0002-9939-1964-0163893-X">The Schnirelmann density of the squarefree integers</a>, Proc. Amer. Math. Soc. 15 (1964), pp. 515-516.

%F a(n) = n - A285735(n).

%o (Scheme)

%o (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))))))))

%o ;; Much faster version:

%o (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))))))

%o (Python)

%o from sympy.ntheory.factor_ import core

%o def issquarefree(n): return core(n) == n

%o def a285734(n):

%o if n==1: return 0

%o j=n//2

%o while True:

%o if issquarefree(j) and issquarefree(n - j): return j

%o else: j-=1

%o print([a285734(n) for n in range(1, 101)]) # _Indranil Ghosh_, May 02 2017

%o (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

%Y Cf. A005117, A008966, A071068, A285718, A285719, A285735, A285736, A286106, A286107.

%K nonn

%O 1,4

%A _Antti Karttunen_, May 02 2017

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