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.

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

Offset: 1

Author

Antti Karttunen, May 02 2017

Keywords

• nonn

Comments

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.

Links

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

Crossrefs

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

Programs

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

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

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))))))

Formula

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