A285719 a(1) = 1, and for n > 1, a(n) = the largest squarefree number k such that n-k is also squarefree.

1, 1, 2, 3, 3, 5, 6, 7, 7, 7, 10, 11, 11, 13, 14, 15, 15, 17, 17, 19, 19, 21, 22, 23, 23, 23, 26, 26, 26, 29, 30, 31, 31, 33, 34, 35, 35, 37, 38, 39, 39, 41, 42, 43, 43, 43, 46, 47, 47, 47, 46, 51, 51, 53, 53, 55, 55, 57, 58, 59, 59, 61, 62, 62, 62, 65, 66, 67, 67, 69, 70, 71, 71, 73, 74, 74, 74, 77, 78, 79, 79, 79, 82, 83, 83, 85, 86, 87, 87, 89, 89, 91, 91

Offset: 1

Author

Antti Karttunen, May 02 2017

Keywords

• nonn

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.

Note: a(n+1) differs from A070321(n) for the first time at n=50, with a(51) = 46, while A070321(50) = 47.

Examples

			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) = 46.
		

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, A008683, A013928, A013929, A070321, A071068, A285718, A285735.

Programs

Mathematica

lsfn[n_]:=Module[{k=n-1},While[!SquareFreeQ[k]||!SquareFreeQ[n-k],k--];k]; Join[{1},Array[ lsfn,100,2]] (* Harvey P. Dale, Apr 27 2023 *)

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
def a285719(n): return n - a285718(n)
print([a285719(n) for n in range(1, 121)]) # Indranil Ghosh, May 02 2017

Scheme

(define (A285719 n) (- n (A285718 n)))
(define (A285719 n) (if (= 1 n) n (let loop ((k (A013928 n))) (if (not (zero? (A008683 (- n (A005117 k))))) (A005117 k) (loop (- k 1))))))

Formula

a(n) = n - A285718(n).