A254307 Least k such that there are n positive integers, all less than or equal to k, such that the sum of the reciprocals of their squares equals 1.

6, 4, 6, 3, 4, 6, 6, 4, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 8, 6, 6, 8, 6, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 7, 8, 8, 8, 9, 8, 8, 9, 8, 8, 9, 9, 8, 9, 9, 8, 9, 9, 9, 9, 10, 9, 9, 10, 9, 10, 10, 9

Offset: 6

Author

Charles R Greathouse IV, Jan 27 2015

Keywords

• nonn

Comments

a(2), a(3), and a(5) are undefined, so this sequence starts at offset 6. Gasarch (2015) shows that a(n) exists for all n >= 6, though this was known (folklore?) previously; he also poses three open questions.

First occurrence of n: 1, 4, 9, 7, 25, 6, 49, 29, 53, 69, 121, 87, 140, 179, 221, ..., . - Robert G. Wilson v, Feb 15 2015

Examples

			a(1) = 1: 1 = 1/1.
a(4) = 2: 1 = 1/4 + 1/4 +1/4 + 1/4.
a(6) = 6: 1 = 1/4 + 1/4 + 1/4 + 1/9 + 1/9 + 1/36.
a(7) = 4: 1 = 1/4 + 1/4 + 1/4 + 1/16 + 1/16 + 1/16 + 1/16.
a(8) = 6: 1 = 1/4 + 1/4 + 1/9 + 1/9 + 1/9 + 1/9 + 1/36 + 1/36.
a(9) = 3: 1 = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9.
		

Links

• Bill Gasarch, The Solution to a problem in a Romanian math problem book (2015)

Crossrefs

Cf. A000058, A063664.

Programs

PARI

/* oo = 10^10; \\ uncomment for earlier pari versions */
ssd(n,total,mn,mx)=my(t,best=oo); if(total<=0,return(0)); if(n==1, return(if(issquare(1/total,&t)&&t>=mn&&t<=mx&&denominator(t)==1,t,0))); for(k=mn, min(sqrtint(n\total),mx), t=ssd(n-1,total-1/k^2,k,mx); if(t,best=min(best,t))); best
a(n)=my(k=sqrtint(n-1),t=oo);while(t==oo,k++;t=ssd(n-1,1-1/k^2,2,k));k

Formula

sqrt(n) <= a(n) < 2*sqrt(n) for n > 8. The lower bound is sharp since a(n^2) = n.