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%I #28 Nov 17 2015 10:12:18
%S 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,
%T 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,
%U 9,10,10,9
%N 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.
%C 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.
%C 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
%H Bill Gasarch, <a href="http://www.cs.umd.edu/~gasarch/BLOGPAPERS/recipsq.pdf">The Solution to a problem in a Romanian math problem book</a> (2015)
%F sqrt(n) <= a(n) < 2*sqrt(n) for n > 8. The lower bound is sharp since a(n^2) = n.
%e a(1) = 1: 1 = 1/1.
%e a(4) = 2: 1 = 1/4 + 1/4 +1/4 + 1/4.
%e a(6) = 6: 1 = 1/4 + 1/4 + 1/4 + 1/9 + 1/9 + 1/36.
%e a(7) = 4: 1 = 1/4 + 1/4 + 1/4 + 1/16 + 1/16 + 1/16 + 1/16.
%e a(8) = 6: 1 = 1/4 + 1/4 + 1/9 + 1/9 + 1/9 + 1/9 + 1/36 + 1/36.
%e a(9) = 3: 1 = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9.
%o (PARI) /* oo = 10^10; \\ uncomment for earlier pari versions */
%o 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
%o a(n)=my(k=sqrtint(n-1),t=oo);while(t==oo,k++;t=ssd(n-1,1-1/k^2,2,k));k
%Y Cf. A000058, A063664.
%K nonn
%O 6,1
%A _Charles R Greathouse IV_, Jan 27 2015
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