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

6,1

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

Links

Table of n, a(n) for n=6..76.

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

Formula

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

Example

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.

Prog

(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

Crossrefs

Cf. A000058, A063664.

Sequence in context: A201408 A153606 A086057 * A346112 A176394 A198235

Adjacent sequences: A254304 A254305 A254306 * A254308 A254309 A254310

Keyword

nonn

Author

Charles R Greathouse IV, Jan 27 2015

Status

approved