[seqfan] Re: An interesting sequence

[Arthur O'Dwyer]

On Sun, Apr 16, 2023 at 1:02 PM Arthur O'Dwyer <arthur.j.odwyer at gmail.com>
wrote:

> (Adding participants-thus-far on the To: line.)
>
> It seems it has been proven ([1]) that all natural numbers above 245 are
> expressible as the sum of 5 distinct squares.
> This is pretty much what you need, since (n is the mean of 5 distinct
> squares) = (nx is the sum of 5 distinct squares). [...]
> [1] — "Darstellung natürlicher Zahlen als Summe von Quadraten" (Franz
> Halter-Koch, 1982) http://matwbn.icm.edu.pl/ksiazki/aa/aa42/aa4212.pdf
>

And that paper seems to give a formula for which numbers will actually
require 5 squares (as opposed to 4 or fewer), which is basically just "any
number of the form 4^h * k for certain ks." I claim that a number n has
a(n)=5 in Yifan's sequence if-and-only-if it is exactly a power of 4 times
one of these six numbers:

0b10011

0b11000

0b100000

0b100001

0b1011

0b110000

I bet there's a proof in Halter-Koch's paper; I have none myself.
Experimentally, this rule is true at least as high as:

a(311296) = 5 (899^2 + 865^2 + 7^2 + 2^2 + 1^2)    (0b1001100'0000'0000'0000
)

a(393216) = 5 (1229^2 + 675^2 + 3^2 + 2^2 + 1^2)   (0b1100000'0000'0000'0000
)

a(524288) = 5 (1555^2 + 451^2 + 3^2 + 2^2 + 1^2)  (0b10000000'0000'0000'0000
)

a(540672) = 5 (1491^2 + 693^2 + 5^2 + 2^2 + 1^2)  (0b10000100'0000'0000'0000
)

a(720896) = 5 (1887^2 + 209^2 + 5^2 + 2^2 + 1^2)  (0b10110000'0000'0000'0000
)

a(786432) = 5 (1545^2 + 1243^2 + 9^2 + 2^2 + 1^2) (0b11000000'0000'0000'0000
)


–Arthur