[seqfan] Re: An interesting sequence

[W. Edwin Clark]

According to
https://math.stackexchange.com/questions/1410157/integers-which-are-the-sum-of-non-zero-squares
every integer greater than 33 is a sum of five positive squares. So for 5n
> 33 we have
that n is the average of 5 squares. So a(n) is at most 5.

On Sun, Apr 16, 2023 at 12:53 PM Zach DeStefano <zachdestefano at gmail.com>
wrote:

> I've confirmed the conjecture for the first 50,000 terms and could easily
> compute further. My simple C++ program doesn't take more than a minute to
> calculate the exact value of these terms, and I would be happy to upload an
> a-file when this sequence has an A number.
>
> The trick I use is maintaining a map from averages (fractions represented
> as a pair of unsigned integers) to sets of counts (subset of {1, 2, 3, 4}
> represented as a bitset) and an array storing a running list of minimum
> squares required for each integer average so far.
>
> I consider each square one at a time and augment the map and array as
> needed. Before considering the square of n, I print all minimum counts for
> k < n^2 / 5 that haven't been emitted.
>
> - Zach
>
> On Sun, Apr 16, 2023 at 10:11 AM Fred Lunnon <fred.lunnon at gmail.com>
> wrote:
>
> >   Update at 2 hours CPU time ---
> >
> >     x = 5  for  n  in  {5, 19, 24, 32, 33, 44, 48, 76, 96} ,
> >
> >     96  =  ( 3^2 + 9^2 + 10^2 + 11^2 + 13^2 )5  ;
> >
> > next standout at  n = 128 .
> >
> > It's getting slow at this point --- I'll post a full table and abandon
> this
> > run soon.
> >
> > WFL
> >
> >
> > On Sun, Apr 16, 2023 at 1:25 PM Fred Lunnon <fred.lunnon at gmail.com>
> wrote:
> >
> > >
> > > Ouch --- thanks.
> > > Now that blunder is fixed, my brute-force Magma lash-up finds
> > >
> > >     x = 5  for  n  in  {5, 19, 24, 32, 33, 44, 48, 76} ;
> > >
> > >     76 = ( 3^2 + 5^2 + 9^2 + 11^2 + 12^2 )/5 .
> > >
> > > No further zeros up to next standout case at  n = 96 .
> > >
> > > WFL
> > >
> > >
> > > On Sun, Apr 16, 2023 at 11:23 AM <jens at voss-ahrensburg.de> wrote:
> > >
> > >>
> > >> a(11) = 5 since 11 = (25 + 16 + 9 + 4 + 1) / 5.
> > >>
> > >> Am 2023-04-16 12:07, schrieb Fred Lunnon:
> > >> > << only a(2), a(3), a(6), a(8), a(12) are 0 >>
> > >> >
> > >> > What about  n = 11  ?!     WFL
> > >> >
> > >> >
> > >> >
> > >> > On Sun, Apr 16, 2023 at 6:07 AM Yifan Xie <xieyifan4013 at 163.com>
> > wrote:
> > >> >
> > >> >> Hi,
> > >> >> a(n) is the smallest positive integer x such that n can be
> expressed
> > >> >> as
> > >> >> the arithmetic mean of x distinct nonzero squares, or 0 if x does
> not
> > >> >> exist. Based on my calculation of a(1) to a(76) by hand, only a(2),
> > >> >> a(3),
> > >> >> a(6), a(8), a(12) are 0 and no terms are larger than 5.
> > >> >> Please consider this sequence, and if possible, provide a program
> for
> > >> >> me.
> > >> >>
> > >> >> Best regards,
> > >> >> Yifan Xie (xieyifan4013 at 163.com)
> > >>
> > >>
> > >> --
> > >> Seqfan Mailing list - http://list.seqfan.eu/
> > >>
> > >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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