[seqfan] Re: An interesting sequence

[W. Edwin Clark]

Oops. I forgot the requirement that the squares be distinct.

On Sun, Apr 16, 2023 at 2:04 PM W. Edwin Clark <wclark at mail.usf.edu> wrote:

> 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/
>> >
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>