[seqfan] Re: An interesting sequence

[Fred Lunnon]

  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)
>>
>>
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>>
>