Notes on A062316 et al

[David Wilson]

Another "duh" moment.  Thank you, that takes the mystery out of it.
The first differences of A062316 are <= 8 because every n == 6 (mod 8)
is in A062316.  The first differences are all == 0 (mod 4) since every
element of A062316 == 2 (mod 4).  Hence all first differences are 4 or 8.

I can show that 4 and 8 each show up infinitely often as differences.

We know the sums of two squares (A001481) are of density 0 over Z+,
so the nonsums (A062316) are of density 1.  This means that A062316,
being those nonsums == 2 (mod 4), has density 1/4.  But if 4 did not
appear infinitely often in A062316, 8 would be the only possible difference
for sufficiently large elements, which would all have to be == 6 (mod 8).
The density of A062316 would then be 1/8, a contradiction.  So 4 shows
up infinitely often as a first difference.  QED.

For difference 8.  Consider the 9 consecutive integers 8k^2 - 2 through
8k^2 + 6 for any k >= 1.
    8k^2 - 2  == 6 (mod 8) is in A062316 (modulus not sum of two squares).
    8k^2 - 1 != 2 (mod 4) is out (modulus not difference of two squares).
    8k^2 != 2 (mod 4) is out.
    8k^2 +1 != 2 (mod 4) is out.
    8k^2 + 2 = (2k-1)^2 + (2k+1)^2 is not in (sum of two squares).
    8k^2 + 3 != 2 (mod 4) is out.
    8k^2 + 4 != 2 (mod 4) is out.
    8k^2 + 5 != 2 (mod 4) is out.
    8k^2 + 6 == 6 (mod 8) is in.
So 8 shows up as the first difference between 8k^2 - 2 and 8k^2 + 6 for
every k >= 1.  QED.


Similar logic can be applied to A022544, the nonsums of two squares.
Since every n = 3 (mod 4) is an element, first differences are <= 4.

I can also show that each of the first differences 1, 2, 3 and 4 shows up
infinitely often in the same way I showed first difference 8 shows up
infinitely often in A062316.

For difference 1,
    8k+6 == 6 (mod 8) is in A022544.
    8k+7 == 7 (mod 8) is in.

For difference 2,
    64k^2 + 48k + 12 == 12 (mod 16) is in.
    64k^2 + 48k + 13 = (8k+3)^2 + 2^2 is out.
    64k^2 + 48k + 14 == 14 (mod 16) is in.

For difference 3,
    16k^2 + 16k + 3 == 3 (mod 8) is in.
    16k^2 + 16k + 4 = (4k+2)^2 + 0^2 is out.
    16k^2 + 16k + 5 = (4k+2)^2 + 1^2 is out.
    16k^2 + 16k + 6 == 6 (mod 8) is in.

For difference 4,
    4k^4 + 8k^3 + 4k^2 - 1 == 7 (mod 8) is in.
    4k^4 + 8k^3 + 4k^2 = (2k^2 + 2k)^2 + 0^2 is out.
    4k^4 + 8k^3 + 4k^2 + 1 = (2k^2 + 2k)^2 + 1^2 is out.
    4k^4 + 8k^3 + 4k^2 + 2 = (2k^2 + 2k - 1)^2 + (2k + 1)^2 is out.
    4k^4 + 8k^3 + 4k^2 + 3 == 3 (mod 8) is in.

----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <ham>; <seqfan at ext.jussieu.fr>
Sent: Saturday, February 26, 2005 10:26 PM
Subject: Re: Notes on A062316 et al


> David Wilson wrote:
>
>> A062316 is, by definition, the intersection of the numbers which are not
>> sums of two squares (A022544) and the numbers which are not the 
>> difference
>> of two squares (A016825). A016825 consists precisely those n == 2 (mod 
>> 4),
>> so the above observation that a(n) is always even can be sharpened to 
>> a(n)
>> == 2 (mod) 4.
>>
>> The observation that the first difference is 4 or 8 appears to be 
>> correct.
>
> It is correct:  The sequence contains all numbers == 6 (mod 8), since
> all squares are == 0, 1, or 4 (mod 8).
>
> Dean Hickerson
> dean at math.ucdavis.edu
>
>
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