[seqfan] Re: A000118: either as many, or thrice as many

[Richard J. Mathar]

pa> I squared A004018 as a power series (i.e. (1*z⁰+4*z^1+4*z^2+0*z^3+...)^2 = 
pa> 1*z⁰+8*z^1+24*z^2+32*z^3+...) and found that it's A000118, which is no surprise. 
pa> a[n]=a[2n] when n is even (when squaring A004018, every other product is 0 
pa> because one of the terms being multiplied is the (4m+3)rd term, which is 
pa> always 0), and it appears that a[2n]=3a[n] when n is odd. How do you prove 
pa> that, when n is odd, 2n has three times as many ways as n of being the sum of 
pa> four squares? ....

So the question is: A000118(2*o) = 3*A000118(o) for all odd o?

Here is a (fuzzy) attempt for to prove this phenomenon:
A000118(n) has representations b(1)^2 +b(2)^2 +b(3)^2+b(4)^2
where b are integers (admitting negative integers), and 
swapping terms b(i) and b(j) adds to the count if b(i)<> b(j),
what called *ordered* sums of squares.
Call b(1)^2+b(2)^2 the *first* part, and b(3)^2+b(4)^2 the *second* part.

Because o is odd, the first part of A000118(o) is even 
and the second part odd, or the first part is odd and the second part is
even.


Because 2*o=e is even, the both parts of A000118(2*o) are even
or both parts of A000118(2*o) are odd.

A000118 is a convolution of A004018 with itself and A004018
is 4 times A002654 and the multiplicative property of A002654(e)
[e some even number] is that A002654(e) = A002654(odd part of e).

If both parts of A000118(2*o) are even we have 16 times the
contribution of A002654(e1)*A002654(e2) where we may write
always A002654(e1/2)*A002654(e2/2), so actually for e=2o
the cases where A000118(2*o) has two even parts gives the same
contribution as A000118(o).

Example: for A000118(14)=A000118(2*7) = (apart from factor 16)
A002654(0)*A002654(14)
+
A002654(1)*A002654(13)
+
A002654(2)*A002654(12)
+
A002654(3)*A002654(11)
+...
A002654(14)*A002654(0)
the even parts are from
A002654(0)*A002654(14)
+
A002654(2)*A002654(12)
+
A002654(4)*A002654(10)
+...
A002654(14)*A002654(0)
=
A002654(0)*A002654(7)
+
A002654(1)*A002654(6)
+
A002654(2)*A002654(5)
+...
A002654(7)*A002654(0)
= (reconvolution, apart from factor 16) -> A000118(7).

So the missing answer is whether the cases where both parts
of A000118(2*o) are odd gives a further contribution of 2*A000118(o) (?)

The basic property seems to be that for b(1)^2+b(2)^2 odd
and b(3)^2+b(4)^2 odd [and therefore parity of b(1) different from parity of b(2),
and also parity of b(3) different from parity of b(4)] one has 4 possible parity
scenarios for b(1) to b(4) from left to right:
+-+-
+--+
-++-
-+-+
and for each of these scenarios one can swap terms such that
after reordering the two parts are even:
+-+- b(1)^2+b(2)^2+b(3)^2+b(4)^2 = b(1)^2+b(3)^2 +b(2)^2+b(4)^2 ++--
+--+ b(1)^2+b(2)^2+b(3)^2+b(4)^2 = b(1)^2+b(4)^2 +b(2)^2+b(3)^2 ++--
-++- b(1)^2+b(2)^2+b(3)^2+b(4)^2 = b(1)^2+b(4)^2 +b(2)^2+b(3)^2 --++
-+-+ b(1)^2+b(2)^2+b(3)^2+b(4)^2 = b(1)^2+b(3)^2 +b(2)^2+b(4)^2 --++
For each of these swapped sums the left and (reordered) right sum
count separately, because the terms hat have been moved had different
parity and were therefore different. Now on the right hand
side one can again apply the halving argument as if the two parts
of A000118 had had two even parts (as above), giving another contribution
of A000118(o), and the we see that on the reordered sums there
is in each case another variant of swapping both terms of both
parts, for another A000118(o).

So in summary, the property that we are having convolutions
with terms that can be spliced in various cases of parities
and that A002654 "discards" the even component while
counting seem to ensure that A000118(2*o) =3*A000118(o).

HTH, Richard