sum of 4 squares being square

[vdmcc]

Hi all,

how do I call the collection of numbers  (a,b,c,d)
such that a^2+b^2+c^2+d^2= integer^2
ref:
---------------------------------------------
ID Number: A002635 (Formerly M0053 and N0018)
Sequence:  1,1,1,1,2,1,1,1,1,2,2,1,2,2,1,1,2,2,3,2,2,2,2,1,1,3,3,3,3,...
Name:      Partitions of n into 4 squares.
Offset:    0
at position n^2 = {1,4,9,16,25,.. we find
{1,2,2,2,3,4,4,2,6,7,6,4,8,10,14,2,11,14,13,7,23,15,17,4,24,21,31,10,25,37,2
8,
  2,46,29,49,14, ..
----------------------------------------------

BUT with the proviso that integer multiples (k*a,k*b,k*c,k*d) are removed ?

It is somewhat like unique integer length "integer vectors" along
"different crystallographic directions" in 4 dimensions.

example: the points counted by A002635 are:
               foursquares=
A002635[ 1]=1  {z[1,0,0,0]},
A002635[ 4]=2  {z[1,1,1,1],z[2,0,0,0]},
A002635[ 9]=2  {z[2,2,1,0],z[3,0,0,0]},
A002635[16]=2  {z[2,2,2,2],z[4,0,0,0]},
A002635[25]=3  {z[4,2,2,1],z[4,3,0,0],z[5,0,0,0]},
A002635[36]=4  {z[3,3,3,3],z[4,4,2,0],z[5,3,1,1],z[6,0,0,0]},
A002635[49]=4  {z[4,4,4,1],z[5,4,2,2],z[6,3,2,0],z[7,0,0,0]},
A002635[64]=2  {z[4,4,4,4],z[8,0,0,0]},
A002635[81]=6
{z[6,5,4,2],z[6,6,3,0],z[7,4,4,0],z[8,3,2,2],z[8,4,1,0],z[9,0,0,0]}

But
clean out the integer multiples (trivials) and you get:
z[1,0,0,0],
z[1,1,1,1],
z[2,2,1,0],
nihil
z[4,2,2,1],
z[4,3,0,0],z[5,3,1,1],
z[4,4,4,1],
z[5,4,2,2],z[6,3,2,0],z[6,5,4,2],
z[7,4,4,0],z[8,3,2,2],z[8,4,1,0],
counted by:
{1,1,1,0,2,1,3,0,4,3,5,0,7,5,10,0,10,6,12,0,18,8,16,0,21,12,25,0,24,18,27,0,
  39,17,43,0,...

If you're into lattuces (:-), then tell me,
does this have a name?

A propos,
ID Number: A014110
Sequence:  1,4,6,4,5,12,12,4,6,16,18,12,8,16,24,12,5,24,30,16,18,28,24,
           12,12,28,42,28,12,36,48,16,6,36,42,36,29,28,48,28,18,48,60,
           28,24,60,48,24,8,44,72,48,30,48,84,36,24,52,54
Name:      Ordered ways of writing n as a sum of 4 squares of natural
numbers.
Formula:   Coefficient of q^n in (1/16)*(1 + theta_3(0, q))^4; or coeff. of
q^n in
           (Sum q^(i^2),i=0..inf)^4.
Keywords:  easy,nonn
Offset:    0
Author(s): Joe Keane (jgk at jgk.org)

at position n^2 = {1,4,9,16,25,.. we find
{4,5,16,5,28,29,44,5,88,59,88,29,116,101,244,5,184,209,224,59,440,221,316,29
,
  436,299,628,101,484,647,548,5,964,491,956,209}
and that corresponds to the "unordered foursquares"
with multiplicity = Permutations of z[a,b,c,d]
but you surely knew that.

Any chance for a G.F. for A002635 ? looks feasible.

wouter.





w.meeussen.vdmcc at vandemoortele.be
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fax +32 (0) 51 33 21 75



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