duplicate names

[Rainer Rosenthal]

Dear SeqFans,

could you please explain to me what the difference is between

                        exactly n ways
            and
                    exactly n different ways?

This question concerns OEIS sequences A000446 and A016032:

A000446 Smallest number that is the sum of 2 squares in exactly n ways.
A016032 Least number which is the sum of two squares in exactly n different ways.

In my opinion both descriptions are equivalent. But the sequences are
not equal and both of them lack a description. Maybe the description
is omitted because the title says it all? But either it doesn't or
somebody made computing errors.

Cheers,
Rainer Rosenthal
r.rosenthal at web.de

P.S.  Short explanation of my interest in these sequences:
I came across these sequences while searching for numbers k such that
k-x^2 is a square for all x in  { a, b, a+b, a-b} for certain a and b.

Having checked all numbers k from 1 to 1,500,000 I found:

    k      a     b
-------------------
 160225   140   216
 640900   280   432
1442025   420   648

The idea behind my search is to find magic 3x3 squares with all
entries square numbers.

Oops - having written this I realize: the second and third row in my
table are simply multiples of the first line. How embarrassing, but
on the other hand: how nice!




Ralf wrote:

> Paul wrote 
> >       Sequences  A120588 - A120607  are examples of 
> > solutions to functional equations of the form: 
> >     r*A(x) = c + b*x + A(x)^n 
> 
> Some more examples of g.f.s satisfying a cubic are below.
> Those that have a closed form all are products of
> binomials so I guess the pattern continues, although I
> couldn't find examples of higher than cubic g.f. solutions
> with closed form in the OEIS.
> 
> 
> ralf
> 
> A001764
> A007863
> A036759
> A036765
> A078531
> A088927...
> 
> and a whole bunch by Emeric Deutsch:
> 
> %F A102403 G.f.=G=G(z) satisfies z^3*G^3+z(1-z)G^2-G+1=0.
> %F A128729 G.f.=G=G(z) satisfies
> z^2*G^3-z(2-z)G^2+(1-z^2)G-1+z+z^2 =0. %F A067955
> a(n)=(1/n)sum(binomial(n, j)binomial((n-3-j)/2, j-1),
> j=1..floor((n-1)/3)) g.f. G(z) satisfies
> (1+z)G^3-zG^2-G+z=0 %F A120984
> a(n)=(1/(n+1)*sum(3^(3j-n)*binomial(n+1,j)*binomial(j
> ,n-2j), j=0..n+1). G.f.=G(z) satisfies
> G=1+3z^2*G^2+z^3*G^3. %F A120985
> a(n)=(1/(n+1))*sum(3^(n-3j)*binomial(n+1,2j+1)*binomial(n-
> 2j,j), j=0..n/2). G.f.=G(z) satisfies G=1+3zG + z^3*G^3.
> %F A128725 G.f.=G=G(z) satisfies
> z^2*G^3-2zG^2+(1+z-z^2)G-1=0. %F A128736 G.f.=G=G(z)
> satisfies zG^3=(1-2z)(G-1)(2G-1).
>
> ralf

We also have the higher-order k-Catalan numbers (not
with g.f. G=G(z) satisfying G=1+zG^k.

For k=4,5,...10 they are in A002293..96, A007556, A062994,
and A062744, where references can be found.

Emeric