What next in 0,1,6,25,96,361,1350,5041?

[Ralf Stephan]

You wrote 
> S_4   = A000290    the squares
> S_5   = A004146    Alternate Lucas Numbers - 2
> S_7   = A054493    A Pellian-related sequence
> S_8   = A001108    a(n)-th triangular number is a square
> S_9   = A049684    F(2n)^2 where F() = Fibonacci numbers

also 
S_6(n) = A001075(n)-1 = A087096(n)-1 = (A016064(n)-1)/2
S_10(n) = (A001091(n+1)-1)/3
S_11 = part. sums of A057081

let's look at the gfs:

S_4: (x^2+x)/(1-x)^3
S_5: (x^2+x)/(1-4x+4x^2-x^3)
S_6: (x^2+x)/(1-5x+5x^2-x^3)
...
S_11: (x^2+x)/(1-10x+10x^2-x^3)

So the general recurrence seems to be
S_k: a(n) - (k-1)a(n-1) + (k-1)a(n-2) - a(n-3) = 0
     a(n) = (k-1)a(n-1) - (k-1)a(n-2) + a(n-3)

Proving the product recurrence should be easy now.
This means you also have
S_1: A011655 (periodic 0,1,1,...)
S_2: A007887 (periodic 0,1,2,1,...)
S_3: |A078070| (periodic 0,1,3,4,3,1,...)

Please submit an array.

Regards,
ralf