# [PURRS-devel] problem with PURRS ? - with regards to OEIS's A136429

Alexander Povolotsky apovolot at gmail.com
Mon Apr 7 02:50:44 CEST 2008

```x(n)=x(n-x(n-1))+x(n-x(n-2))

PURRS output "this recurrence is malformed"

On Sat, Apr 5, 2008 at 10:27 PM, Alexander Povolotsky
<apovolot at gmail.com> wrote:
> Hi,
>
>  Here is the OEIS sequence description:
>  A136429
>  a(n) = sum( F(k+1)^2 F(n-k+1)^2, k = 0..n ) where F(n) = Fibonacci
>  number (A000045).
>
>         1, 2, 9, 26, 84, 250, 747, 2182, 6323, 18132, 51624, 146004, 410677,
>  1149578, 3204477, 8899502, 24634620, 67990414, 187154271, 513939214,
>  1408246247, 3851081256, 10512259920, 28647203880, 77946605545,
>  211782868754
>
>  OFFSET
>
>  0,2
>
>  FORMULA
>
>  G.f.: (1-x)^2/((1+x)^2(1-3x+x^2)^2).
>
>   Recurrence: a(n+6) = 4 a(n+5) - 10 a(n+3) + 4 a(n+1) - a(n).
>
>  AUTHOR
>
>  Emanuele Munarini (emanuele.munarini(AT)polimi.it), Apr 01 2008
>
>  So I tried PURRS
>  http://www.cs.unipr.it/purrs/
>  PURRS Demo Results
>  Exact solution for x(n) = -x(6+n)+4*x(5+n)-10*x(3+n)+4*x(1+n)
>  for the initial conditions
>  x(0) = 1
>  x(1) = 2
>  x(2) = 9
>  x(3) = 26
>  x(4) = 84
>  x(5) = 250
>  x(n) = -(-1)^n*n-2/5*(3/2+1/2*sqrt(5))^n+9/5*(-1)^n+4/5*(3/2+1/2*sqrt(5))^n*sqrt(5)-4/5
>  *(3/2-1/2*sqrt(5))^n*sqrt(5)-2/5*(3/2-1/2*sqrt(5))^n
>  for each n >= 0
>  Then I have defined sequence in PARI using close formula generated by PURRS
>  (21:16) gp > a(n)=-(-1)^n*n-2/5*(3/2+1/2*sqrt(5))^n+9/5*(-1)^n+4/5*(3/2+1/2*sqrt(5))^n*sqrt(5)-4/5
>  *(3/2-1/2*sqrt(5))^n*sqrt(5)-2/5*(3/2-1/2*sqrt(5))^n
>
>  But it doesn't even give initial conditions ... ?
>  (21:16) gp > a(0)
>  %5 = 1.000000000000000000000000000
>  (21:16) gp > a(1)
>  %6 = 2.000000000000000000000000000
>  (21:17) gp > a(3)
>  %7 = 26.00000000000000000000000000
>  (21:17) gp > a(4)
>  %8 = 63.00000000000000000000000000
>  (21:17) gp > a(5)
>  %9 = 174.0000000000000000000000000
>  (21:19) gp > a(6)
>  %10 = 443.0000000000000000000000000
>
>  Did I make a mistake in above or ... ?
>
>  Ciao,
>  Regards,
>  Alex
>

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