Recurrence relation with not constant coefficients - math

Question

I have this recurrence relation
L^2 G[p]= 2(p-1)(2p-1)G[p-1] + ((p-1)(p-2)+a^2) G[p-2], where L and a are parameters.
Does anyone could help me to find the solution? Thanks

Answer 1

This looks like quite a complicated calculation, especially when no start values are given.
To get some more insight, one could use sympy, Python's symbolic math library to print the formulas for small values of p:
from sympy import symbols
def func_G(p):
if p == 0:
return G0
elif p == 1:
return G1
else:
return (2 * (p - 1) * (2 * p - 1) * func_G(p - 1) + ((p - 1) * (p - 2) + a ** 2) * func_G(p - 2)) / L ** 2
a, L, G0, G1 = symbols('a L G0 G1')
for p in range(8):
print(p, ':', func_G(p).simplify())
Prints out:
0 : G0
1 : G1
2 : (G0*a**2 + 6*G1)/L**2
3 : (20*G0*a**2 + G1*L**2*(a**2 + 2) + 120*G1)/L**4
4 : (840*G0*a**2 + 42*G1*L**2*(a**2 + 2) + 5040*G1 + L**2*(a**2 + 6)*(G0*a**2 + 6*G1))/L**6
5 : (60480*G0*a**2 + 3024*G1*L**2*(a**2 + 2) + 362880*G1 + 72*L**2*(a**2 + 6)*(G0*a**2 + 6*G1) + L**2*(a**2 + 12)*(20*G0*a**2 + G1*L**2*(a**2 + 2) + 120*G1))/L**8
6 : (G0*L**4*a**6 + 26*G0*L**4*a**4 + 120*G0*L**4*a**2 + 10960*G0*L**2*a**4 + 90720*G0*L**2*a**2 + 6652800*G0*a**2 + 158*G1*L**4*a**4 + 2620*G1*L**4*a**2 + 5040*G1*L**4 + 398400*G1*L**2*a**2 + 1209600*G1*L**2 + 39916800*G1)/L**10
7 : (248*G0*L**4*a**6 + 7488*G0*L**4*a**4 + 38880*G0*L**4*a**2 + 1770240*G0*L**2*a**4 + 15966720*G0*L**2*a**2 + 1037836800*G0*a**2 + G1*L**6*a**6 + 44*G1*L**6*a**4 + 444*G1*L**6*a**2 + 720*G1*L**6 + 28224*G1*L**4*a**4 + 526080*G1*L**4*a**2 + 1088640*G1*L**4 + 62513280*G1*L**2*a**2 + 199584000*G1*L**2 + 6227020800*G1)/L**12