**POLYA032: Sum of cubes of Pellian Numbers **(proposed
by Jose Luis Diaz-Barrero, 11/22/02).

Let P_n denote the n-th Pellian number; i.e. P_0 = 0, P_1=1,
and for n >= 2, P_n= 2P_{n-1} + P_{n-2}.

Find a close expression, if possible, in terms of Fibonacci numbers
for sum_{k=1}^n (P_n)^3, the sum of the cubes of the first n Pellian
numbers.

Comment by Proposer: This problem was motivated
by the fact that there exists in the literature ([1],[2]) close expressions
for the sum of cubes of Fibonacci and Lucas numbers in terms of the Fibonacci
numbers, namely,

(1) \sum_{k=1}^n (F_k)^3 = 1/10(F_{3n+2}
+ 6(-1)^{n-1}F_n + 5),

and

(2) \sum_{k=1}^n (L_k)^3 = 1/2(F_{3n+3} +
F_{3n+1} + 12(-1)^n F_n + 6(-1)^{n-1}F_{n-1} + 3).

Until now I was unable to find ( if there exists) a similar
expression for the sum given in the preceding statement.

[1] Cooper and R. Kennedy, Problem 3. *Missouri
Journal of Mathematical Sciences*, 1 (1988). 29.

[2] J. L. Diaz-Barrero, Problem 139. *Missouri
Journal of Mathematical Sciences,* 14 (2002) 210.

Discussions: **[JPE] (11/22/02):** We
can easily find a closed form of

S(n) = sum(P(k)^3,k=0..n) in terms of P(n).

P(n) = ((1+root(2))^n - (1-root(2))^n)/2/root(2);

hence

(1) : 8 P(k)^3 = P(3k) - 3 (-1)^k P(k)

Now, with sum of geometric sequences, we get

2 sum((-1)^k P(k), k=0..n) = (-1)^n (P(n+1)-P(n)) - 1

and

14 sum(P(3k),k=0..n) = 5 P(3n+1) + 3 P(3n) - 5

It follows now from (1) that

S(n) = (5/112) P(3n+1) + (3/112) P(3n)+(-1)^n (3/16)
(P(n)-P(n+1)) + 1/7.

**[SCL] (11/22/02):** The Fibonacci
numbers and the Lucas numbers are virtually interchangeable --they are
just aspects of the same recurrence but with different boundary conditions.
The Pellian numbers grow like (1+sqrt(2))^n. Their cubes grow like (1+sqrt(2))^{3n},
and it should be easy to find an expression similar to the type you want
with the exception that the closed expression is in terms of Pellian numbers.
I haven't done this step, but I expect it is an easy exercise. The above
comments make it fairly obvious that I don't expect a closed expression
in terms of Fibonacci numbers. If there were such an expression, the lead
term would be somewhere around F_{q}, where q=3s/t, with s=log(1+sqrt(2)),
and t=log((1+sqrt(5))/2).

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