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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),
(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);
(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
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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