✦ Exploratory · beyond the papers, not peer reviewed

Pell numbers: does the trick work again?

The Pentametric formula works because a simple rule, the remainder of n on division by 5, tells it whether to add or subtract the correction at each Lucas number. The Pell numbers obey an identity of exactly the same shape. Can a rule based on remainders choose the sign for them too? This page searches, and so far the answer is no.

The two identities side by side

	Fibonacci and Lucas	Pell
sequences	F: 0, 1, 1, 2, 3, 5, 8 … L: 2, 1, 3, 4, 7, 11, 18 …	P: 0, 1, 2, 5, 12, 29, 70 … q: 1, 1, 3, 7, 17, 41, 99 …
rule	x_k+1 = x_k + x_k−1	x_k+1 = 2x_k + x_k−1
identity	L_k² − 5F_k² = 4(−1)^k	q_k² − 2P_k² = (−1)^k
square from the companion	F_k² = (L_k² − 4(−1)^k)/5	P_k² = (q_k² − (−1)^k)/2
a rule for the sign from n alone?	yes: the remainder mod 5 (Legendre symbol)	searching…

q_k (half the Pell–Lucas numbers) plays the role of L_k. The sign (−1)^k depends on whether k is even or odd, which the rule must recover from q_k alone.

Test a modulus

A rule “by remainder mod m” works if no two terms with the same remainder have opposite signs.

Why it matters

For the Lucas numbers, the remainder mod 5 cycles 2, 1, 3, 4, 2, 1, 3, 4 … and the squares of these remainders (4, 1, 4, 1 …) alternate exactly with the sign (−1)^k. That coincidence is what lets one formula, (n² + 4·n5)/5, work for every n and land on F_k² at the Lucas numbers. For the Pell companion numbers no such modulus turns up: for every m tested, some q_k and q_j with k even and j odd leave the same remainder, so no rule based on remainders can tell them apart.

Exploratory: the search is a computation over full periods (k ≥ 2), not a proof. A proof that no modulus can ever work, or a modulus beyond the search range, would settle the question.