slug: parisse-stirling-fibonacci-alternating-sum bibkey: parisse2024hypersequences doi: 10.5281/zenodo.13331499 url: https://math.colgate.edu/~integers/y70/y70.pdf triage: theorem motivation_gids:
- D5/S3/Combinatorics/ParisseStirlingFibonacciAlternatingSum.result
The Alternating Stirling Weighted Sum of Even-Index Fibonacci and Lucas Numbers
Problem
Parisse, Integers 24 (2024), Article A70, Section 4, states two conjectures. The first, after Equation (4.7):
Conjecture 1. For all ℓ ∈ N_0, we have
∑_{m=0}^{ℓ} (-1)^m m! S(ℓ+1, m+1) F_{2(m+1)} = (-1)^ℓ ∑_{m=0}^{ℓ} m! S(ℓ, m) F_{m+2} .
The second, after Equation (4.9):
Conjecture 2. For all ℓ ∈ N_0, we have
∑_{m=0}^{ℓ} (-1)^m m! S(ℓ+1, m+1) L_{2(m+1)} = (-1)^ℓ ∑_{m=0}^{ℓ} m! S(ℓ, m) L_{m+2} .
Here S(j, m) is the Stirling number of the second kind, F the Fibonacci
numbers with F_1 = F_2 = 1, and L the Lucas numbers with L_0 = 2, L_1 = 1.
Both are the constant term of the article’s own formula for the weighted sums
∑_{k=0}^{n} k^ℓ F_k and ∑_{k=0}^{n} k^ℓ L_k, rewritten through Equation (3.14),
c_{ℓ,m}(0) = (-1)^m m! S(ℓ+1, m+1).
The article’s remarks fix the reading of the stacked bracket symbols. The
right-hand side of Conjecture 1 is named there as A000557, whose recorded formula
is a(n) = ∑_{k=0..n} k! Stirling2(n, k) Fibonacci(k+2); the right-hand side of
Conjecture 2, times -1, is named as A263968, whose recorded formula is
a(n) = (-1)^{n+1} ∑_{k=0..n} k! Lucas(k+2) Stirling2(n, k). Both readings are
therefore Stirling numbers of the second kind, not binomial coefficients.
Motivation
The frozen theorem
D5/S3/Combinatorics/ParisseStirlingFibonacciAlternatingSum.result settles both.
Neither identity depends on the initial values of the sequence, so the recorded
proof establishes the common statement for an arbitrary integer sequence
satisfying u_{n+2} = u_n + u_{n+1} and reads off the two conjectures as the
Fibonacci and Lucas instances.
Gap
Issue 9415 records the screen carried out before write-up. The author’s own sequel, Integers 26 (2026), Article A36, published 2/20/26, was opened in full and contains no occurrence of the word conjecture. The OEIS entries A000557 and A263968 were opened; both link this article, and neither carries the alternating identity among its formulas. A web pass over the two sequence numbers and the article title returned no proof; the nearest recent item, arXiv:2511.10797 on weighted sums of Lucas sequences, mentions neither Parisse nor Stirling numbers. Citation-index result pages were not exhaustively reachable, so this is a bounded negative finding.
Route
Let Q_n(X) = ∑_m m! S(n, m) X^m. This is the classical Fubini polynomial, also
called the ordered Bell polynomial; Q_n(1) is the number of ordered partitions of an
n-element set. The Stirling recurrence S(n+1, m) = m S(n, m) + S(n, m-1) becomes the
differential recurrence
Q_0 = 1 , Q_{n+1} = X(1 + X) Q_n' + X Q_n .
The first ingredient is the functional equation
X · Q_n(-1 - X) = (-1)^n (1 + X) Q_n(X) for n ≥ 1 ,
proved by induction: differentiating the statement at n and combining it with
the composite of the recurrence gives the statement at n + 1 in one linear step.
The same equation drops out of the exponential generating function
∑_n Q_n(x) t^n / n! = 1/(1 - x(e^t - 1)), which is the standard one for the Fubini
polynomials. Substituting x → -1-x and multiplying by x gives
x / ((1+x) e^t - x); substituting t → -t and multiplying by 1+x gives
(1+x) e^t / ((1+x) e^t - x). The two differ by the constant -1, which is exactly the
n = 0 term, and that is why the equation carries the hypothesis n ≥ 1. The recorded
proof runs the induction rather than the generating function, which would need the formal
exponential series; the computation above is an independent check of the statement.
The second ingredient is the pairing L_c(p) = ∑_k p_k u_{k+c} of a polynomial
with a sequence satisfying u_{n+2} = u_n + u_{n+1}. It satisfies
L_c(X p) = L_{c+1}(p) by reindexing, L_c((1+X) p) = L_{c+2}(p) by the
recurrence, and
L_1((1 + X)^m) = u_{2m+1} ,
which is the binomial index-doubling identity ∑_i C(m, i) u_{i+c} = u_{2m+c},
itself an induction that uses the recurrence alone.
The two ingredients meet as follows. Writing c_k for the coefficients of Q_ℓ,
the Stirling recurrence gives m! S(ℓ+1, m+1) = c_{m+1} + c_m, so the left-hand
side of the conjecture is ∑_m (-1)^m (c_{m+1} + c_m) u_{2m+2}. Reindexing the
first part by j = m + 1 and using u_{2j+2} - u_{2j} = u_{2j+1} collapses this
to ∑_k c_k (-1)^k u_{2k+1}, which is exactly L_1(Q_ℓ(-1 - X)). The boundary
term at j = 0 drops out because Q_ℓ has no constant term for ℓ ≥ 1, which
also gives Q_ℓ = X P. Cancelling X in the functional equation turns the
composite into (-1)^ℓ (1 + X) P, and then
L_1(Q_ℓ(-1-X)) = (-1)^ℓ L_1((1+X) P) = (-1)^ℓ L_3(P) = (-1)^ℓ L_2(Q_ℓ) ,
the last expression being the right-hand side of the conjecture. The case ℓ = 0
is both sides equal to u_2.
Falsifier
A different value of S(ℓ+1, m+1) for small ℓ, or a sign error in the
functional equation, would break the identity at once. The first common values of
the two sides of Conjecture 1 are 1, -2, 8, -50, 416, -4322, 53888, -783890,
whose absolute values are A000557; the first values for Conjecture 2 are
3, -4, 18, -112, 930, -9664, which up to sign are A263968. The functional
equation can be checked at n = 2: Q_2 = X + 2X^2, and
X(-1 - X + 2(1 + X)^2) = X(1 + 3X + 2X^2) = (1 + X)(X + 2X^2).
Evidence
Both identities were checked numerically for ℓ = 0 through ℓ = 125 with no
mismatch before the formalisation was attempted.
The generalisation to an arbitrary sequence satisfying u_{n+2} = u_n + u_{n+1}
is not a strengthening for its own sake: the initial values never enter the
argument, and carrying them would require a separate treatment of the boundary
term for each sequence. The Lucas case is where this shows, since L_0 = 2 is not
zero and the boundary term is killed instead by the vanishing constant term of
Q_ℓ.
Triage
theorem; Tier 1 named external open questions, preregistered in issue 9415
before the probe. The admission basis is escape-witness; the classification is
proof_shape: content. The module reports utility: none: no declaration in it
is a bounded enumeration, a checker, a numeric reduction or a certified instance,
and the statement is a universally quantified identity rather than a finite
computation.
ASSUMED-UNVERIFIED
The literature screen is bounded: the article, its 2026 sequel, the OEIS entries A000557 and A263968, and a web pass over the two sequence numbers and the article title were opened; citation-index result pages were not exhaustively reachable, so no worldwide priority claim is made.
The numerical range ℓ ≤ 125 is a pre-formalisation check only and is superseded
by the recorded proof; it is stated here because it is what motivated the attempt,
not as evidence for the identity.