Some Properties of the Multiple Binomial Transform and the Hankel Transform of Shifted Sequences

Jiaqiang Pan School of Biomedical Engineering and Instrumental Science Zhejiang University Hangzhou 210027 China

                     panshw@mail.hz.zj.cn

Abstract In this paper, the author studies the multiple binomial transform and the Hankel transform of shifted sequences of an integer sequence, particularly a linear homogeneous recurrence sequence, and some of their properties.

1 Notation

In this paper, we generally use function symbols, like a(t), b(t), etc., to express integer sequences, where t ∈ ℕ_{0} = {0, 1, 2, …}. However sometimes, to employ matrix tools in deduction process, we also denote the integer sequences by using (infinite-dimensional) vector symbols, like a = (a(0), a(1), a(2), a(3), ⋯, ⋯)^{T}, b = (b(0), b(1), b(2), b(3), ⋯, ⋯)^{T}, etc.

2 Multiple binomial transforms of shifted sequences

Definition 1 (Shifting integer sequences). Let a(t) be an integer sequence and σ be the shift operator. Then we define the pth-order shifted sequence a_{(p)}(t)), (p = 0, 1, 2, …), of a(t), as follows:

a_{(p)}(t) = σ^{p}(a) = a(t + p), t = 0, 1, 2, …, (1)

Note that in the case p = 0, a_{(0)}(t) = σ^{0}(a) = a(t).

Definition 2 (Multiple binomial transforms). Let a(t) be an integer sequence. Then according to Pan [1], we define the n-fold binomial transform of a(t), and denote its image sequence by B_{n}(a) or a^{(n)}(t), as follows:

a^{(1)}(t) = B_{1}(a) = ∑_{k=0}^{t} \binom{t}{k}a(k),  a^{(n)}(t) = B_{n}(a) = \overbrace{B_{1}(B_{1}(⋯ (B_{1}}^{n−fold}(a)))),  (2)

where n = 0, 1, 2, . . .. Note that in the case n = 0, B_{0}(a) = a^{(0)}(t) = a(t), that is, the transform B_{0} just is the identity transform.

Definition 3 (Inverse multiple binomial transform). Let a(t) be an integer sequence. Then according to Pan [1], we define the m-fold inverse binomial transform of a(t), and denote its image sequence by B_{−m}(a) or a^{(−m)}(t), as follows:

a^{(−1)}(t) = B_{−1}(a) = ∑_{k=0}^{t}(−1)^{t−k}\binom{t}{k}a(k), a^{(−m)}(t) = B_{−m}(a) = \overbrace{B_{−1}(B_{−1}(⋯ (B_{−1}}^{m−fold}(a)))),

(3)

where m = 1, 2, . . ..

Remark 4. We can express (2) in the matrix form: a^{(1)} = B_{1}a, where the transform matrix B_{1} is an infinite-order lower-triangular matrix, as follows:

B_{1} = ( \begin{matrix}( & {0 \atop 0} & ) \\ \binom{1}{0} & \binom{1}{1} \\ \binom{2}{0} & \binom{2}{1} & \binom{2}{2} \\ \binom{3}{0} & \binom{3}{1} & \binom{3}{2} & \binom{3}{3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) = ( \begin{matrix}1 \\ 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 3 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,  (4)

and

a^{(n)} = (a^{(n)}(0), a^{(n)}(1), a^{(n)}(2), ⋯, ⋯)^{T} = B_{n}a = B_{1}^{n}a,  (5)

where n = 0, 1, 2, 3, . . .. The transform matrix of the n-fold binomial transform B_{n} (= B_{1}^{n}) is always a lower-triangular transform matrix with each of the diagonal elements being one.

Remark 5. We can also express (3) in matrix form, as a^{(−1)} = B_{−1}a, where the transform matrix B_{−1} is an infinite-order lower-triangular matrix, as

B^{−1} = ( \begin{matrix}( & {0 \atop 0} & ) \\ −\binom{1}{0} & \binom{1}{1} \\ \binom{2}{0} & −\binom{2}{1} & \binom{2}{2} \\ −\binom{3}{0} & \binom{3}{1} & −\binom{3}{2} & \binom{3}{3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) = ( \begin{matrix}1 \\ −1 & 1 \\ 1 & −2 & 1 \\ −1 & 3 & −3 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,  (6)

and

a^{(−m)} = (a^{(−m)}(0), a^{(−m)}(1), a^{(−m)}(2), ⋯, ⋯)^{T} = B_{−m}a = B_{−1}^{m}a,  (7)

where m = 1, 2, 3, . . .. The transform matrix B_{−m} (= B_{−1}^{m}) is also always a lower-triangular transform matrix with each of the diagonal elements being one. We see that B_{1}B_{−1} = B_{−1}B_{1} = E, where E is the infinite-order unit matrix. It is the matrix form of well-known inversion relation: ∑_{k=i}^{t}(−1)^{t−k}\binom{t}{k}\binom{k}{i} = ∑_{k=i}^{t}(−1)^{k−i}\binom{t}{k}\binom{k}{i} = δ_{ti}, where t, i = 0, 1, 2, ….

Remark 6. We view the n-fold binomial or inverse binomial transform B_{n}, (n = 0, ±1, ±2, ±3, …), to be one simple transform of integer sequences, because such inversion relations as B_{2}B_{−2} = B_{−2}B_{2} = E, B_{3}B_{−3} = B_{−3}B_{3} = E hold, and so forth. For example, for 2-fold binomial and inverse binomial transforms, the transform matrices are respectively

B_{2} = ( \begin{matrix}1 \\ 2 & 1 \\ 4 & 4 & 1 \\ 8 & 12 & 6 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix} ) , B_{−2} = ( \begin{matrix}1 \\ −2 & 1 \\ 4 & −4 & 1 \\ −8 & 12 & −6 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix} ) ,  (8)

Now, let us give the multiple binomial transforms of the shifting sequences a_{(p)}(t), (p = 0, 1, 2, …), of an integer sequence a(t).

Theorem 7. Let a(t) be an integer sequence. Then

B_{n}(a_{(p)}) = (σ − n)^{p}(B_{n}(a)) = (σ − n)^{p}(a^{(n)}) = ∑_{k=0}^{p}(−n)^{p−k}\binom{p}{k}σ^{k}(a^{(n)}),  (9)

where n = 0, ±1, ±2, . . ..

Proof. Use the mathematical induction. When n = ±1 and p = 1,

B_{1}(σ(a)) = ∑_{k=0}^{t} \binom{t}{k}a(k + 1) = ∑_{k=1}^{t+1} \binom{t}{k − 1}a(k) = ∑_{k=1}^{t+1} \binom{t + 1}{k}a(k) − ∑_{k=1}^{t+1} \binom{t}{k}a(k)

= ∑_{k=0}^{t+1} \binom{t + 1}{k}a(k) − a(0) − [∑_{k=0}^{t} \binom{t}{k}a(k) − a(0)] = σ(B_{1}(a)) − B_{1}(a) = (σ − 1)(B_{1}(a)),

and

B_{−1}(σ(a)) = ∑_{k=0}^{t}(−1)^{t−k}\binom{t}{k}a(k + 1) = ∑_{k=1}^{t+1}(−1)^{t+1−k}\binom{t}{k − 1}a(k)

= ∑_{k=0}^{t+1}(−1)^{t+1−k} [\binom{t + 1}{k} − \binom{t}{k}] a(k) = ∑_{k=0}^{t+1}(−1)^{t+1−k}\binom{t + 1}{k}a(k) + ∑_{k=0}^{t}(−1)^{t−k}\binom{t}{k}a(k)

= σ(B_{−1}(a)) + B_{−1}(a) = (σ + 1)(B_{−1}(a)).

If for n = ±k(k is some positive integer), B_{±k}(σ(a)) = (σ ∓ k)(B_{±k}(a)) holds, then for n = ±(k + 1), B_{±(k+1)}(σ(a)) = B_{±1}(σ(B_{±k}(a))) ∓ kB_{±1}(B_{±k}(a)) = (σ ∓ 1)(B_{±(k+1)}(a)) ∓ kB_{±(k+1)}(a) = (σ ∓ (k + 1))(B_{±(k+1)}(a)) also holds. Hence, for any integer n, B_{n}(σ(a)) = (σ − n)(B_{n}(a)) holds. On the other hand, if for p = m(m is some positive integer) that B_{n}(σ^{m}(a)) = (σ − n)^{m}(B_{n}(a)) holds, then when p = m + 1, we get that B_{n}(σ^{m+1}(a)) = (σ − n)^{m}(B_{n}(σ(a))) = (σ − n)^{m}((σ − n)(B_{n}(a))) = (σ − n)^{m+1}(B_{n}(a)). Hence, for any positive integer n and p, B_{n}(σ^{p}(a)) = (σ −n)^{p}(B_{n}(a)). Special cases that n = 0 and/or p = 0 are trivial.

Corollary 8. Let a(t) be an integer sequence, and P (σ) be an integer-coefficient polynomial in σ. Then

B_{n}(P (σ)(a)) = P (σ − n)(B_{n}(a)) = P (σ − n)(a^{(n)}),  (10)

where n = 0, ±1, ±2, ….

Proof. Let P (σ) be a integer-coefficient polynomial of degree p (p = 0, 1, 2, …) in σ: P (σ) = ∑_{k=0}^{p} c_{k}σ^{k}, where c_{k}s are (p + 1) integers. From Theorem 7, we have that B_{n}(P (σ)(a)) = B_{n}(∑_{k=0}^{p} c_{k}σ^{k}(a)) = ∑_{k=0}^{p} c_{k}B_{n}(σ^{k}(a)) = ∑_{k=0}^{p} c_{k}(σ − n)^{k}(B_{n}(a)) = P (σ − n)(B_{n}(a)) = P (σ − n)(a^{(n)}).

Remark 9. By using Corollary 8, we can more succinctly prove the following known property of recurrence sequences (see [1, Thm. 17]). Let a(t) be a linear homogeneous recurrence sequence of order q with the recurrence equation

P (σ)(a) = ∑_{k=0}^{q} b_{k}σ^{q−k}(a) = 0,  (11)

where b_{0} = 1, b_{1}, b_{2}, …, b_{q} are q given integers. Then its q complex characteristic values λ_{k}, k = 1, 2, …, q, are the roots of polynomial (algebraic) equation:

P (λ) = ∑_{k=0}^{q} b_{k}λ^{q−k} = 0.  (12)

On the other hand, by taking transformation B_{n} of the two sides of (11), and then employing Corollary 8, we find that sequences a^{(n)}(t), (n = 0, ±1, ±2, …), satisfy recurrence equation:

P (σ − n)(a^{(n)}) = 0.  (13)

This implies that q complex characteristic values λ_{k}^{(n)}, (k = 1, 2, …, q), of a^{(n)}(t) are the roots of the algebraic equation:

P (λ^{(n)} − n) = ∑_{k=0}^{q} b_{k}(λ^{(n)} − n)^{q−k} = 0.  (14)

Comparing (12) with (14), we find that λ_{k}^{(n)} − n = λ_{k}, namely

λ_{k}^{(n)} = λ_{k} + n, (k = 1, 2, …, q).  (15)

3 Shifted sequences and the Hankel transform

Layman proved the invariance of the Hankel transform under applications of the binomial transform or its inverse transform (see [2]). For an integer sequence, the n-fold binomial (or inverse binomial) transform is the same as the n times successive binomial (or inverse binomial) transform operation, Pan [1] pointed out that the invariance of the Hankel transform holds under applications of the n-fold binomial (or n-fold invert binomial) transform. Now by using Theorem 7, we give a more direct and succinct proof of the invariance, as follows. Remark 10. By using Definition 1, we express the Hankel matrix H_{a} of sequence a(t) as

H_{a} = ( a σ(a) σ^{2}(a) σ^{3}(a) ⋯) = ( a a_{(1)} a_{(2)} a_{(3)} ⋯) ,  (16)

and Hankel matrix H_{a^{(n)}} of integer sequence a^{(n)}(t) as

H_{a^{(n)}} = ( a^{(n)} σ(a^{(n)}) σ^{2}(a^{(n)}) σ^{3}(a^{(n)}) ⋯) ,  (17)

According to Theorem 7, we have that

B_{n}H_{a} = ( B_{n}a B_{n}a_{(1)} B_{n}a_{(2)} B_{n}a_{(3)} ⋯) = ( a^{(n)} (σ − n)(a^{(n)}) (σ − n)^{2}(a^{(n)}) (σ − n)^{3}(a^{(n)}) ⋯) . (18)

Comparing (18) with (17), we see that the upper-left (t+1)×(t+1) (t = 0, 1, 2, …) sub-matrix of B_{n}H_{a} has the same determinant to the upper-left sub-matrix of the Hankel matrix H_{a^{(n)}} of sequence a^{(n)}(t). On the other hand, the determinant of the upper-left (t + 1) × (t + 1) (t = 0, 1, 2, …) sub-matrix of matrix B_{n}H_{a} is equal to the determinant of the upper-left (t + 1) × (t + 1) (t = 0, 1, 2, …) sub-matrix of matrix H_{a}, because the determinant of any upper-left sub-matrices of matrix B_{n} (n = ±1, ±2, ±3, …) is always equal to one. In other words, the sequences a and a^{(n)} both have the same Hankel transform, for any integer n. Remark 11. This result gives an affirmative answer to one of Layman’s two questions raised in [2]: Are there other interesting transforms, T, of an integer sequence S, in addition to the Binomial and Invert transforms, with the property that the Hankel transform of S is the same as the Hankel transform of the T transform of S? For example, T = B_{2} or B_{−2}, which have transform matrices listed in (8).

Next, we investigate the Hankel transform of recurrence sequences. The following theorem gives a basic property of the Hankel transform of recurrence sequences.

Theorem 12. Let a(t) be a linear homogeneous recurrence sequence of order q, with recurrence equation (11). Then the Hankel transform h_{a}(t) of sequence a(t) is a finite sequence with length q, that is, for t ≥ q, h_{a}(t) ≡ 0.

Proof. We see from (16) and (11) that if multiplying the first, the second, …, the q-th column vectors of the Hankel matrix H_{a} by b_{q}, b_{q−1}, …, b_{1} respectively, and then adding them to the (q + 1)th column σ^{q}(a), we cause the (q + 1)-th column to be a zero-column. This operation does not change the determinants of principal sub-matrices of H_{a}. On the other hand, for a infinite-order square matrix with its (q + 1)-th column being a zero-column, determinants of the principal sub-matrices of order q + 1, q + 2, q + 3, …, namely h(q), h(q + 1), h(q + 2), …, are always equal to zeros. That is, the Hankel transform h(t) is a finite integer sequence with the length of q.

Corollary 13. All of the n-fold binomial transforms a^{(n)}(t) (n = 0, ±1, ±2, ±3, …) of a q-order recurrence sequence a(t) have identical Hankel transform with the length of q.

Remark 14. For example, as recurrence sequences of order 2 and 3, the Fibonacci sequence F (t) (\underline{A000045} in [3]) and its multiple binomial transforms \underline{A001906}, \underline{A093131}, \underline{A039834}, etc. (see Pan [1]) all have the same Hankel transform with length 2: h_{F} (0) = 1, h_{F} (1) = 1, and the Tribonacci sequence T (t) (\underline{A000073} in [3]) and its multiple binomial transforms \underline{A115390}, etc. (see Pan [1]) all have the same Hankel transform with length 3: h_{T} (0) = 3, h_{T} (1) = 8, h_{T} (2) = −44.

Finally, we give special relations of the Hankel transforms of a^{(n)}(t), (n = 0, ±1, ±2, …), and a_{(p)}(t), (p = 0, 1, 2, …), with the general term formula of the recurrent sequences a(t), respectively.

Theorem 15. Let a(t) be a linear homogeneous recurrence sequence of order q, with the general-term formula: a(t) = ∑_{i=1}^{q} c_{i}λ_{i}^{t}, t ∈ ℕ_{0}. Then the Hankel transforms h_{a^{(n)}}(t), (n = 0, ±1, ±2, …), are such that

h_{a^{(n)}}(t) = ∑_{(i_{1},i_{2},··· ,i_{t+1})} ∏_{k=1}^{t+1}(c_{i_{k}}λ_{i_{k}}^{k−1}) ∏_{1≤k<m≤(t+1)}(λ_{i_{k}} − λ_{i_{m}}), t = 0, 1, …, q − 1,  (19)

where the summation is over the q!/(q −t−1)! different (t+1)-permutations (i_{1}, i_{2}, ⋯, i_{t+1}) of set {1, 2, …, q}. Particularly, the first term h_{a^{(n)}}(0) = ∑_{i=1}^{q} c_{i} = a(0), and the qth (last) term h_{a^{(n)}}(q − 1) = ∏_{i=1}^{q} c_{i} ∏_{1≤i<j≤q}(λ_{i} − λ_{j})^{2}.

Proof. Denoting j-order vectors (1, λ_{i}, λ_{i}^{2}, ⋯, λ_{i}^{j−1}) by λ(i, j), and (j × j) Vandermonde square-matrices (λ(i_{1}, j), λ(i_{2}, j), …, λ(i_{j}, j)) by 𝕍(i_{1}, i_{2}, ⋯, i_{j}) respectively, where i ∈ {1, 2, …, q}, and (i_{1}, i_{2}, ⋯, i_{j}) is a j-permutation of set {1, 2, …, q}, (1 ≤ j ≤ q), we find that the t-th term of Hankel transform h_{a}(t) of a(t), that is, the determinant of upper-left (t + 1) × (t + 1) sub-matrix of Hankel matrix (16), is

h_{a}(t) = det [ ∑_{i=1}^{q} c_{i}λ(i, t + 1) ∑_{i=1}^{q} c_{i}λ_{i}λ(i, t + 1) ⋯ ∑_{i=1}^{q} c_{i}λ_{i}^{t}λ(i, t + 1) ]

= ∑_{(i_{1},i_{2},··· ,i_{(t+1)})} ( ∏_{k=1}^{t+1}(c_{i_{k}}λ_{i_{k}}^{k−1})) det 𝕍(i_{1}, i_{2}, ⋯, i_{t+1}),

where the summation is over q!/(q−t−1)! different (t+1)-permutations (i_{1}, i_{2}, ⋯, i_{t+1}) of set {1, 2, …, q}. The Vandermonde determinant det 𝕍(i_{1}, i_{2}, ⋯, i_{t+1}) equals ∏_{1≤k<m≤(t+1)}(λ_{i_{k}}− λ_{i_{m}}). Because h_{a^{(n)}}(t) = h_{a}(t), (19) holds. In case t = 0, we see that h_{a^{(n)}}(0) = h_{a}(0) = ∑_{i=1}^{q} c_{i} = a(0); in the case t = q − 1, we have that

h_{a^{(n)}}(q − 1) = h_{a}(q − 1) = det [ ∑_{i=1}^{q} c_{i}λ(i, q) ∑_{i=1}^{q} c_{i}λ_{i}λ(i, q) ⋯ ∑_{i=1}^{q} c_{i}λ_{i}^{q−1}λ(i, q) ] ,

The matrix in the right side of the above equality just equals a product of three square matrices: 𝕍(1, 2, ⋯, q) · diag{c_{1}, c_{2}, …, c_{q}} · 𝕍^{T} (1, 2, ⋯, q). Hence, we have that

h_{a^{(n)}}(q−1) = det 𝕍(1, ⋯, q)×det diag{c_{1}, …, c_{q}}×det 𝕍^{T} (1, ⋯, q) = ∏_{i=1}^{q} c_{i} ∏_{1≤i<j≤q}(λ_{i}−λ_{j})^{2}

Theorem 16. Let a(t) be a linear homogeneous recurrence sequence of order q, with a general-term formula: a(t) = ∑_{i=1}^{q} c_{i}λ_{i}^{t}, t ∈ ℕ_{0}. Then the Hankel transform h_{a_{(p)}}(t) of the shifted sequence a_{(p)}, (p = 0, 1, 2, …), of sequence a(t) are given by

h_{a_{(p)}}(t) = ∑_{(i_{1},i_{2},··· ,i_{t+1})} ∏_{k=1}^{t+1}(c_{i_{k}}λ_{i_{k}}^{k−1+p}) ∏_{1≤k<m≤(t+1)}(λ_{i_{k}} − λ_{i_{m}}), t = 0, 1, …, q − 1, (20)

where summarizing is over q!/(q − t − 1)! different (t + 1)-permutations (i_{1}, i_{2}, ⋯, i_{t+1}) of set {1, 2, …, q}. Particularly, the first term h_{a_{(p)}}(0) = ∑_{i=1}^{q} c_{i}λ_{i}^{p}, and the q-th (last) term h_{a_{(p)}}(q − 1) = ∏_{i=1}^{q}(c_{i}λ_{i}^{p}) ∏_{1≤i<j≤q}(λ_{i} − λ_{j})^{2}.

Proof. The general term of a_{(p)}(t) is a_{(p)}(t) = ∑_{i=1}^{q} c_{i}λ_{i}^{t+p} = ∑_{i=1}^{q} d_{i}λ_{i}^{t}, t ∈ ℕ_{0}, where d_{i} = c_{i}λ_{i}^{p} (i = 1, 2, …, q). We see from Theorem 15 that the Hankel transform h_{a_{(p)}}(t) of sequence a_{(p)} (note that it is also a recurrence sequence of order q) is

h_{a_{(p)}}(t) = ∑_{(i_{1},i_{2},··· ,i_{t+1})} ∏_{k=1}^{t+1}(d_{i_{k}}λ_{i_{k}}^{k−1}) ∏_{1≤k<m≤(t+1)}(λ_{i_{k}} − λ_{i_{m}}),

where summarizing is over q!/(q − t − 1)! different (t + 1)-permutations (i_{1}, i_{2}, ⋯, i_{t+1}) of set {1, 2, …, q}. Replacing d_{1}, d_{2}, …, d_{q} by c_{1}λ_{1}^{p}, c_{2}λ_{2}^{p}, …, c_{1}λ_{q}^{p} respectively, we obtain (20). From Theorem 15, we obtain that h_{a_{(p)}}(0) = ∑_{i=1}^{q} d_{i} = ∑_{i=1}^{q} c_{i}λ_{i}^{p}, and

h_{a_{(p)}}(q − 1) = ∏_{i=1}^{q} d_{i} ∏_{1≤i<j≤q}(λ_{i} − λ_{j})^{2} = ∏_{i=1}^{q}(c_{i}λ_{i}^{p}) ∏_{1≤i<j≤q}(λ_{i} − λ_{j})^{2}

.

Remark 17. We take the generalized Lucas sequence s(t) = 3, 1, 3, 7, 11, 21, 39, … (sequence \underline{A001644} in [3]) as an example used for verification. The third order recurrent sequence has a general term formula that s(t) = λ_{1}^{t} + λ_{2}^{t} + λ_{3}^{t} (Note that c_{1} = c_{2} = c_{3} = 1), where three characteristic values λ_{i} (i = 1, 2, 3) are the roots of algebraic equation λ^{3} − λ^{2} − λ − 1 = 0. They are that

λ_{1} = \frac{1}{3}(1 + α + β), λ_{2} = \frac{1}{3}(1 + ω_{1}α + ω_{2}β), λ_{2} = \frac{1}{3}(1 + ω_{2}α + ω_{1}β).

where two real numbers α = \sqrt[3]{19 + \sqrt{297}}, β = \sqrt[3]{19 − \sqrt{297}}; and 1, ω_{1}, ω_{2} are three complex cubic roots of 1. Hence, noting that ω_{1} + ω_{2} = −1 and ω_{1}ω_{2} = 1, we get that the Hankel transform of s(t) (and any of its multiple binomial transforms) has the three terms:

h_{s}(0) = c_{1} + c_{2} + c_{3} = 1 + 1 + 1 = 3,

h_{s}(1) = c_{1}c_{2}λ_{2}(λ_{2} −λ_{1})+c_{2}c_{1}λ_{1}(λ_{1} −λ_{2})+c_{1}c_{3}λ_{3}(λ_{3} −λ_{1})+c_{3}c_{1}λ_{1}(λ_{1} −λ_{3})+c_{2}c_{3}λ_{3}(λ_{3} −λ_{2})

+c_{3}c_{2}λ_{2}(λ_{2} − λ_{3}) = (λ_{1} − λ_{2})^{2} + (λ_{1} − λ_{3})^{2} + (λ_{2} − λ_{3})^{2} = 2αβ = 8,

h_{s}(2) = c_{1}c_{2}c_{3}(λ_{1} − λ_{2})^{2}(λ_{1} − λ_{3})^{2}(λ_{2} − λ_{3})^{2} = (λ_{1} − λ_{2})^{2}(λ_{1} − λ_{3})^{2}(λ_{2} − λ_{3})^{2}

= −\frac{1}{27}(α^{2} + β^{2} + αβ)^{2}(α − β)^{2} = −\frac{1}{27}(α^{3} + β^{3} + 16)(α^{3} + β^{3} − 16) = −44.

4 Acknowledgement

I would like to thank the referee for his/her useful suggestions and improvements.

References

[1] J.-Q. Pan, Multiple binomial transforms and families of integer sequences, J. Integer Seq., 13 (2010), Article 10.4.2.

[2] J. W. Layman, The Hankel transform and some of its properties, J. Integer Seq., 4 (2001), Article 01.1.5.

[3] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, published electronically

   at http://oeis.org/.

2010 Mathematics Subject Classification: Primary 11B65; Secondary 11B75. Keywords: shifted sequences, multiple binomial transforms, Hankel transform.

(Concerned with sequences \underline{A000045}, \underline{A000073}, \underline{A001906}, \underline{A001644}, \underline{A039834}, \underline{A093131}, and \underline{A115390}.)

Received January 22 2011; revised version received February 22 2011. Published in Journal of Integer Sequences, March 25 2011.

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