Links Between Sums Over Paths in Bernoulli’s Triangles and the Fibonacci Numbers

Denis Neiter and Amsha Proag École Polytechnique Route de Saclay 91128 Palaiseau France

                denis.neiter@polytechnique.org
                 amsha.proag@polytechnique.org

Abstract We investigate paths in Bernoulli’s triangles, and derive several relations linking

the partial sums of binomial coefficients to the Fibonacci numbers.

1 Introduction

Binomial coefficients appear in many identities, some of which are closely connected to the Fibonacci sequence [1, 2]. Pascal’s triangle has been explored for links to the Fibonacci sequence as well as to generalized sequences [3]. The partial sums of the binomial coefficients are less well known, although a number of identities have been found regarding sums of their powers [4, 5] and polynomials [6]. To add to the existing corpus, we review Bernoulli’s second and third-order triangles for relations pertaining to sums of the binomial coefficients. We contribute several relations that link the Fibonacci numbers to binomial partial sums.

2 Notation, definitions and preliminary lemma

We let ℕ^{∗} denote the positive natural numbers, i.e., {1, 2, …} and we let ℕ_{0} refer to ℕ^{∗}∪{0}. For p, q ∈ ℤ, p ⩽ q, we let ⟦p, q⟧ denote the integers between p and q, i.e., ⟦p, q⟧ = {k ∈ ℤ | p ⩽ k ⩽ q\}. For p ∈ ℤ we also let p^{+} denote \frac{1}{2} (p + |p|).

We let (F_{n})_{n∈ℕ_{0}} denote the Fibonacci numbers (\underline{A000045} [7]).

Definition 1. Let B_{n,k} denote the sum of the first k binomial coefficients, i.e.,

∀n, k ∈ ℕ_{0}, B_{n,k} = ∑_{q=0}^{k} \binom{n}{q}.

Definition 1 implies that ∀n ∈ ℕ_{0}, ∀k ⩾ 1, B_{n,k} = B_{n,k−1} + \binom{n}{k} and ∀k ⩾ n, B_{n,k} = 2^{n}.

We can therefore derive the following recurrence relation:

∀n, k ⩾ 1, B_{n,k} = ∑_{q=0}^{k} \binom{n}{q}

= ∑_{q=0}^{k} \binom{n − 1}{q} + ∑_{q=0}^{k} \binom{n − 1}{q − 1}

= ∑_{q=0}^{k} \binom{n − 1}{q} + ∑_{q=0}^{k−1} \binom{n − 1}{q},
B_{n,k} = B_{n−1,k} + B_{n−1,k−1}.  (1)

We recognize the structure of Pascal’s rule for the binomial coefficients with the boundary

values B_{n,0} = 1 and B_{n,n} = ∑_{q=0}^{n} \binom{n}{q} = 2^{n}.

Definition 2. Let Bernoulli’s triangle denote the triangle formed by (B_{n,k})_{n∈ℕ_{0},k∈⟦0,n⟧}. Bernoulli’s triangle is illustrated by Figure 1.

Definition 3. For n, k ∈ ℕ_{0}, let B_{n,k}^{[1]} = \binom{n}{k}. Then for every m ⩾ 1 we define

∀n, k ∈ ℕ_{0}, B_{n,k}^{[m+1]} = ∑_{q=0}^{k} B_{n,q}^{[m]}.

For m ⩾ 1, let Bernoulli’s mth-order triangle B^{[m]} denote the triangle formed by the family (B_{n,k}^{[m]})_{n∈ℕ_{0},k∈⟦0,n⟧}.

The first and second-order triangles B^{[1]} and B^{[2]} correspond respectively to Pascal’s triangle and Bernoulli’s triangle. For m ⩾ 1, the elements of B^{[m]} verify the following recurrence relation:

∀n ⩾ 2, ∀k ∈ ⟦1, n − 1⟧, B_{n,k}^{[m]} = B_{n−1,k}^{[m]} + B_{n−1,k−1}^{[m]}.  (2)

The proof of Relation (2) is the same as that of Relation (1). The corresponding boundary

values are ∀n ∈ ℕ_{0}, ∀m ⩾ 1, B_{n,0}^{[m]} = 1 and ∀m ⩾ 2, B_{n,n}^{[m]} = ∑_{q=0}^{n} B_{n,q}^{[m−1]}.

Definition 4. In Bernoulli’s mth-order triangle B^{[m]}, for c, l ∈ ℤ, n_{0} ∈ ℕ_{0}, k_{0} ∈ ⟦0, n_{0}⟧, let the path following direction (c, l) from (n_{0}, k_{0}) denote the sequence (B_{n_{0}+kl,k_{0}−kc}^{[m]})_{k∈ℕ_{0}} (where l > 0 corresponds to increasing row numbers from n_{0}, and c < 0 to increasing column numbers from k_{0}). For a given pair (n_{0}, k_{0}), (B_{n_{0}+kl,k_{0}−kc}^{[m]})_{k∈ℕ_{0}} contains a finite number of elements when either l ⩽ 0, or l > 0 and c ∉ ⟦−l, 0⟧.

In this work, we focus our attention on sums of elements along the two types of paths defined below. Definition 5. Let m ∈ ℕ^{∗}, n ∈ ℕ_{0}.

(i) Let S_{n}^{[m]} (c, l) denote the sum over the path following direction (c, l) (c > 0, l < 0) from (n, n), i.e.,

S_{n}^{[m]} (c, l) = ∑_{k=0}^{⌊n/c⌋} B_{n+kl,n−kc}^{[m]}.

We shall also use

S̄_{n}^{[m]} (c, l) = 2B_{n,n}^{[m]} − S_{n}^{[m]} (c, l) = B_{n,n}^{[m]} − ∑_{k=1}^{⌊n/c⌋} B_{n+kl,n−kc}^{[m]}.

k 0 1 2 3 4 5 6 7 8 9
n
0  1
1  1 2
2  1 3 4
3  1 4 7 8
4  1 5 11 15 16
5  1 6 16 26 31 32
6  1 7 22 42 57 63 64
7  1 8 29 64 99 120 127 128
8  1 9 37 93 163 219 247 255 256
9  1 10 46 130 256 382 466 502 511 512

Figure 1: Bernoulli’s triangle (\underline{A008949} [7]): the element in row n and column k corresponds to B_{n,k}.

(ii) Let T_{n}^{[m]} (c, l) denote the sum over the path following direction (c, l) (c < 0, l < 0) from

(n, 0), i.e.,

T_{n}^{[m]} (c, l) = ∑_{k=0}^{⌊−n/(c+l)⌋} B_{n+kl,−kc}^{[m]}.

Lemma 6. Let u ∈ ℤ^{ℕ_{0}} and ∀n ∈ ℕ^{∗}, v_{n} = u_{n} − 2u_{n−1}.

Then ∀n ∈ ℕ_{0}, u_{n} = 2^{n}u_{0} + ∑_{k=1}^{n} 2^{n−k}v_{k}.

Proof. From the definition of (v_{n})_{n∈ℕ^{∗}}, we have

∑_{k=1}^{n} 2^{n−k}v_{k} = ∑_{k=1}^{n} 2^{n−k}(u_{k} − 2u_{k−1})
= ∑_{k=1}^{n} 2^{n−k}u_{k} − ∑_{k=1}^{n} 2^{n−k+1}u_{k−1}
= ∑_{k=1}^{n} 2^{n−k}u_{k} − ∑_{k=0}^{n−1} 2^{n−k}u_{k}
= u_{n} − 2^{n}u_{0}.

Applying Lemma 6 to u_{n} = F_{n+2} (i.e., v_{n} = −F_{n−1}) or u_{n} = F_{2n+4} (i.e., v_{n} = F_{2n+1}) yields the following relations for n ∈ ℕ_{0}:

F_{n+2} = 2^{n} − ∑_{k=1}^{n} 2^{n−k}F_{k−1},  (3)
F_{2n+4} = 3 · 2^{n} + ∑_{k=1}^{n} 2^{n−k}F_{2k+1}.  (4)

Note that Benjamin and Quinn have reported Relation (3) [8, Id. 10].

3 Bernoulli’s second-order triangle

3.1 The path following direction (2, −1) from (n, n)

We first consider the path following direction (2, −1) from (n, n) in Bernoulli’s triangle B^{[2]}. As illustrated by Figure 2, the sequence (S̄_{n}^{[2]} (2, −1))_{n∈ℕ_{0}} can be related to the Fibonacci sequence by S̄_{n}^{[2]} (2, −1) = F_{n+2}. Given that S̄_{n}^{[2]} (2, −1) = 2B_{n,n} − S_{n}^{[2]} (2, −1), we find that S_{n}^{[2]} (2, −1) = 2^{n+1} − F_{n+2}, which we formally express as Theorem 7 below.

1  1 = 1
1 2  2 = 2
1 3 4  3 = 4 − 1
1 4 7 8  5 = 8 − 3
1 5 11 15 16  8 = 16 − 7 − 1
1 6 16 26 31 32  13 = 32 − 15 − 4
1 7 22 42 57 63 64  21 = 64 − 31 − 11 − 1
1 8 29 64 99 120 127 128  34 = 128 − 63 − 26 − 5
1 9 37 93 163 219 247 255 256  55 = 256 − 127 − 57 − 16 − 1
1 10 46 130 256 382 466 502 511 512  89 = 512 − 255 − 120 − 42 − 6

Figure 2: Four instances of the path following direction (2, −1) from (n, n) in Bernoulli’s triangle are highlighted in distinct colors. Each term of S̄_{n}^{[2]} (2, −1) corresponds to a Fibonacci number.

Theorem 7.

∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{n−2k} \binom{n − k}{q} = 2^{n+1} − F_{n+2}.  (5)

Proof. We seek to prove that

∀n ∈ ℕ_{0}, S_{n}^{[2]} (2, −1) = 2^{n+1} − F_{n+2}.

We have S_{0}^{[2]} (2, −1) = 1 = 2^{0+1} − F_{0+2}, so the formula is valid for n = 0.

Let n ∈ ℕ^{∗}. Then S_{n}^{[2]} (2, −1) − 2S_{n−1}^{[2]} (2, −1) = ∑_{k=0}^{⌊n/2⌋} B_{n−k,n−2k} − 2 ∑_{k=0}^{⌊(n−1)/2⌋} B_{n−1−k,n−1−2k}.

Assume an even n, e.g., n = 2s. Then

S_{n}^{[2]} (2, −1) − 2S_{n−1}^{[2]} (2, −1) = ∑_{k=0}^{s} B_{2s−k,2s−2k} − 2 ∑_{k=0}^{s−1} B_{2s−1−k,2s−1−2k}

= B_{s,0} + ∑_{k=0}^{s−1}(B_{2s−k,2s−2k} − 2B_{2s−1−k,2s−1−2k}).

From Relation (1) and Definition 1, we have

∀p, q ⩾ 1, B_{p,q} − 2B_{p−1,q−1} = B_{p−1,q} − B_{p−1,q−1} = \binom{p − 1}{q}.  (6)

Therefore

S_{n}^{[2]} (2, −1) − 2S_{n−1}^{[2]} (2, −1) = 1 + ∑_{k=0}^{s−1} \binom{2s − k − 1}{2s − 2k}

= 1 + ∑_{k=0}^{s−1} \binom{2s − k − 1}{k − 1}

= ∑_{k=0}^{s−1} \binom{2s − k − 2}{k}.

Given that ∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} \binom{n − k}{k} = F_{n+1} [8, Id. 4], we have

S_{n}^{[2]} (2, −1) − 2S_{n−1}^{[2]} (2, −1) = F_{(2s−2)+1} = F_{n−1}.

The above relation, illustrated by Figure 3, can be proven for odd values of n using the same method. Applying Lemma 6 to u_{n} = S_{n}^{[2]} (2, −1) and using Relation (3), we obtain

∀n ∈ ℕ_{0}, S_{n}^{[2]} (2, −1) = 2^{n} + ∑_{k=1}^{n} 2^{n−k}F_{k−1} = 2^{n+1} − F_{n+2}. 1  1 = 1

1 2  2 = 2  = 2 · 1 + F_{0}
1 3 4  5 = 4 + 1  = 2 · 2 + F_{1}
1 4 7 8  11 = 8 + 3  = 2 · 5 + F_{2}
1 5 11 15 16  24 = 16 + 7 + 1  = 2 · 11 + F_{3}
1 6 16 26 31 32  51 = 32 + 15 + 4  = 2 · 24 + F_{4}
1 7 22 42 57 63 64  107 = 64 + 31 + 11 + 1 = 2 · 51 + F_{5}
1 8 29 64 99 120 127 128 222 = 128 + 63 + 26 + 5 = 2 · 107 + F_{6}

Figure 3: Fibonacci numbers resulting from S_{n}^{[2]} (2, −1) − 2S_{n−1}^{[2]} (2, −1), forming \underline{A027934} [7].

3.2 The path following direction (c, 1 − c) from (n, n)

Definition 8. For c ⩾ 2, n ⩾ 1, let λ_{n}(c) = S_{n}^{[2]} (c, 1 − c) − 2S_{n−1}^{[2]} (c, 1 − c).

The sequence (λ_{n}(c))_{n∈ℕ^{∗}} corresponds to \underline{A000930} for c = 3 (see Figure 4), \underline{A003269} for c = 4 and \underline{A003520} for c = 5 [7]. We observe that (λ_{n}(c))_{n∈ℕ^{∗}} satisfies the following linear recurrence relation.

Theorem 9.

∀n ∈ ⟦1, c − 1⟧, λ_{n}(c) = 0; λ_{c}(c) = 1 and ∀n > c, λ_{n}(c) = λ_{n−1}(c) + λ_{n−c}(c).

The sequence (λ_{n}(c))_{n∈ℕ^{∗}} is a generalization the Fibonacci sequence, which corresponds to the c = 2 case (as shown above, ∀n ∈ ℕ^{∗}, λ_{n}(2) = F_{n−1}).

Proof. Let c ⩾ 2, n ⩾ 1. From Definition 5, we have

λ_{n}(c) = ∑_{k=0}^{⌊n/c⌋} B_{n+k(1−c),n−kc} − 2 ∑_{k=0}^{⌊(n−1)/c⌋} B_{n−1+k(1−c),n−1−kc}.

We first ascertain the initial values of (λ_{n}(c))_{n∈ℕ^{∗}}. For n < c, ⌊n/c⌋ = ⌊(n − 1)/c⌋ = 0 and λ_{n}(c) = B_{n,n} −2B_{n−1,n−1} = 2^{n} −2·2^{n−1} = 0. For n = c, λ_{c}(c) = B_{c,c} +B_{1,0} −2B_{c−1,c−1} = 2^{c} + 1 − 2 · 2^{c−1} = 1.

We then prove the recurrence relation for n > c. We have

λ_{n−1}(c) = ∑_{k=0}^{⌊(n−1)/c⌋} B_{n−1+k(1−c),n−1−kc} − 2 ∑_{k=0}^{⌊(n−2)/c⌋} B_{n−2+k(1−c),n−2−kc}

1  1 = 1
1 2  2 = 2  = 2 · 1 + 0
1 3 4  4 = 4  = 2 · 2 + 0
1 4 7 8  9 = 8 + 1  = 2 · 4 + 1
1 5 11 15 16  19 = 16 + 3  = 2 · 9 + 1
1 6 16 26 31 32  39 = 32 + 7  = 2 · 19 + 1
1 7 22 42 57 63 64  80 = 64 + 15 + 1  = 2 · 39 + 2
1 8 29 64 99 120 127 128 163 = 128 + 31 + 4 = 2 · 80 + 3

Figure 4: Instances of S_{n}^{[2]} (3, −2) in B^{[2]} form \underline{A099568} [7] and the difference S_{n}^{[2]} (3, −2) − 2S_{n−1}^{[2]} (3, −2) yields the terms of (λ_{n}(3))_{n∈ℕ^{∗}}, corresponding to \underline{A000930} [7].

and

λ_{n−c}(c) = ∑_{k=0}^{⌊(n−c)/c⌋} B_{n−c+k(1−c),n−c−kc} − 2 ∑_{k=0}^{⌊(n−c−1)/c⌋} B_{n−c−1+k(1−c),n−c−1−kc}.

The summation upper bounds of λ_{n}(c), λ_{n−1}(c) and λ_{n−c}(c) depend on the remainder of the Euclidean division of n by c. Let n = cq + r > c with q = ⌊n/c⌋ ∈ ℕ^{∗}, r ∈ ⟦0, c − 1⟧. There are three cases to consider, which are shown in Table 1.

⌊n/c⌋ ⌊(n − 1)/c⌋ ⌊(n − 2)/c⌋ ⌊(n − c)/c⌋ ⌊(n − c − 1)/c⌋
r = 0  q  q − 1  q − 1  q − 1  q − 2
r = 1  q  q  q − 1  q − 1  q − 1
1 < r < c q  q  q  q − 1  q − 1

Table 1: Summation upper bounds for the various values of r = n − c ⌊n/c⌋.

For r = 0, we have

λ_{n}(c) = ∑_{k=0}^{q} B_{cq+k(1−c),cq−kc} − 2 ∑_{k=0}^{q−1} B_{cq−1+k(1−c),cq−1−kc}

= B_{cq+q(1−c),cq−qc} + ∑_{k=0}^{q−1} (B_{cq+k(1−c),cq−kc} − 2B_{cq−1+k(1−c),cq−1−kc}) ,

and Relation (6) yields

λ_{n}(c) = B_{q,0} + ∑_{k=0}^{q−1} \binom{cq − 1 + k(1 − c)}{cq − kc}

= 1 + ∑_{k=0}^{q−1} \binom{cq − 1 + k(1 − c)}{k − 1}

= 1 + ∑_{k=1}^{q−1} \binom{cq − 1 + k(1 − c)}{k − 1}.

Similarly,

λ_{n−1}(c) = ∑_{k=0}^{q−1} B_{cq−1+k(1−c),cq−1−kc} − 2 ∑_{k=0}^{q−1} B_{cq−2+k(1−c),cq−2−kc}

= ∑_{k=0}^{q−1} \binom{cq − 2 + k(1 − c)}{k − 1},

and

λ_{n−c}(c) = ∑_{k=0}^{q−1} B_{cq−c+k(1−c),cq−c−kc} − 2 ∑_{k=0}^{q−2} B_{cq−c−1+k(1−c),cq−c−1−kc}

= \underbrace{B_{cq−c+(q−1)(1−c),cq−c−(q−1)c}}_{B_{q−1,0}} + ∑_{k=0}^{q−2}(\underbrace{B_{cq−c+k(1−c),cq−c−kc} − 2B_{cq−c−1+k(1−c),cq−c−1−kc}}_{\binom{cq−c−1+k(1−c)}{cq−c−kc}})

= 1 + ∑_{k=0}^{q−2} \binom{cq − c − 1 + k(1 − c)}{k − 1}

= 1 + ∑_{k=1}^{q−1} \binom{cq − 2 + k(1 − c)}{k − 2}.

Therefore we have

λ_{n−1}(c) + λ_{n−c}(c) = 0 + ∑_{k=1}^{q−1} \binom{cq − 2 + k(1 − c)}{k − 1} + ∑_{k=1}^{q−1} \binom{cq − 2 + k(1 − c)}{k − 2} + 1

= ∑_{k=1}^{q−1} \binom{cq − 1 + k(1 − c)}{k − 1} + 1
= λ_{n}(c).

This proves Theorem 9 for the r = 0 case. We can use the same method for r = 1 or r > 1, since only the summation upper bounds are modified in those cases.

We can write the explicit expression of λ_{n}(c) as λ_{n}(c) = ∑_{i=0}^{⌊(n−c)/(c−1)⌋} \binom{n − c + i(1 − c)}{i}.

We obtain, using Lemma 6, S_{n}^{[2]}(c, 1 − c) = 2^{n} + ∑_{k=1}^{n} 2^{n−k}λ_{k}(c), which leads to the following

relation.

Corollary 10.

∑_{k=0}^{⌊n/c⌋} ∑_{q=0}^{n−kc} \binom{n − (c − 1)k}{q} = 2^{n} + ∑_{k=1}^{n} 2^{n−k} ∑_{i=0}^{⌊(k−c)/(c−1)⌋} \binom{k − c − (c − 1)i}{i}.

3.3 The path following direction (−1, −1) from (n, 0)

The path following direction (−1, −1) from (n, 0) has a connection to the Fibonacci sequence that appears in the difference between successive terms of T_{n}^{[2]}(−1, −1), as illustrated by Figure 5. Not only do we notice that

∀p ⩾ 1, T_{2p}^{[2]}(−1, −1) = T_{2p−1}^{[2]}(−1, −1) + F_{2p+1},

but also that

∀p ⩾ 1, T_{2p+1}^{[2]}(−1, −1) = T_{2p}^{[2]}(−1, −1) + T_{2p−1}^{[2]}(−1, −1).

Proof. To prove these two recurrence relations, we rely on Relation (1). For odd indices, we have

∀p ⩾ 1, T_{2p+1}^{[2]}(−1, −1) = ∑_{k=0}^{p} B_{2p+1−k,k}

= B_{2p+1,0} + ∑_{k=1}^{p} (B_{2p−k,k−1} + B_{2p−k,k})

= B_{2p,0} + ∑_{k=1}^{p} B_{2p−k,k} + ∑_{k=0}^{p−1} B_{2p−1−k,k}

= ∑_{k=0}^{p} B_{2p−k,k} + ∑_{k=0}^{p−1} B_{2p−1−k,k}
= T_{2p}^{[2]}(−1, −1) + T_{2p−1}^{[2]}(−1, −1).

For even indices,

∀p ⩾ 1, T_{2p}^{[2]}(−1, −1) = ∑_{k=0}^{p} B_{2p−k,k}

= ∑_{k=0}^{p} ∑_{q=0}^{k} \binom{2p − k}{q}

= ∑_{k=0}^{p} (∑_{q=0}^{k−1} \binom{2p − k}{q} + \binom{2p − k}{k})

= ∑_{k=0}^{p−1} ∑_{q=0}^{k} \binom{2p − 1 − k}{q} + ∑_{k=0}^{p} \binom{2p − k}{k}.

Since ∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} \binom{n − k}{k} = F_{n+1} [8, Id. 4], we have ∑_{k=0}^{p} \binom{2p − k}{k} = F_{2p+1}. Hence,

∀p ⩾ 1, T_{2p}^{[2]}(−1, −1) = T_{2p−1}^{[2]}(−1, −1) + F_{2p+1}.

These two relations make it possible to derive a recurrence relation that pertains only to the odd subsequence:

∀p ⩾ 1, T_{2p+1}^{[2]}(−1, −1) − 2T_{2p−1}^{[2]}(−1, −1) = F_{2p+1}.

1  1 = 1
1 2  1 = 1
1 3 4  3 = 1 + 2  = 1 + F_{3}
1 4 7 8  4 = 1 + 3  = 3 + 1
1 5 11 15 16  9 = 1 + 4 + 4  = 4 + F_{5}
1 6 16 26 31 32  13 = 1 + 5 + 7  = 9 + 4
1 7 22 42 57 63 64  26 = 1 + 6 + 11 + 8  = 13 + F_{7}
1 8 29 64 99 120 127 128  39 = 1 + 7 + 16 + 15  = 26 + 13
1 9 37 93 163 219 247 255 256 73 = 1 + 8 + 22 + 26 + 16 = 39 + F_{9}

Figure 5: Instances of T_{n}^{[2]} (−1, −1) in Bernoulli’s triangle and differences from the preceding terms.

Therefore, using Lemma 6 and Relation (4), we have

∀p ⩾ 1, T_{2p+1}^{[2]}(−1, −1) = 2^{p} \underbrace{T_{1}^{[2]}(−1, −1)}_{1} + ∑_{k=1}^{p} 2^{p−k}F_{2k+1}
= 2^{p} + F_{2p+4} − 3 · 2^{p}
= F_{2p+4} − 2^{p+1}.

We feed back the odd subsequence into the expression for the even subsequence as follows:

∀p ⩾ 1, T_{2p}^{[2]}(−1, −1) = T_{2p−1}^{[2]}(−1, −1) + F_{2p+1}
= F_{2p+2} − 2^{p} + F_{2p+1}
= F_{2p+3} − 2^{p}.

We have thus proven that ∀n ∈ ℕ_{0}, T_{n}^{[2]}(−1, −1) = F_{n+3} − 2^{⌊(n+1)/2⌋}, which we can rewrite as the following theorem. Theorem 11.

∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{k} \binom{n − k}{q} = F_{n+3} − 2^{⌊(n+1)/2⌋}.

Remark 12. Among the paths of Bernoulli’s triangle following other directions, one can find other sequences that follow linear recurrence relations. For example, (S̄_{n}^{[2]}(3, −1))_{n∈ℕ_{0}} corresponds to \underline{A005251} [7] and (S̄_{n}^{[2]}(4, −1))_{n∈ℕ_{0}} to \underline{A138653} [7]. We believe that further sequences related to partial sums of binomial coefficients are yet to be uncovered in B^{[2]}.

4 Bernoulli’s third-order triangle

We now consider Bernoulli’s third-order triangle B^{[3]} (see Figure 6). From Definition 3,

∀n ∈ ℕ_{0}, k ∈ ⟦0, n⟧, B_{n,k}^{[3]} = ∑_{q=0}^{k} B_{n,q} = ∑_{q=0}^{k} ∑_{r=0}^{q} \binom{n}{r}.

Recall that B_{n,0}^{[3]} = 1 and ∀n ⩾ 2, ∀k ∈ ⟦1, n − 1⟧, B_{n,k}^{[3]} = B_{n−1,k}^{[3]} + B_{n−1,k−1}^{[3]}. Furthermore, B_{n,n}^{[3]} = (n + 2)2^{n−1}.

k 0 1 2 3 4 5  6  7  8  9
n
0  1
1  1 3
2  1 4 8
3  1 5 12 20
4  1 6 17 32 48
5  1 7 23 49 80 112
6  1 8 30 72 129 192 256
7  1 9 38 102 201 321 448 576
8  1 10 47 140 303 522 769 1024 1280
9  1 11 57 187 443 825 1291 1793 2304 2816

Figure 6: Bernoulli’s third-order triangle (\underline{A193605} [7]): the element in row n and column k corresponds to B_{n,k}^{[3]}.

4.1 The path following direction (2, −1) from (n, n)

Definition 5 states that S̄_{n}^{[3]} (2, −1) = B_{n,n}^{[3]} − ∑_{k=1}^{⌊n/2⌋} B_{n−k,n−2k}^{[3]}. Figure 7 suggests the following recurrence relation:

S̄_{n}^{[3]} (2, −1) − 2S̄_{n−1}^{[3]} (2, −1) = F_{n}.  (7)

Proof. Let n ∈ ℕ^{∗}. We have

S̄_{n}^{[3]} (2, −1) − 2S̄_{n−1}^{[3]} (2, −1) = B_{n,n}^{[3]} − ∑_{k=1}^{⌊n/2⌋} B_{n−k,n−2k}^{[3]} − 2B_{n−1,n−1}^{[3]} + 2 ∑_{k=1}^{⌊(n−1)/2⌋} B_{n−1−k,n−1−2k}^{[3]}.

1  1 = 1
1 3  3 = 3  = 2 · 1 + F_{1}
1 4 8  7 = 8 − 1  = 2 · 3 + F_{2}
1 5 12 20  16 = 20 − 4  = 2 · 7 + F_{3}
1 6 17 32 48  35 = 48 − 12 − 1  = 2 · 16 + F_{4}
1 7 23 49 80 112  75 = 112 − 32 − 5  = 2 · 35 + F_{5}
1 8 30 72 129 192 256  158 = 256 − 80 − 17 − 1  = 2 · 75 + F_{6}
1 9 38 102 201 321 448 576  329 = 576 − 192 − 49 − 6  = 2 · 158 + F_{7}

Figure 7: Instances of S̄_{n}^{[3]} (2, −1) in B^{[3]} that illustrate Relation (7).

Firstly, B_{n,n}^{[3]} − 2B_{n−1,n−1}^{[3]} = (n + 2)2^{n−1} − 2(n + 1)2^{n−2} = 2^{n−1}. Let us assume that n is even, e.g., n = 2s (the method is the same for odd values of n).

We have

S̄_{n}^{[3]} (2, −1) − 2S̄_{n−1}^{[3]} (2, −1) = 2^{2s−1} − ∑_{k=1}^{s} B_{2s−k,2s−2k}^{[3]} + 2 ∑_{k=1}^{s−1} B_{2s−1−k,2s−1−2k}^{[3]}

= 2^{2s−1} − \underbrace{B_{s,0}^{[3]}}_{1} − ∑_{k=1}^{s−1} (B_{2s−k,2s−2k}^{[3]} − 2B_{2s−1−k,2s−1−2k}^{[3]}) .

From Definition 3 and Relation (2), we have, for m ⩾ 2,

∀p, q ⩾ 1, B_{p,q}^{[m]} − 2B_{p−1,q−1}^{[m]} = B_{p−1,q}^{[m]} − B_{p−1,q−1}^{[m]} = B_{p−1,q}^{[m−1]},

hence,

S̄_{n}^{[3]} (2, −1) − 2S̄_{n−1}^{[3]} (2, −1) = 2^{2s−1} − 1 − ∑_{k=1}^{s−1} B_{2s−1−k,2s−2k}.

Moreover,

∑_{k=1}^{s−1} B_{2s−1−k,2s−2k} = ∑_{k=2}^{s} B_{2s−k,2s−2k+2}
= ∑_{k=2}^{s} (B_{2s−k,2s−2k} + \binom{2s − k}{2s − 2k + 1} + \binom{2s − k}{2s − 2k + 2})
= ∑_{k=2}^{s} B_{2s−k,2s−2k} + ∑_{k=2}^{s} \binom{2s − k + 1}{2s − 2k + 2}
= S_{2s}^{[2]} (2, −1) − \underbrace{B_{2s,2s}}_{2^{2s}} −B_{2s−1,2s−2} + ∑_{k=2}^{s} \binom{2s − k + 1}{k − 1}.

Theorem 7 states that S_{2s}^{[2]} (2, −1) = 2^{2s+1} − F_{2s+2}.

We also have B_{2s−1,2s−2} = B_{2s−1,2s−1} − \binom{2s − 1}{2s − 1} = 2^{2s−1} − 1. Furthermore,

∑_{k=2}^{s} \binom{2s + 1 − k}{k − 1} = ∑_{k=1}^{s} \binom{2s − k + 1}{k − 1} − \binom{2s}{0}
= ∑_{k=0}^{s−1} \binom{2s − k}{k} − 1
= ∑_{k=0}^{s} \binom{2s − k}{k} − \binom{s}{s} − 1
= F_{2s+1} − 2.

Finally,

S̄_{n}^{[3]} (2, −1) − 2S̄_{n−1}^{[3]} (2, −1) = 2^{2s−1} − 1 − (2^{2s+1} − F_{2s+2} − 2^{2s} − (2^{2s−1} − 1) + F_{2s+1} − 2)
= F_{2s+2} − F_{2s+1}
= F_{2s}
= F_{n}.

Lemma 6 and Relation (3) lead to an explicit expression of S̄_{n}^{[3]} (2, −1), as follows:

S̄_{n}^{[3]} (2, −1) = 2^{n}S̄_{0}^{[3]} (2, −1) + ∑_{k=1}^{n} 2^{n−k}F_{k}

= 2^{n} + ∑_{k=1}^{n+1} 2^{n+1−k}F_{k−1}
= 2^{n} + (2^{n+1} − F_{n+3})
= 3 · 2^{n} − F_{n+3}.

From the explicit form of S̄_{n}^{[3]} (2, −1), we deduce that S_{n}^{[3]} (2, −1) = 2B_{n,n}^{[3]} + F_{n+3} − 3 · 2^{n} and obtain the following theorem.

Theorem 13.

∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{n−2k} ∑_{r=0}^{q} \binom{n − k}{r} = F_{n+3} + (n − 1)2^{n}.

4.2 The path following direction (−1, −1) from (n, 0)

In B^{[3]}, the path following direction (−1, −1) from (n, 0) does not appear at first glance to have any obvious connection to the Fibonacci sequence; however, the differences of consecutive terms of T_{n}^{[3]}(−1, −1) follow a pattern similar to T_{n}^{[2]}(−1, −1). Indeed, the odd subsequence of the difference sequence has the same behavior as in B^{[2]}, while the even subsequence has a connection to the T_{n}^{[2]}(−1, −1) numbers of even n (see Figure 8).

More precisely, we notice that ∀p ⩾ 1, T_{2p}^{[3]}(−1, −1) = T_{2p−1}^{[3]}(−1, −1) + T_{2p}^{[2]}(−1, −1) and T_{2p+1}^{[3]}(−1, −1) = T_{2p}^{[3]}(−1, −1) + T_{2p−1}^{[3]}(−1, −1).

Proof. We prove these two relations directly at order m, using Relation (2) and proceeding as in the derivation of Theorem 11 (section 3.3). For odd terms, we have

∀m ⩾ 2, ∀p ⩾ 1, T_{2p+1}^{[m]}(−1, −1) = ∑_{k=0}^{p} B_{2p+1−k,k}^{[m]}

= B_{2p+1,0}^{[m]} + ∑_{k=1}^{p} (B_{2p−k,k−1}^{[m]} + B_{2p−k,k}^{[m]})

= B_{2p,0}^{[m]} + ∑_{k=1}^{p} B_{2p−k,k}^{[m]} + ∑_{k=0}^{p−1} B_{2p−1−k,k}^{[m]}

= ∑_{k=0}^{p} B_{2p−k,k}^{[m]} + ∑_{k=0}^{p−1} B_{2p−1−k,k}^{[m]},
T_{2p+1}^{[m]}(−1, −1) = T_{2p}^{[m]}(−1, −1) + T_{2p−1}^{[m]}(−1, −1).  (8)

1  1 = 1
1 3  1 = 1
1 4 8  4 = 1 + 3  = 1 + T_{2}^{[2]}(−1, −1)
1 5 12 20  5 = 1 + 4  = 4 + 1
1 6 17 32 48  14 = 1 + 5 + 8  = 5 + T_{4}^{[2]}(−1, −1)
1 7 23 49 80 112  19 = 1 + 6 + 12  = 14 + 5
1 8 30 72 129 192 256  45 = 1 + 7 + 17 + 20 = 19 + T_{6}^{[2]}(−1, −1)
1 9 38 102 201 321 448 576 64 = 1 + 8 + 23 + 32 = 45 + 19

Figure 8: Differences between successive terms of T_{n}^{[3]} (−1, −1) in Bernoulli’s third-order triangle.

For even terms,

∀m ⩾ 2, ∀p ⩾ 1, T_{2p}^{[m]}(−1, −1) = ∑_{k=0}^{p} B_{2p−k,k}^{[m]}
= ∑_{k=0}^{p} ∑_{q=0}^{k} B_{2p−k,q}^{[m−1]}
= ∑_{k=0}^{p} (∑_{q=0}^{k−1} B_{2p−k,q}^{[m−1]} + B_{2p−k,k}^{[m−1]})
= ∑_{k=0}^{p−1} ∑_{q=0}^{k} B_{2p−1−k,q}^{[m−1]} + ∑_{k=0}^{p} B_{2p−k,k}^{[m−1]}
= ∑_{k=0}^{p−1} B_{2p−1−k,k}^{[m]} + T_{2p}^{[m−1]}(−1, −1),
T_{2p}^{[m]}(−1, −1) = T_{2p−1}^{[m]}(−1, −1) + T_{2p}^{[m−1]}(−1, −1).  (9)

From Relations (8) and (9), we derive a general expression of T_{n}^{[3]}(−1, −1) for n ∈ ℕ_{0}. Isolating the odd subsequence leads to

∀p ⩾ 1, T_{2p+1}^{[3]}(−1, −1) − 2T_{2p−1}^{[3]}(−1, −1) = T_{2p}^{[2]}(−1, −1).

Therefore, using Lemma 6 and Relation (4), we obtain

∀p ⩾ 1, T_{2p+1}^{[3]}(−1, −1) = 2^{p} \underbrace{T_{1}^{[3]}(−1, −1)}_{1} + ∑_{k=1}^{p} 2^{p−k} \underbrace{T_{2k}^{[2]}(−1, −1)}_{F_{2k+3}−2^{k}}
= 2^{p} + ∑_{k=1}^{p} 2^{p−k}F_{2k+3} − ∑_{k=1}^{p} 2^{p}
= 2^{p} + F_{2p+6} − 2^{p+3} − p2^{p}
= F_{2p+6} − (p + 7)2^{p}.

Hence,

∀p ⩾ 1, T_{2p}^{[3]}(−1, −1) = T_{2p−1}^{[3]}(−1, −1) + T_{2p}^{[2]}(−1, −1)
= F_{2p+4} − (p + 6)2^{p−1} + F_{2p+3} − 2^{p}
= F_{2p+5} − (p + 8)2^{p−1}.

Finally, we reach the following relation:

∀n ∈ ℕ_{0}, T_{n}^{[3]}(−1, −1) = F_{n+5} − 2^{⌊(n+1)/2⌋} (\frac{1}{2} ⌊(n + 1)/2⌋ + \frac{7}{2} + \frac{(−1)^{n}}{2} ) .

Theorem 14.

∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{k} ∑_{r=0}^{q} \binom{n − k}{r} = F_{n+5} − 2^{p} (\frac{1}{2}p + \frac{7}{2} + \frac{(−1)^{n}}{2} ) ,

where p = ⌊(n + 1)/2⌋. Remark 15. Paths in B^{[3]} following other directions yield interesting sequences in a similar manner to those in B^{[2]}. For example, (S̄_{n}^{[3]}(3, −1) − 2S̄_{n−1}^{[3]}(3, −1))_{n∈ℕ^{∗}} corresponds to \underline{A005314} [7]. Further work should be able to uncover additional sequences by following other paths in B^{[3]}.

5 Higher-order triangles

5.1 The path following direction (−1, −1) from (n, 0)

Using Relations (8) and (9) for m = 4 and m = 5, we proceed as we did for T_{n}^{[3]}(−1, −1) to determine the sums over the path following direction (−1, −1) from (n, 0) in B^{[4]} and B^{[5]}.

We obtain the following identities:

∀n ∈ ℕ_{0}, ∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{k} ∑_{r=0}^{q} ∑_{s=0}^{r} \binom{n − k}{s} = F_{n+7} − 2^{p} (\frac{1}{8}p^{2} + \frac{17}{8}p + 10 + (−1)^{n} (\frac{1}{4}p + 2)) ,

and ∑_{k=0}^{⌊n/2⌋} ∑_{q=0}^{k} ∑_{r=0}^{q} ∑_{s=0}^{r} ∑_{t=0}^{s} \binom{n − k}{t} =

F_{n+9} − 2^{p} (\frac{1}{48}p^{3} + \frac{5}{8}p^{2} + \frac{317}{48}p + 27 + (−1)^{n} (\frac{1}{16}p^{2} + \frac{19}{16}p + 6)) ,

where p = ⌊\frac{n+1}{2}⌋.

The expressions of T_{n}^{[m]}(−1, −1) for m ∈ ⟦2, 5⟧ contain a Fibonacci term and a second term composed of the product of a power of 2 by a polynomial, which suggests a general connection between T_{n}^{[m]}(−1, −1) and the Fibonacci sequence. We write a general expression of T_{n}^{[m]}(−1, −1) in the following theorem.

Theorem 16.

∀m ∈ ℕ^{∗}, ∀n ∈ ℕ_{0}, ∑_{i_{1}=0}^{⌊n/2⌋} ∑_{i_{2}=0}^{i_{1}} ⋯ ∑_{i_{m}=0}^{i_{m−1}} \binom{n − i_{1}}{i_{m}} = F_{n+2m−1} − 2^{p} (Q^{[m]} (p) + (−1)^{n}R^{[m]} (p)) ,

where p = ⌊\frac{n+1}{2}⌋, Q^{[m]} and R^{[m]} are polynomials with coefficients in ℚ and of degree (m−2)^{+} and (m − 3)^{+} respectively.

Proof. We seek to prove that ∀m ⩾ 1,

∀n ∈ ℕ_{0}, T_{n}^{[m]}(−1, −1) = F_{n+2m−1} − 2^{p} (Q^{[m]} (p) + (−1)^{n}R^{[m]} (p)) ,

where p = ⌊\frac{n+1}{2}⌋, Q^{[m]} and R^{[m]} are polynomials with coefficients in ℚ and of degree (m−2)^{+} and (m − 3)^{+} respectively.

We proceed by induction on m. For m = 1, we have [8, Id. 4]

∀n ∈ ℕ_{0}, T_{n}^{[1]}(−1, −1) = ∑_{k=0}^{⌊n/2⌋} \binom{n − k}{k} = F_{n+1},

which is in line with the above formulation with Q^{[1]} = R^{[1]} = 0.

Similarly, we already have the result for m = 2 with Q^{[2]} = 1 and R^{[2]} = 0 from Theorem 11:

∀n ∈ ℕ_{0}, T_{n}^{[2]}(−1, −1) = F_{n+3} − 2^{⌊(n+1)/2⌋}.

Assume that the result is true for a given m ⩾ 2, i.e., there exist two polynomials Q^{[m]} and R^{[m]} with coefficients in ℚ and respectively of degree (m − 2)^{+} and (m − 3)^{+}, such that ∀n ∈ ℕ_{0}, T_{n}^{[m]}(−1, −1) = F_{n+2m−1} − 2^{p} (Q^{[m]} (p) + (−1)^{n}R^{[m]} (p)), where p = ⌊\frac{n+1}{2}⌋. We will now prove the result for m + 1.

First, from Relations (8) and (9) at order m + 1, we have

∀s ⩾ 1, T_{2s+1}^{[m+1]}(−1, −1) = 2T_{2s−1}^{[m+1]}(−1, −1) + T_{2s}^{[m]}(−1, −1).

Using Lemma 6 applied to u_{s} = T_{2s+1}^{[m+1]}(−1, −1) (i.e., v_{s} = T_{2s}^{[m]}(−1, −1)), we obtain

∀s ⩾ 1, T_{2s+1}^{[m+1]}(−1, −1) = 2^{s} \underbrace{T_{1}^{[m+1]}(−1, −1)}_{1} + ∑_{k=1}^{s} 2^{s−k}T_{2k}^{[m]}(−1, −1).

Since T_{1}^{[m]}(−1, −1) = 1, we may write that

∀s ⩾ 1, T_{2s−1}^{[m+1]}(−1, −1) = 2^{s−1} + ∑_{k=1}^{s−1} 2^{s−1−k}T_{2k}^{[m]}(−1, −1).

Recall that ∀k ∈ ℕ_{0}, T_{2k}^{[m]}(−1, −1) = F_{2k+2m−1}−2^{k} (Q^{[m]}(k) + R^{[m]}(k)). Lemma 6 applied

to u_{s} = F_{2s+2m+2} (i.e., v_{s} = F_{2s+2m−1}) yields ∑_{k=1}^{s} 2^{s−k}F_{2k+2m−1} = F_{2s+2m+2} − 2^{s}F_{2m+2}, hence,

∑_{k=1}^{s−1} 2^{s−1−k}F_{2k+2m−1} = \frac{1}{2} (F_{2s+2m+2} − 2^{s}F_{2m+2} − F_{2s+2m−1}) = F_{2s+2m} − 2^{s−1}F_{2m+2}.

Thus, we have

∀s ⩾ 1, T_{2s−1}^{[m+1]}(−1, −1) = 2^{s−1} + ∑_{k=1}^{s−1} 2^{s−1−k} (F_{2k+2m−1} − 2^{k} (Q^{[m]}(k) + R^{[m]}(k)))

= 2^{s−1} + F_{2s+2m} − 2^{s−1}F_{2m+2} − 2^{s−1} ∑_{k=1}^{s−1} (Q^{[m]}(k) + R^{[m]}(k)) ,
T_{2s−1}^{[m+1]}(−1, −1) = F_{(2s−1)+2(m+1)−1} − 2^{s−1} (F_{2m+2} − 1 + A(s)) ,  (10)

where A(X) = ∑_{k=1}^{X−1} (Q^{[m]}(k) + R^{[m]}(k)). For any α ∈ ℕ_{0}, the polynomial ∑_{k=1}^{X−1} k^{α} is of degree

α + 1 in X. Therefore, since Q^{[m]} + R^{[m]} is of degree (m − 2)^{+} and m ⩾ 2, A is of degree (m − 1)^{+}.

Relation (10) gives the expression of T_{n}^{[m+1]}(−1, −1) for all odd n. We may thus retrieve the even subsequence from Relation (9) to obtain, for s ⩾ 1,

T_{2s}^{[m+1]}(−1, −1) = T_{2s−1}^{[m+1]}(−1, −1) + T_{2s}^{[m]}(−1, −1)
= F_{(2s−1)+2(m+1)−1} − 2^{s−1} (F_{2m+2} − 1 + A(s))
+F_{2s+2m−1} − 2^{s} (Q^{[m]}(s) + R^{[m]}(s)) ,
T_{2s}^{[m+1]}(−1, −1) = F_{2s+2(m+1)−1} − 2^{s−1} (F_{2m+2} − 1 + A(s) + 2Q^{[m]}(s) + 2R^{[m]}(s)) .(11)

We introduce the polynomials Q^{[m+1]}, R^{[m+1]} as follows:

Q^{[m+1]} = \frac{1}{2} (A + F_{2m+2} − 1 + Q^{[m]} + R^{[m]}) and R^{[m+1]} = \frac{1}{2} (Q^{[m]} + R^{[m]}) .

It is clear that Q^{[m+1]}, R^{[m+1]} ∈ ℚ[X]. In addition, Q^{[m+1]} is of the same degree as A, i.e., (m − 1)^{+} and R^{[m+1]} is of the same degree as Q^{[m]}, i.e., (m − 2)^{+}.

Therefore, we can synthesize Relations (10) and (11) as follows:

T_{n}^{[m+1]} = F_{n+2(m+1)−1} − 2^{p} (Q^{[m+1]}(p) + (−1)^{n}R^{[m+1]}(p)) ,

where p = ⌊\frac{n+1}{2}⌋, Q^{[m+1]} and R^{[m+1]} are respectively ((m + 1) − 2)^{+}th and ((m + 1) − 3)^{+}th degree polynomials with coefficients in ℚ. This concludes the proof by induction.

5.2 Towards additional formulae

The present work has reviewed Bernoulli’s second and third-order triangles and found multiple connections to the Fibonacci sequence, expressed in Theorems 7 and 11 to 14. Theorem 16 generalizes Theorems 11 to 14 to higher orders. Furthermore, Theorem 9 uncovers a relation between partial sums of binomial coefficients in Bernoulli’s triangle and sequences satisfying an additive recurrence relation that generalizes the Fibonacci sequence.

The methodology followed in this article (i.e., the study of paths over Bernoulli’s triangles) reveals sequences that are formed by the partial sums of binomial coefficients. More generally, these paths could yield further fruit by mapping various functions f over the elements

of Bernoulli’s triangles in order to study the corresponding sequences ∑_{k} f(B_{n_{0}+kl,k_{0}−kc}^{[m]}).

6 Acknowledgments

We thank Olivier Bordellès and Harry Robertson for careful proofreading. We are grateful to the anonymous referee for numerous suggestions. Denis Neiter would also like to thank Na Wang and Richard André-Jeannin for their support during the writing of this article.

References

[1] M. K. Azarian, Fibonacci identities as binomial sums, Int. J. Contemp. Math. Sciences 7 (2012), 1871–1876.

[2] M. K. Azarian, Identities involving Lucas or Fibonacci and Lucas numbers as binomial sums, Int. J. Contemp. Math. Sciences 7 (2012), 2221–2227.

[3] V. E. Hoggatt, Jr., Convolution triangles for generalized Fibonacci numbers, Fibonacci Quart. 8 (1970) 158–171.

[4] N. J. Calkin, A curious binomial identity, Discrete Math. 131 (1994), 335–337.

[5] M. Hirschhorn, Calkin’s binomial identity, Discrete Math. 159 (1996), 273–278.

[6] B. He, Some identities involving the partial sum of q-binomial coefficients, Electron. J. Combin. 21 (2014), # P3.17.

[7] N. J. A. Sloane (ed.), The Online Encyclopedia of Integer Sequences, published electronically at http://oeis.org, 2016.

[8] A. T. Benjamin and J. J. Quinn, Proofs That Really Count, Mathematical Association of America, 2003.

2010 Mathematics Subject Classification: Primary 11B39; Secondary 05A19. Keywords: Fibonacci number, binomial coefficient, Pascal’s triangle, Bernoulli’s triangle.

(Concerned with sequences \underline{A000045}, \underline{A000930}, \underline{A003269}, \underline{A003520}, \underline{A005251}, \underline{A005314}, \underline{A008949}, \underline{A027934}, \underline{A099568}, \underline{A138653}, and \underline{A193605}.)

Received June 10 2016; revised version received September 23 2016; October 5 2016. Published in Journal of Integer Sequences, October 10 2016.

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