Padovan-like sequences and Bell polynomials

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Padovan-like sequences and Bell polynomials N. Gogin a , A. Myll¨ari b,∗ a Abo ˚ Akademi University, Finland b St. George’s University, Grenada

Received 14 February 2014; received in revised form 9 May 2015; accepted 13 August 2015

Abstract We study a class of Padovan-like sequences that can be generated using special matrices of the third order. We show that terms of any sequence of this class can be expressed via Bell polynomials and their derivatives using as arguments terms of another such sequence with smaller indices. Computer algebra system (CAS) Mathematica was used for cumbersome calculations and hypothesis-testing. c 2015 Published by Elsevier B.V. on behalf of International Association for Mathematics and Computers in Simulation (IMACS). ⃝

Keywords: Integer sequences; Padovan sequence; Fibonacci sequence; Partial Bell polynomials; Padovan polynomials

1. Introduction Integer sequences appear in many branches of science. The Fibonacci and Padovan sequences have long been used in purely theoretical problems of mathematics (such as in the famous solution of 10th Hilbert problem by Yu.V. Matiyasevich [10]) as well as in many applications (see. e.g. Wikipedia page on Fibonacci heap [7]), and purely technical disciplines, such as architecture [11]. Recently, a new and elegant result of Viswanath [12] related to the properties of random Fibonacci numbers, stimulated the interest of researchers to the properties of these sequences in terms of the behavior of stochastic dynamical systems [2,4]. On the other hand, continues the study of algebraic relations of these sequences with combinatorial polynomials (Fibonacci, Padovan, Chebyshev, Kravchuk) naturally associated with them. Fibonacci numbers are known for more than two thousand years, Padovan numbers are much younger—they were introduced only recently [9]. Below, we will study Padovan-like sequences that can be generated using special matrices of the third order. We will find expressions for terms of one sequence in terms of the other sequence via Bell polynomials (also known in the combinatorial theory as “partition polynomials”). So-called matrix method of generation of such sequences and their various generalizations is widely used in many of the problems mentioned above. As an example one can mention recent papers [3,14,15], in which this method was used to study various algebraic properties (generating function, generalized Binet formula, sums, etc.) of these sequences. In contrast to this papers, also using the (different) matrix method, we establish a purely algebraic ∗ Corresponding author.

E-mail address: [email protected] (A. Myll¨ari). http://dx.doi.org/10.1016/j.matcom.2015.08.008 c 2015 Published by Elsevier B.V. on behalf of International Association for Mathematics and Computers in Simulation (IMACS). 0378-4754/⃝

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(polynomial) relationship between terms of various sequences. This allows us to construct a polynomial algorithm that “recalculates” terms of any of the sequences of this family to another sequence of the same family using CAS. CAS Mathematica was used for cumbersome calculations and hypothesis testing. Everywhere in what follows all matrices have dimension 3 × 3 and are being denoted by uppercase letters except the matrix   0 1 0 e = 0 0 0 0 0 0 that plays a special role in our considerations. We write [X ]i j to denote the (i, j)-entry, 1 ≤ i, j ≤ 3 in a matrix X . The following formulas for matrix e and any matrices X, Y, X 1 , . . . , X s are obvious: e2 = 0;

[X e]22 = [X ]21 ; [e]11 = [X e]11 = 0; [X eY ]11 = [X ]11 · [Y ]21 ;   s  [eX 1 eX 2 e . . . eX s e]11 = 0. [X r ]21 · e; eX 1 eX 2 e . . . eX s e =

(1) (2)

r =1

Furthermore, a (strong) composition of an integer n into k parts, is a way of writing n as the ordered sum of k strictly positive integers [1,6], but if zero terms are still allowed the composition is said to be weak. The numbers of these compositions (respectively) are:     n−1 n+k−1 Comp(n, k) = , Compw(n, k) = (3) k−1 k−1 and these numbers are set to be 0 for n = 0. A partition of an integer n into k parts, is a way of writing n as (unordered) sum of k strictly positive integers [2]. It is easy to see that to any partition 1 j1 2 j2 3 j3 . . . (n −k +1) jn−k+1 of n into k parts, where j1 + j2 + j3 +· · ·+ jn−k+1 = k k! and 1 · j1 + 2 · j2 + 3 · j3 + · · · + (n − k + 1) · jn−k+1 = n, there corresponds exactly j1 ! j2 ! j3 !... jn−k+1 ! compositions of n into k parts via ordering the summands in the partition. So, the sum of monomials xi1 xi2 . . . xik , where the sequences {i 1 , i 2 , . . . , i k } are running over all compositions of n into k parts and where {xr }r ≥1 is the (infinite) set of commuting variables, is equal to the analogous sum over all partitions of n into k parts:   k! jn−k+1 j j xi1 xi2 . . . xik = x11 x22 . . . xn−k+1 j ! j ! . . . j ! 1 2 n−k+1 Part Comp k! Bn,k (x1 · 1!, x2 · 2!, . . . , xn−k+1 · (n − k + 1)!) n! ∗ = Bn,k (x1 , x2 , . . . , xn−k+1 ), =

(4)

where Bn,k (x1 , x2 , . . . , xn−k+1 ) =



  jn−k+1  x  j1  x  j2 xn−k+1 n! 1 2 ... j1 ! j2 ! . . . jn−k+1 ! 1! 2! (n − k + 1)!

(5)

are the partial Bell polynomials, and where the sum is running over all (n − k + 1) sets { j1 , j2 , . . . , jn−k+1 } of nonnegative (and non-zero) integers such that j1 + j2 + j3 + · · · + jn−k+1 = k and 1 · j1 + 2 · j2 + 3 · j3 + · · · + (n − k + 1) · jn−k+1 = n (see [5]).   Mathematica 9 contains function Bell Y n, k, {x1 , x2 , . . . , xn−k+1 } which gives the partial Bell polynomial Bn,k (x1 , x2 , . . . , xn−k+1 ), but here we use in fact a slightly modified form of these polynomials ∗ Bn,k (x1 , x2 , . . . , xn−k+1 ) =

k! Bn,k (x1 · 1!, x2 · 2!, . . . , xn−k+1 · (n − k + 1)!), n!

according to formula (4). Example 1. The full set of compositions of n = 5 into k = 3 parts is {{3, 1, 1} , {2, 2, 1} , {2, 1, 2} , {1, 3, 1} , {1, 2, 2} , {1, 1, 3}}

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and formula (4) becomes ∗ x3 x1 x1 + x2 x2 x1 + x2 x1 x2 + x1 x3 x1 + x1 x2 x2 + x1 x1 x3 = 3x1 x22 + 3x12 x3 = B5,3 (x1 , x2 , x3 ) 3! = Bell Y [5, 3, {1x1 , 2x2 , 6x3 }] 5! 1 (60x1 x22 + 60x12 x3 ). = 20

We need also to mention a formula  (X + Y )n = X k1 Y m 1 . . . X ks Y m s ,

n > 0,

(6)

1≤s≤n k1 +m 1 +···+ks +m s =n

where X and Y are matrices and for every s, 1 ≤ s ≤ n the summation is taken over all weak compositions of n into 2s parts. Finally, some concluding remarks:  (a) we write [z k ]g = gk to denote the k-th coefficient in the formal power series g(z) = k≥0 gk z k ;  (b) binomial coefficient qp with non-integer p everywhere below is set to be equal 0. 2. Generating matrices for Padovan-like sequences For every α ∈  0 Aα = 1 0

Z let

 α 1 0 0 1 0   un and let (for now) vn be the first column of matrix Anα , n ≥ 0. Then the equality An+1 = Aα Anα obviously implies α wn

that u n+1 = αvn + wn ,

vn+1 = u n ,

wn+1 = vn

(7)

and a simple induction shows that u n+1 = αu n−1 + u n−2 ,

u 0 = 1,

u 1 = 0,

u2 = α

(8)

and 

un Anα = u n−1 u n−2

 u n−1 u n−2  . u n−3

u n+1 un u n−1

(9)

A sequence satisfying recurrence (8) will be referred as a “Padovan-like sequence” with a corresponding generating  matrix Aα and parameter α, note that u n = Anα 11 for all n ≥ 0. Recurrence (8) implies by simple induction the formula n

un =

 2  k=0

k n − 2k



· α 3k−n ,

n ≥ 0.

(10)

Indeed, 

n

αu n + u n−1

 2 



n−1  2 

 k · α 3k−n+1 = α· ·α + n − 1 − 2k k=0 k=0  n     2  k k  = α· α 3k−n + n − 2k n − 2k − 1 k=0 k n − 2k



3k−n

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Table 1 Examples of sequences generated by the matrix Aα . α

OEIS number

Initial terms

Comments

α=1 α=2 α=3

A000931 A008346 A052931

1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, . . . 1, 0, 2, 1, 4, 4, 9, 12, 22, 33, 56, 88, 145, . . . 1, 0, 3, 1, 9, 6, 28, 27, 90, 109, 297, 417, . . .

Padovan sequence: u n = pn u n = f n = Fibonacci(n) + (−1)n –

α=0

A079978

1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, . . .

α = −1 α = −2

A077961 A077965

1, 0, −1, 1, 1, −2, 0, 3, −2, −3, 5, 1, −8, 4, 9, . . . 1, 0, −2, 1, 4, −4, −7, 12, 10, −31, −8, 72, −15, −152, 102, . . .

u n = tn 1, n ≡ 0 (mod 3) = 0, n ≡ 1, 2 (mod 3) n = 3 0 – –

  k+1  (n + 2) − 2(k + 1) k=0  n     2 +1  r  = α· α 3(r −1)−n (n + 2) − 2r r =1   n+2    2 r    −1 3r −(n+2) α α = α·  (n + 2) − 2r r =1

n  2  3k−n  α = α·

= u n+2 . In Table 1 we give some examples of Padovan-like sequences with their corresponding numbers in On-Line Encyclopedia of Integer Sequences (OEIS, see https://oeis.org/). Following the article A079978 in OEIS the sequence A079978 will for brevity be referred here as “1/999sequence” since the sequence of decimal digits of 1/999 is 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, . . . . Sequence A008346 can be considered (and named) as Pell–Padovan sequence [3]. 3. Padovan-like sequences and Bell polynomials Now let β ∈ Z and let vn be the n-th term of the Padovan-like sequence relating to matrix Aβ , and let wn be the n-th term of the Padovan-like sequence relating to Aα+β , i.e.    n wn wn+1 wn−1 0 α+β 1 wn wn−2  = 1 0 0 Anα+β = wn−1 wn−2 wn−1 wn−3 0 1 0 = (Aα + βe)n = (Aβ + αe)n , where wn+1 = (α + β)wn−1 + wn−2 , w0 = 1, w1 = 0, w2 = α + β. Then applying formula (6) to X = Aα and Y = βe, one sees immediately that due to the identity e2 = 0, in the right side of (6) all m i must be equal to 0 or 1, and further, all terms with at least one ki = 0 equal 0. So, in this case the summation is taken over all (strong) compositions, or more exactly, for n ≥ 3 the terms in the right side of (6) can have precisely one of the following four forms: (I) : β s−1 Akα1 e Akα2 e . . . e Akαs , 1 ≤ ki ≤ n − 1, k1 + · · · + ks + (s − 1) = n,   i.e. k1 + · · · + ks = n − s + 1, 1 ≤ s ≤ n+1 2 . (for s = 1 this term equals Anα ) (II) : β s Akα1 e Akα2 e . . . e Akαs e, 1 ≤ ki ≤ n − 1, k1 + · · · + ks + s = n,

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  i.e. k1 + · · · + ks = n − s, 1 ≤ s ≤ n2 . (III) : β s e Akα1 e Akα2 e . . . e Akαs , 1 ≤ ki ≤ n − 1, k1 + · · · + ks + s = n,   i.e. k1 + · · · + ks = n − s, 1 ≤ s ≤ n2 . (IV) : β s+1 e Akα1 e Akα2 e . . . e Akαs e, 1 ≤ ki ≤ n − 1, k1 + · · · + ks + (s + 1) = n,   i.e. k1 + · · · + ks = n − s − 1, 1 ≤ s ≤ n−1 2 , where 1 ≤ i ≤ s and 00 = 1 by definition. 

Since we are interested only in wn = Anα+β [β

s

Akα1 e Akα2 e . . . e Akαs e]11

= [β

s+1

 11

  = (Aα + βe)n 11 , formulas (1), (2) and (9) imply that

e Akα1 e Akα2 e . . . e Akαs e]11

= 0,

[β s−1 Akα1 e Akα2 e . . . e Akαs ]11 = β s−1 [Akα1 (e Akα2 e . . . e)Akαs ]11 s−1  = β s−1 [Akα1 e Akαs ]11 [Akαr ]21 r =2   s−1  s−1 = β u k1 · u kr −1 · u ks −1 r =2   s  s−1 = β u k1 · u kr −1 , r =2 k

[β s e Akα1 e Akα2 e . . . e Akαs ]11 = β s [(e Akα1 e Akα2 e . . . Aαs−1 e)Akαs ]11   s−1 s   s = β u kr −1 · u ks −1 = β s u kr −1 , r =1

r =1

thus   wn = Anα+β =

  = (Aα + βe)n 11 11    β s−1 Akα1 e Akα2 e . . . e Akαs

 s



ki =n−s+1

+



11

s



i

 1≤s≤



ki =n−s

11

i

 =

  β s e Akα1 e Akα2 e . . . e Akαs



n+1 2



β s−1 



u k1

s 



u kr −1 +

r =2

ki =n−s+1



 1≤s≤

βs

n

i



2



s 

ki =n−s

r =1

 u kr −1 ,

(11)

i

where for every s and 1 ≤ i ≤ s, the summation in the first (respectively second) inner sum is taken over all compositions of n − s + 1 (respectively n − s) into s parts. For n = 1 and n = 2 formula (11) is evident: w1 = β 0 (β · 0 + u 1 ) = 0,

w2 = β 0 (βu 0 + u 2 ) = β + α.

Example 2. Let α = β = 1. Then u n = pn , wn = f n = Fibonacci(n) + (−1)n (see Table 1), so, by (11) one must have wn = f n = Fibonacci(n) + (−1)n   s    = pk 1 pkr −1 + 1≤s≤



n+1 2

 i

ki =n−s+1

r =2

  1≤s≤

n  2

i



s 

ki =n−s

r =1

 pkr −1 .

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Let us take n = 7 and check this equality in detail: For s = 1 the sets of all compositions of n − s + 1 = 7 and n − s = 6 into s = 1 parts are {7} and {6} respectively, so, in this case the first inner sum is s1 = p7 = 3, the second is s2 = p6−1 = 2, and hence the resulting sum is S1 = s1 + s2 = 3 + 2 = 5. For s = 2 the sets of all compositions of n − s + 1 = 6 and n − s = 5 into s = 2 parts are {{1, 5}, {2, 4}, {3, 3}, {4, 2}, {5, 1}} and {{1, 4}, {2, 3}, {3, 2}, {4, 1}} respectively, so, in this case s1 = p1 p4 + p2 p3 + p3 p2 + p4 p1 + p5 p0 = 0 ∗ 1 + 1 ∗ 1 + 1 ∗ 1 + 1 ∗ 0 + 2 ∗ 1 = 4 s2 = p0 p3 + p1 p2 + p2 p1 + p3 p0 = 1 ∗ 1 + 0 ∗ 1 + 1 ∗ 0 + 1 ∗ 1 = 2 S2 = s1 + s2 = 4 + 2 = 6. For s = 3 the sets of all composition of n − s + 1 = 5 and n − s = 4 into s = 3 parts are {{1, 1, 3}, {1, 2, 2}, {1, 3, 1}, {2, 1, 2}, {2, 2, 1}, {3, 1, 1}} and {{1, 1, 2}, {1, 2, 1}, {2, 1, 1}} respectively, so, in this case s1 = p1 p0 p2 + p1 p1 p1 + p1 p2 p0 + p2 p0 p1 + p2 p1 p0 + p3 p0 p0 = 0 + 0 + 0 + 0 + 0 + 1 = 1 s2 = p0 p0 p1 + p0 p1 p0 + p1 p0 p0 = 0 + 0 + 0 = 0 S3 = s1 + s2 = 1 + 0 = 1.   = 4 the sets of all composition of n − s + 1 = 4 and n − s = 3 into s = 4 parts are {1, 1, 1, 1} and {} For s = 7+1 2 respectively, so, in this case s1 = p1 p0 p0 p0 = 0 s2 = 0 S4 = s1 + s2 = 0 + 0 = 0. The final result is S1 + S2 + S3 + S4 = 5 + 6 + 1 + 0 = 12 = f 7 in full accordance with formula (11). Now we are going to express the right side of (11) through Bell polynomials: Formulas (4) and (5) show straightforwardly that the second summand in (11) can be written as follows:   n s 2     ∗ u kr −1 = βs β s Bn−s,s (u 0 , u 1 , . . . , u n−2s ). 1≤s≤

n



2

ki =n−s

r =1

s=1

i

(∞) ∂ To express in the same way the first summand in (11), we will use differential operator D = r =1 xr +1 ∂ xr , correctly defined on the set of all polynomials in variables {xr }r ≥1 . Then it is not difficult to verify that 

  1≤s≤



n+1 2



β s−1 



ki =n−s+1

u k1

s 

r =2

 u kr −1

n+1



2  β s−1 ∗ (D Bn−s+1,s )(u 0 , u 1 , . . . , u n−2s+1 ) = s s=1

i

and thus we have proved the following: Proposition 1. If {u n }, {vn }, {wn } are Padovan-like sequences with parameters α, β, α + β, respectively, then for all n ≥ 1  n+ε 

1  2  β s−ε ε ∗ wn = (D Bn−s+ε,s )(u 0 , u 1 , . . . , u n−2s+ε ) sε ε=0 s=1  n+ε 

1  2  α s−ε ε ∗ = (D Bn−s+ε,s )(v0 , v1 , . . . , vn−2s+ε ), sε ε=0 s=1 ∗ (x , x , . . . , x 1 where Bn,k 1 2 n−k+1 ) are modified Bell polynomials (see (4)), D = D = identity operator.

(12) (∞)

∂ r =1 xr +1 ∂ xr

and D0 = id is

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Example 3. Let α = 2, β = −1, n = 8 and let us check in detail how formula (12) works to obtain p8 from the numbers { f m }0≤m≤8 : For ε = 0 the corresponding modified Bell polynomials in the inner sum of (12) are exactly x7 , x32 + 2x2 x4 + 2x1 x5 , 3x1 x22 + 3x12 x3 , x14 , so this sum equals −( f 6 ) + ( f 22 + 2 f 1 f 3 + 2 f 0 f 4 ) − (3 f 0 f 12 + 3 f 02 f 2 ) + ( f 04 ) = −9 + 12 − 6 + 1 = −2. For ε = 1 after a little cumbersome calculation one finds that the corresponding polynomials are as follows: x9 , x1 x7 + 2x2 x6 + 2x3 x5 + x42 , 4x1 x2 x4 + 2x1 x32 + x12 x5 + 3x22 x3 , 3x12 x22 + x13 x3 , so in this case the sum equals f 8 − ( f 0 f 6 + 2 f 1 f 5 + 2 f 2 f 4 + f 32 ) + · · · − (3 f 02 f 12 + f 03 f 2 ) = 22 − 26 + 12 − 2 = 6 and finally we get −2 + 6 = 4 = p8 . In the general case we have, of course, the equality  n+ε 

1  2  (−1)s−ε ε ∗ pn = (D Bn−s+ε,s )( f 0 , f 1 , . . . , f n−2s+ε ), sε ε=0 s=1

n≥1

(13)

and its “inverse”: f n = Fibonacci(n) + (−1)n  n+ε 

1  2  1 ε ∗ = (D Bn−s+ε,s )( p0 , p1 , . . . , pn−2s+ε ), ε s ε=0 s=1

n ≥ 1.

(14)

Example 4. For the cases α = 2, β = −2 and α = 1, β = −1 we get two identities which we believe are new:  n+ε 

n 1  2  (−2)s−ε ε ∗ 3 , (D B )( f , f , . . . , f ) = 0 1 n−2s+ε n−s+ε,s ε s 0 ε=0 s=1

n ≥ 1,

(15)

n ≥ 1,

(16)

 n+ε 

n 1  2  (−1)s−ε ε ∗ (D Bn−s+ε,s )( p0 , p1 , . . . , pn−2s+ε ) = 3 , ε 0 s ε=0 s=1 where n 3

 =

0

1, n ≡ 0 (mod 3) 0, n ≡ 1, 2 (mod 3)

(see Table 1). 4. Padovan-like sequences via “1/999-sequence” Since Aα = A0 + αe one can apply formula (12)  n to express terms of any Padovan-like sequence via the terms of the simplest in this class “1/999-sequence” tn = 03 . In order to consider this question in full detail we return again to formula (11) replacing {u r } by {tr } and β by α: 

un =

n+1 2



 s=1

 α s−1



tk 1

k1 +···+ks =n−s+1

s  i=2

 tki −1 +

n 2 

s=1

αs



s 

tki −1 ,

n ≥ 1.

(17)

k1 +···+ks =n−s i=1

It is clear that in the inner sum of the second summand in (17) the congruencies k1 ≡ 1 (mod 3) must hold for all i, 2 ≤ i ≤ s, hence this summand can be written as n 2 

s=1

α s · Comp(n − s, s|1),

n≥1

(18)

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where Comp(n − s, s|1) denotes the number of all compositions of n − s into s parts each of which is congruent to 1 modulo 3. s Furthermore, it is clear that for s ≥ 2 the product tk1 i=2 tki −1 is equal to 1 if and  only if when k1 ≡ 0 (mod 3) and ki ≡ 1 (mod 3) for all i, 2 ≤ i ≤ s, and if s = 1, then this term equals tn =

n 3

0

.

Hence the first summand in (17) can be written as 

n 3

0

+

n+1



2 

α s−1 · Comp(n − s + 1, s|0, 1)

(19)

s=2

where Comp(n − s + 1, s|0, 1) denotes the number of all compositions of n − s + 1 into s parts the first of which is congruent to 0 modulo 3 and all the rest are congruent to 1 modulo 3. So, we get 

n un =

3

0

+

n+1 2





n

α s−1 · Comp(n − s + 1, s|0, 1) +

s=2

2 

α s · Comp(n − s, s|1).

(20)

s=1

Lemma 1. For n ≥ 1 the following formulas are valid: 1. Comp(n − s, s|1) = Comp(n, s|2). n  −1 3 2. Comp(n, s|0) = s−1   n+2s −1 3 Comp(n, s|1) = s−1  n+s  3 −1 . Comp(n, s|2) = s−1

(21)

3. Comp(n − s + 1, s|0, 1) = Comp(n + s − 1, s|0).

(23)

(22)

Proof. 1. If k1 + k2 + · · · + ks = n − s, where all ki ≡ 1 (mod 3), 1 ≤ i ≤ s, then adding 1 to each ki in the left side and s to the right side of this equality, one gets a composition of n into s parts (k1 + 1) + (k2 + 1) + · · · + (ks + 1) = n, in which all addends are congruent to 2 modulo 3. Clearly, there is a one-to-one correspondence between these sets of compositions. 2. We begin with a proof of the first equality of (22): If n ≥ 1 and Comp(n, s|0) ̸= 0, then clearly 3|n. Dividing left and right sides of such a composition by 3 one gets a composition of n/3 into s parts and this is again a one-to-one correspondence. The first formula in (3), with replacement of n by n/3 completes this proof. The proof of the second (resp. third) equality of (22) is quite similar to that of item 1: In every composition of n into s parts each being congruent to 1 (resp. 2) modulo 3, one must add to its left and right sides s 2’s (resp. 1’s) to get a composition of n + 2s (resp. n + s) into s parts each of which is congruent to 0 modulo 3, and then apply the proved first equality of (22). 3. The proof is again quite similar to that of item 1: one must add s −1 2’s to the left and right sides of the |0, 1)-type composition. Now, applying to (20) formulas (21)–(23), we find that for s ≥ 1 [α s ]u n = Comp(n − (s + 1) + 1, (s + 1)|0, 1) + Comp(n − s, s|1)  n+s   n+s   n+s  −1 −1 3 3 3 = + = s s−1 s

(24)

N. Gogin, A. Myll¨ari / Mathematics and Computers in Simulation (

)

9

–

and finally we get n

un =

2 

αs

 n+s 

s=0

3

s

,

n ≥ 0.

The polynomials n   n+l 2  3 Πn (z) = · zl , l l=0

(25)

n≥0

(26)

satisfy the following recurrent equation Π0 (z) = 1,

Π1 (z) = 0,

Π2 (z) = z,

Πn (z) = zΠn−2 (z) + Πn−3 (z),

n≥3

(27)

and are known as Padovan polynomials [8]. So, formula (25) means that u n = Πn (α), and implies in particular the known formulas pn = Πn (1), f n = Πn (2) [8] and, of course, similar formulas are valid for sequences A052931, A079978, A077961, and A077965. Considering parameters α and β in formula (12) as independent variables α = x, β = y, we get the following immediate corollary of Proposition 1: Proposition 2.  n+ε 

1  2  y s−ε ε ∗ (D Bn−s+ε,s )(Π0 (x), Π1 (x), . . . , Πn−2s+ε (x)). 1. Πn (x + y) = sε ε=0 s=1

(28)

1 dm ∗ Πn (x) = Bn−m,m (Π0 (x), Π1 (x), . . . , Πn−2m (x)) m! d x m 1 ∗ + (D Bn−m,m+1 )(Π0 (x), Π1 (x), . . . , Πn−2m−1 (x)), m ≥ 1. (29) m+1 Proof of item 1, as was stated previously, is evident from formula (12) whereas item 2is a simple consequence of the Taylor formula applied to (28). 2.

Example 5. In (29) let n = 6, m = 1. The first seven Padovan polynomials are Π0 (x) = 1, Π1 (x) = 0, ∗ Π2 (x) = x, Π3 (x) = 1, Π4 (x) = x 2 , Π5 (x) = 2x, Π6 (x) = 1 + x 3 , and B6−1,1 (x1 , . . . , x5 ) = x5 , 1 1 ∗ 2 2 (D B6−1,1 )(x 1 , . . . , x 4 ) = 2 D(2x 1 x 4 + 2x 2 x 3 ) = x 1 x 5 + 2x 2 x 4 + x 3 . Hence by (29) one has: d Π6 (x) = Π4 (x) + (Π0 (x)Π4 (x) + 2Π1 (x)Π3 (x) + Π22 (x)) dx = x 2 + (1 · x 2 + 2 · 0 · 1 + x 2 ) d (1 + x 3 ). = 3x 2 = dx Proposition 3. For n ≥ 2 the following formulas are valid: n

1. u n−2 =

2 

αr −1 Comp(n, r |2),

n≥2

(30)

r =1 n

2. pn−2 =

2 

Comp(n, r |2),

n ≥ 2.

(31)

r =1

Proof. 1. The third formula (22) shows that  n+s  3 Comp(n + 2, s + 1|2) = , n≥0 s

(32)

10

N. Gogin, A. Myll¨ari / Mathematics and Computers in Simulation (

)

–

thus n

un =

2 

αs

 n+s  3

s

s=0 n

=

2 

α s Comp(n + 2, s + 1|2)

s=0   n+2

=

2 

αr −1 Comp(n + 2, r |2),

n≥0

(33)

r =1

and changing here n to n − 2 we get the claimed results. 2. Formula (31) is evident from (30) with α = 1. Remark 1. Formula (31) was reported by Vladetta Jovovic in OEIS [13], whereas (30) generalizes this result. Remark 2. Since, as mentioned above, u n = Πn (α), the following formula for Padovan polynomials is obvious from (33): 

Πn (z) =

n+2 2





z r −1 Comp(n + 2, r |2),

n ≥ 0.

(34)

r =1

(34) shows the combinatorial meaning of the coefficients of Padovan polynomials. References [1] G.E. Andrews, The Theory of Partitions, Addison-Wesley, PC, 1976. [2] A. Benavoli, L. Chisci, A. Farina, Fibonacci sequence, golden section, Kalman filter and optimal control, Signal Process. 89 (2009) 1483–1488. [3] G. Bilgici, Generalized order-k Pell–Padovan-like numbers by matrix methods, Pure Appl. Math. J. 2 (6) (2013) 174–178. [4] N. Gogin, A. Myll¨ari, On the average growth rate of random compositions of Fibonacci and Padovan recurrences, in: V.P. Gerdt, E.W. Mayr, Ev.V. Vorozhtsov (Eds.), Computer Algebra in Scientific Computing, in: Lecture Notes in Computer Science, vol. 5743, 2009, pp. 240–246. [5] http://en.wikipedia.org/wiki/Bell polynomials. [6] http://en.wikipedia.org/wiki/Composition (number theory). [7] http://en.wikipedia.org/wiki/Fibonacci heap. [8] http://en.wikipedia.org/wiki/Padovan polynomials. [9] http://en.wikipedia.org/wiki/Padovan sequence. [10] Yu.V. Matiyasevich, Hilbert’s Tenth Problem, MIT Press, Cambridge, Massachusetts, 1993. [11] R. Padovan, Dom Hans Van Der Laan and the plastic number, in: K. Williams, J.F. Rodrigues (Eds.), Nexus IV: Architecture and Mathematics, Kim Williams Books, Fucecchio (Florence), 2002, pp. 181–193. [12] D. Viswanath, Random Fibonacci sequences and the number 1.13198824..., Math. Comp. 69 (231) (2000) 1131–1155. [13] J. Vladetta, Sloane’s A000931: Padovan Sequence, the On-Line Encyclopedia of Integer Sequences, OEIS Foundation. [14] F. Yilmaz, D. Bozkurt, Some properties of Padovan sequence by matrix method, Ars Combin. 104 (2012) 149–160. [15] N. Yilmaz, N. Taskara, Matrix sequences in terms of Padovan and Perrin numbers, J. Appl. Math. 2013 (2013).