A note on binomial transform of the generalized fifth order Jacobsthal numbers

Author(s): Yüksel Soykan1, Erkan Taşdemir2, Vedat Irge1
1Department of Mathematics, Art and Science Faculty, Zonguldak Bülent Ecevit University, 67100, Zonguldak, Turkey.
2Pınarhisar Vocational School, Kırklareli University, 39300, Kırklareli, Turkey.
Copyright © Yüksel Soykan, Erkan Taşdemir, Vedat Irge. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we define the binomial transform of the generalized fifth order Jacobsthal sequence and as special cases, the binomial transform of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas sequences will be introduced. We investigate their properties in details.

Keywords: Binomial transform; Fifth order Jacobsthal sequence; Fifth order Jacobsthal numbers; Binomial transform of fifth order Jacobsthal sequence; Binomial transform of fifth order Jacobsthal-Lucas sequence.

1. Introduction and preliminaries

In this paper, we introduce the binomial transform of the generalized fifth order Jacobsthal sequence and we investigate, in detail, two special cases which we call them the binomial transform of the fifth order Jacobsthal and fifth order Jacobsthal-Lucas sequences. We investigate their properties in the next sections. In this section, we present some properties of the generalized (r,s,t,u,v) sequence (generalized Pentanacci) sequence.

The generalized (r,s,t,u,v) sequence (the generalized Pentanacci sequence or 5-step Fibonacci sequence)

{Wn}n0={Wn(W0,W1,W2,W3,W4;r,s,t,u,v)}n0 is defined by the fifth-order recurrence relations
Wn=rWn1+sWn2+tWn3+uWn4+vWn5,&nbsp &nbsp &nbsp W0=a,W1=b,W2=c,W3=d,W4=e,
(1)
where the initial values W0,W1,W2,W3,W4 are arbitrary complex (or real) numbers and r,s,t,u,v are real numbers. Pentanacci sequence has been studied by many authors and more detail can be found in the extensive literature dedicated to these sequences, see for example [1,2,3,4,5]. The sequence {Wn}n0 can be extended to negative subscripts by defining Wn=uvW(n1)tvW(n2)svW(n3)rvW(n4)+1vW(n5) for n=1,2,3,. Therefore, recurrence (1) holds for all integer n.

As {Wn} is a fifth order recurrence sequence (difference equation), it’s characteristic equation is

x5rx4sx3tx2uxv=0,
(2)
whose roots are α,β,γ,δ,λ. Note that we have the following identities: α+β+γ+δ+λ=r,αβ+αλ+αγ+βλ+αδ+βγ+λγ+βδ+λδ+γδ=s,αβλ+αβγ+αλγ+αβδ+αλδ+βλγ+αγδ+βλδ+βγδ+λγδ=t,αβλγ+αβλδ+αβγδ+αλγδ+βλγδ=uαβγδλ=v. Generalized Pentanacci numbers can be expressed, for all integers n, using Binet’s formula.

Theorem 1. (Binet’s formula of generalized (r,s,t,u,v) numbers (generalized Pentanacci numbers))

Wn=p1αn(αβ)(αγ)(αδ)(αλ)+p2βn(βα)(βγ)(βδ)(βλ)+p3γn(γα)(γβ)(γδ)(γλ)+p4δn(δα)(δβ)(δγ)(δλ)+p5λn(λα)(λβ)(λγ)(λδ),
(3)
where p1=W4(β+γ+δ+λ)W3+(βλ+βγ+λγ+βδ+λδ+γδ)W2(βλγ+βλδ+βγδ+λγδ)W1+(βλγδ)W0,p2=W4(α+γ+δ+λ)W3+(αλ+αγ+αδ+λγ+λδ+γδ)W2(αλγ+αλδ+αγδ+λγδ)W1+(αλγδ)W0,p3=W4(α+β+δ+λ)W3+(αβ+αλ+βλ+αδ+βδ+λδ)W2(αβλ+αβδ+αλδ+βλδ)W1+(αβλδ)W0,p4=W4(α+β+γ+λ)W3+(αβ+αλ+αγ+βλ+βγ+λγ)W2(αβλ+αβγ+αλγ+βλγ)W1+(αβλγ)W0,p5=W4(α+β+γ+δ)W3+(αβ+αγ+αδ+βγ+βδ+γδ)W2(αβγ+αβδ+αγδ+βγδ)W1+(αβγδ)W0.

Usually, it is customary to choose r,s,t,u,v so that the Eq. (2) has at least one real (say α) solutions. Eq. (3) can be written in the following form: Wn=A1αn+A2βn+A3γn+A4δn+A5λn, where A1=p1(αβ)(αγ)(αδ)(αλ),A2=p2(βα)(βγ)(βδ)(βλ),A3=p3(γα)(γβ)(γδ)(γλ),A4=p4(δα)(δβ)(δγ)(δλ),A5=p5(λα)(λβ)(λγ)(λδ). Next, we give the ordinary generating function n=0Wnxn of the sequence Wn.

Lemma 1. Suppose that fWn(x)=n=0Wnxn is the ordinary generating function of the generalized (r,s,t,u,v) sequence {Wn}n0. Then, n=0Wnxn is given by

n=0Wnxn=W0+(W1rW0)x+(W2rW1sW0)x2+(W3rW2sW1tW0)x3+(W4rW3sW2tW1uW0)x41rxsx2tx3ux4vx5.
(4)

We next find Binet formula of generalized (r,s,t,u,v) numbers {Wn} by the use of generating function for Wn.

Theorem 2. (Binet’s formula of generalized (r,s,t,u,v) numbers)

Wn=q1αn(αβ)(αγ)(αδ)(αλ)+q2βn(βα)(βγ)(βδ)(βλ)+q3γn(γα)(γβ)(γδ)(γλ)+q4δn(δα)(δβ)(δγ)(δλ)+q5λn(λα)(λβ)(λγ)(λδ),
(5)
where q1=W0α4+(W1rW0)α3+(W2rW1sW0)α2+(W3rW2sW1tW0)α+(W4rW3sW2tW1vW0),q2=W0β4+(W1rW0)β3+(W2rW1sW0)β2+(W3rW2sW1tW0)β+(W4rW3sW2tW1vW0),q3=W0γ4+(W1rW0)γ3+(W2rW1sW0)γ2+(W3rW2sW1tW0)γ+(W4rW3sW2tW1vW0),q4=W0δ4+(W1rW0)δ3+(W2rW1sW0)δ2+(W3rW2sW1tW0)δ+(W4rW3s3W2tW1vW0),q5=W0λ4+(W1rW0)λ3+(W2rW1sW0)λ2+(W3rW2sW1tW0)λ+(W4rW3sW2tW1vW0).

Matrix formulation of Wn can be given as [6]:
(Wn+4Wn+3Wn+2Wn+1Wn)=(rstuv10000010000010000010)n(W4W3W2W1W0).
(6)
In fact, Kalman give the formula in the following form (WnWn+1Wn+2Wn+3Wn+4)=(01000001000001000001rstuv)n(W0W1W2W3W4). Next, we consider two special cases of the generalized (r,s,t,u,v) sequence {Wn} which we call them (r,s,t,u,v) and Lucas (r,s,t,u,v) sequences. (r,s,t,u,v) sequence{Gn}n0 and Lucas (r,s,t,u,v) sequence {Hn}n0 are defined, respectively, by the fifth-order recurrence relations
Gn+5=rGn+4+sGn+3+tGn+2+uGn+1+vGn,G0=0,G1=1,G2=r,G3=r2+s,G4=r3+2sr+t,
(7)
Hn+5=rHn+4+sHn+3+tHn+2+uHn+1+vHn,H0=5,H1=r,H2=2s+r2,H3=r3+3sr+3t,H4=r4+4r2s+4tr+2s2+4u.
(8)
The sequences {Gn}n0  and {Hn}n0 can be extended to negative subscripts by defining Gn=uvG(n1)tvG(n2)svG(n3)rvG(n4)+1vG(n5),Hn=uvH(n1)tvH(n2)svH(n3)rvH(n4)+1vH(n5), for n=1,2,3, respectively. Therefore, recurrences (7) and (8) hold for all integers n.

For more details on the generalized (r,s,t,u,v) numbers, see [4].

Some special cases of (r,s,t,u,v) sequence {Gn(0,1,r,r2+s,r3+2sr+t;r,s,t,u,v)} and Lucas (r,s,t,u,v) sequence {Hn(5,r,2s+r2,r3+3sr+3t,r4+4r2s+4tr+2s2+4u;r,s,t,u,v)} are as follows:

  1. Gn(0,1,1,2,4;1,1,1,1,1)=Pn, Pentanacci sequence,
  2. Hn(5,1,3,7,15;1,1,1,1,1)=Qn, Pentanacci-Lucas sequence,
  3. Gn(0,1,2,5,13;2,1,1,1,1)=Pn, fifth-order Pell sequence,
  4. Hn(5,2,6,17,46;2,1,1,1,1)=Qn, fifth-order Pell-Lucas sequence.
For all integers n, (r,s,t,u,v) and Lucas (r,s,t,u,v) numbers (using initial conditions in (3) or (5)) can be expressed using Binet’s formulas as Gn=αn+3(αβ)(αγ)(αδ)(αλ)+βn+3(βα)(βγ)(βδ)(βλ)+γn+3(γα)(γβ)(γδ)(γλ)+δn+3(δα)(δβ)(δγ)(δλ)+λn+3(λα)(λβ)(λγ)(λδ),Hn=αn+βn+γn+δn+λn, respectively.

Lemma 1 gives the following results as particular examples (generating functions of (r,s,t,u,v), Lucas (r,s,t,u,v) and modified (r,s,t,u,v) numbers).

Corollary 1. Generating functions of (r,s,t,u,v), Lucas (r,s,t,u,v) and modified (r,s,t,u,v) numbers are n=0Gnxn=x1rxsx2tx3ux4vx5,n=0Hnxn=54rx3sx22tx3ux41rxsx2tx3ux4vx5, respectively.

The following theorem shows that the generalized Pentanacci sequence Wn at negative indices can be expressed by the sequence itself at positive indices.

Theorem 3. For nZ, for the generalized Pentanacci sequence (or generalized (r,s,t,u,v)-sequence or 5-step Fibonacci sequence) we have the following: Wn=124vn(W0Hn44WnHn3+3W0H2n2+12Hn2W2n6W0Hn2H2n6W0H4n8WnH3n12H2nW2n24HnW3n+24W4n+8W0HnH3n+12WnHnH2n)=vn(W4nHnW3n+12(Hn2H2n)W2n16(Hn3+2H3n3H2nHn)Wn+124(Hn4+3H2n26Hn2H2n6H4n+8H3nHn)W0).

Proof. For the proof, see [5], Theorem 1.

Using Theorem 3, we have the following corollary, see [5], Corollary 4.

Corollary 2. For nZ, we have Hn=124vn(Hn4+3H2n26Hn2H2n6H4n+8H3nHn).

Note that Gn and Hn can be given as follows by using G0=0 and H0=5 in Theorem 3: Gn=vn(G4nHnG3n+12(Hn2H2n)G2n16(Hn3+2H3n3H2nHn)Gn),Hn=124vn(Hn4+3H2n26Hn2H2n6H4n+8H3nHn), respectively.

Next, we consider the case r=1, s=1,t=1,u=1,v=2 and in this case we write Vn=Wn. A generalized fifth order Jacobsthal sequence {Vn}n0={Vn(V0,V1,V2,V3,V4)}n0 is defined by the fifth order recurrence relations

Vn=Vn1+Vn2+Vn3+Vn4+2Vn5
(9)
with the initial values V0=c0,V1=c1,V2=c2, V3=c3, V4=c4 not all being zero.

The sequence {Vn}n0 can be extended to negative subscripts by defining

Vn=12V(n1)12V(n2)12V(n3)12V(n4)+12V(n5) for n=1,2,3,. Therefore, recurrence (9) holds for all integer n. For more information on the generalized fifth order Jacobsthal numbers, see [7].

The first few generalized fifth order Jacobsthal numbers with positive subscript and negative subscript are given in the Table 1

Table 1. A few generalized fifth order Jacobsthal numbers.
n Vn Vn
0 V0
1 V1 12V412V112V212V312V0
2 V2 34V314V114V214V014V4
3 V3 78V218V118V018V318V4
4 V4 1516V1116V0116V2116V3116V4
5 2V0+V1+V2+V3+V4 3132V0132V1132V2132V3132V4
6 2V0+3V1+2V2+2V3+2V4 3164V43364V13364V23364V33364V0
7 4V0+4V1+5V2+4V3+4V4 95128V333128V133128V233128V033128V4

Eq. (3) can be used to obtain Binet’s formula of generalized fifth order Jacobsthal numbers. Generalized fifth order Jacobsthal numbers can be expressed, for all integers n, using Binet’s formula

Vn=p1αn(αβ)(αγ)(αδ)(αλ)+p2βn(βα)(βγ)(βδ)(βλ)+p3γn(γα)(γβ)(γδ)(γλ)+p4δn(δα)(δβ)(δγ)(δλ)+p5λn(λα)(λβ)(λγ)(λδ), where p1=V4(β+γ+δ+λ)V3+(βλ+βγ+λγ+βδ+λδ+γδ)V2(βλγ+βλδ+βγδ+λγδ)V1+(βλγδ)V0,p2=V4(α+γ+δ+λ)V3+(αλ+αγ+αδ+λγ+λδ+γδ)V2(αλγ+αλδ+αγδ+λγδ)V1+(αλγδ)V0,p3=V4(α+β+δ+λ)V3+(αβ+αλ+βλ+αδ+βδ+λδ)V2(αβλ+αβδ+αλδ+βλδ)V1+(αβλδ)V0,p4=V4(α+β+γ+λ)V3+(αβ+αλ+αγ+βλ+βγ+λγ)V2(αβλ+αβγ+αλγ+βλγ)V1+(αβλγ)V0,p5=V4(α+β+γ+δ)V3+(αβ+αγ+αδ+βγ+βδ+γδ)V2(αβγ+αβδ+αγδ+βγδ)V1+(αβγδ)V0. As {Vn} is a fifth order recurrence sequence (difference equation), it’s characteristic equation is
x5x4x3x2x2=(x2)(x4+x3+x2+x+1)=0.
(10)
The roots α,β,γ,δ and λ of Eq. (10) are given by: α=2,β=14(51)+25+104i,γ=14(51)25+104i,δ=14(5+1)+25+104i,λ=14(5+1)25+104i. Note that we have the following identities: α+β+γ+δ+λ=1,αβ+αλ+αγ+βλ+αδ+βγ+λγ+βδ+λδ+γδ=1,αβλ+αβγ+αλγ+αβδ+αλδ+βλγ+αγδ+βλδ+βγδ+λγδ=1,αβλγ+αβλδ+αβγδ+αλγδ+βλγδ=1,αβγδλ=2. Now we consider four special cases of the sequence {Vn}. Fifth-order Jacobsthal sequence {Jn}n0, fifth order Jacobsthal-Lucas sequence {jn}n0, adjusted fifth order Jacobsthal sequence {Sn}n0 and modified fifth order Jacobsthal-Lucas sequence {Rn}n0 are defined, respectively, by the fifth order recurrence relations
Jn+5=Jn+4+Jn+3+Jn+2+Jn+1+2Jn,J0=0,J1=1,J2=1,J3=1,J4=1,
(11)
jn+5=jn+4+jn+3+jn+2+jn+1+2jn, j0=2,j1=1,j2=5,j3=10,j4=20,
(12)
Sn+5=Sn+4+Sn+3+Sn+2+Sn+1+2Sn,S0=0,S1=1,S2=1,S3=2,S4=4,
(13)
Rn+5=Rn+4+Rn+3+Rn+2+Rn+1+2Rn,R0=5,R1=1,R2=3,R3=7,R4=19.
(14)
The sequences {Jn}n0, {jn}n0, {Sn}n0 and {Rn}n0 can be extended to negative subscripts by defining Jn=12J(n1)12J(n2)12J(n3)J(n4)+12J(n5),jn=12j(n1)12j(n2)12j(n3)12j(n4)+12j(n5),Sn=12S(n1)12S(n2)12S(n3)12S(n4)+12S(n5),Rn=12R(n1)12R(n2)12R(n3)12R(n4)+12R(n5), for n=1,2,3, respectively. Therefore, recurrences (11)-(14) hold for all integer n.

Next, we present the first few values of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers with positive and negative subscripts in the following Table 2:

Table 2. The first few values of the special fifth order numbers with positive and negative subscripts .
n 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Jn 0 1 1 1 1 4 9 17 33 65 132 265 529 1057
Jn 1 0 12 34 18 1716 132 3164 95128 33256 545512 331024 9912048
jn 2 1 5 10 20 40 77 157 314 628 1256 2509 5021 10042
jn 1 12 14 118 1316 1332 1364 13128 371256 397512 3971024 3972048 3974096
Sn 0 1 1 2 4 8 17 33 66 132 264 529 1057 2114
Sn 0 0 0 12 14 18 116 132 3164 33128 33256 33512 331024
Rn 5 1 3 7 15 36 63 127 255 511 1028 2047 4095 8191
Rn 12 34 78 1516 12932 6364 127128 255256 511512 40971024 20472048 40954096 81918192

For all integers n, Binet formulas of fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers are

Jn=(α3α2)αn(αβ)(αγ)(αδ)(αλ)+(β3β2)βn(βα)(βγ)(βδ)(βλ)+(γ3γ2)γn(γα)(γβ)(γδ)(γλ)+(δ3δ2)δn(δα)(δβ)(δγ)(δλ)+(λ3λ2)λn(λα)(λβ)(λγ)(λδ), jn=(α4+4α3+4α2+4α+4)αn1(αβ)(αγ)(αδ)(αλ)+(β4+4β3+4β2+4β+4)βn1(βα)(βγ)(βδ)(βλ)+(γ4+4γ3+4γ2+4γ+4)γn1(γα)(γβ)(γδ)(γλ)+(δ4+4δ3+4δ2+4δ+4)δn1(δα)(δβ)(δγ)(δλ)+(λ4+4λ3+4λ2+4λ+4)λn1(λα)(λβ)(λγ)(λδ), Sn=αn+3(αβ)(αγ)(αδ)(αλ)+βn+3(βα)(βγ)(βδ)(βλ)+γn+3(γα)(γβ)(γδ)(γλ)+δn+3(δα)(δβ)(δγ)(δλ)+λn+3(λα)(λβ)(λγ)(λδ),Rn=αn+βn+γn+δn+λn respectively.

Binet formulas of fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers can be given in the following forms:

Jn=431αn1155((65+5)+225+5(6+5)i)βn+1155((65+5)+225+5(6+5)i)γn+1155((655)+2255(56)i)δn+1155((655)+2255(5+6)i)λn, jn=3831αn+11240(12(2075)+5+5(1112+310)i)βn+11240(12(2075)5+5(1112+310)i)γn+11240(12(20+75)+55(1112310)i)δn+11240(12(20+75)+55(1112+310)i)λn, Sn=831αn+11240(4(20+75)25+5(37i+i5))βn+11240(4(20+75)+25+5(37i+i5))γn+11240(4(20+75)+25+5(21195)i)δn+11240(4(20+75)+25+5(21+195)i)λn,Rn=αn+βn+γn+δn+λn. Next, we give the ordinary generating function n=0Vnxn of the sequence Vn.

Lemma 2. Suppose that fVn(x)=n=0Vnxn is the ordinary generating function of the generalized fifth order Jacobsthal sequence {Vn}n0. Then, n=0Vnxn is given by n=0Vnxn=V0+(V1V0)x+(V2V1V0)x2+(V3V2V1V0)x3+(V4V3V2V1V0)x41xx2x3x42x5.

The previous Lemma gives the following results as particular examples: generating function of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas are fJn(x)=n=0Jnxn=xx32x41xx2x3x42x5,fjn(x)=n=0jnxn=2x+2x2+2x3+2x41xx2x3x42x5,fSn(x)=n=0Snxn=x1xx2x3x42x5,fRn(x)=n=0Rnxn=54x3x22x3x41xx2x3x42x5, respectively.

2. Binomial transform of the generalized fifth order Jacobsthal sequence Vn

In [8], p. 137, Knuth introduced the idea of the binomial transform. Given a sequence of numbers (an), its binomial transform (a^n) may be defined by the rule a^n=i=0n(ni)ai, with inversion an=i=0n(ni)(1)nia^i, or, in the symmetric version a^n=i=0n(ni)(1)i+1ai, withinversion an=i=0n(ni)(1)i+1a^i. For more information on binomial transform, see, for example, [9,10,11,12] and references therein. For recent works on binomial transform of well-known sequences, see for example, [13,14,15,16,17,18,19,20,21,22,23,24,25].

In this section, we define the binomial transform of the generalized fifth order Jacobsthal sequence Vn and as special cases the binomial transform of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas sequences will be introduced.

Definition 1. The binomial transform of the generalized fifth order Jacobsthal sequence Vn is defined by bn=V^n=i=0n(ni)Vi.

The few terms of bn are b0=i=00(0i)Vi=V0,b1=i=01(1i)Vi=V0+V1,b2=i=02(2i)Vi=V0+2V1+V2,b3=i=03(3i)Vi=V0+3V1+3V2+V3,b4=i=04(4i)Vi=V0+4V1+6V2+4V3+V4. Translated to matrix language, bn has the nice (lower-triangular matrix) form (b0b1b2b3b4)=(1000011000121001331014641)(V0V1V2V3V4). As special cases of bn=V^n, the binomial transforms of the fifth order Jacobsthal and fifth order Jacobsthal-Lucas sequences are defined as follows: The binomial transform of the fifth order Jacobsthal sequence Jn is J^n=i=0n(ni)Ji, and the binomial transform of the fifth order Jacobsthal-Lucas sequence jn is j^n=i=0n(ni)ji, The binomial transform of the adjusted fifth order Jacobsthal sequence Sn is S^n=i=0n(ni)Si, and the binomial transform of the modified fifth order Jacobsthal-Lucas sequence Rn is R^n=i=0n(ni)Ri.

Lemma 3. For n0, the binomial transform of the generalized fifth order Jacobsthal sequence Vn satisfies the following relation: bn+1=i=0n(ni)(Vi+Vi+1).

Proof. We use the following well-known identity: (n+1i)=(ni)+(ni1). Note also that (n+10)=(n0)=1 and (nn+1)=0. Then bn+1=V0+i=1n+1(n+1i)Vi=V0+i=1n+1(ni)Vi+i=1n+1(ni1)Vi=V0+i=1n(ni)Vi+i=0n(ni)Vi+1=i=0n(ni)Vi+i=0n(ni)Vi+1=i=0n(ni)(Vi+Vi+1). This completes the proof.

Remark 1. From the Lemma 3, we see that bn+1=bn+i=0n(ni)Vi+1.

The following theorem gives recurrent relations of the binomial transform of the generalized fifth order Jacobsthal sequence.

Theorem 4. For n0, the binomial transform of the generalized fifth order Jacobsthal sequence Vn satisfies the following recurrence relation:

bn+5=6bn+413bn+3+14bn+27bn+1+3bn.
(15)

Proof. To show (15), writing bn+5=r1×bn+4+s1×bn+3+t1×bn+2+u1×bn+1+v1×bn and taking the values n=0,1,2,3,4 and then solving the system of equations b5=r1×b4+s1×b3+t1×b2+u1×b1+v1×b0b6=r1×b5+s1×b4+t1×b3+u1×b2+v1×b1b7=r1×b6+s1×b5+t1×b4+u1×b3+v1×b2b8=r1×b7+s1×b6+t1×b5+u1×b4+v1×b3b9=r1×b8+s1×b7+t1×b6+u1×b5+v1×b4 we find that r1=6,s1=13,t1=14,u1=7,v1=3.

The sequence {bn}n0 can be extended to negative subscripts by defining bn=73b(n1)143b(n2)+133b(n3)2b(n4)+13b(n5) for n=1,2,3,. Therefore, recurrence (15) holds for all integer n.

Note that the recurence relation (15) is independent from initial values. So,

J^n+5=6J^n+413J^n+3+14J^n+27J^n+1+3J^n,j^n+5=6j^n+413j^n+3+14j^n+27j^n+1+3j^n,S^n+5=6S^n+413S^n+3+14S^n+27S^n+1+3S^n,R^n+5=6R^n+413R^n+3+14R^n+27R^n+1+3R^n.

The first few terms of the binomial transform of the generalized fifth order Jacobsthal sequence with positive subscript and negative subscript are given in the following Table 3.

Table 3. A few binomial transform (terms) of the generalized fifth order Jacobsthal sequence.
n bn bn
0 V0 V0
1 V0+V1 13(V02V1+V22V3+V4)
2 V0+2V1+V2 19(11V0+2V1+2V2+11V37V4)
3 V0+3V1+3V2+V3 127(47V034V1+47V27V37V4)
4 V0+4V1+6V2+4V3+V4 181(115V0+115V1128V2+277V3128V4)
5 3V0+6V1+11V2+11V3+6V4 1243(1411V0533V1+682V2+682V3533V4)
6 15V0+15V1+23V2+28V3+23V4 1729(1411V04421V1+5056V24421V3+1411V4)
7 61V0+53V1+61V2+74V3+74V4 12187(29207V0+776V1+776V2+29207V316720V4)

The first few terms of the binomial transform numbers of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas sequences with positive subscript and negative subscript are given in the following Table 4.

Table 4. A few binomial transform (terms).
n 0 1 2 3 4 5 6 7 8 9 10 11
J^n 0 1 3 7 15 34 89 262 807 2489 7590 22914
J^n 23 89 127 13681 298243 2375729 140392187 143926561 37524719683 78859059049 6474437177147
j^n 2 3 9 30 96 297 900 2700 8076 24165 72393 217077
j^n 53 49 8527 8581 1859243 7691729 207402187 2438146561 24381419683 501154759049 20777630177147
S^n 0 1 3 8 22 63 186 558 1682 5067 15235 45739
S^n 13 29 2927 2981 619243 2563729 69142187 812726561 8127219683 167051559049 6925876177147
R^n 5 6 10 24 70 221 700 2169 6590 19806 59295 177469
R^n 73 359 18827 32581 5347243 8020729 1027882187 4986356561 92510219683 1452605559049 21789082177147

Eq. (3) can be used to obtain Binet’s formula of the binomial transform of generalized fifth order Jacobsthal numbers. Binet’s formula of the binomial transform of generalized fifth order Jacobsthal numbers can be given as

bn=C1θ1n(θ1θ2)(θ1θ3)(θ1θ4)(θ1θ5)+C2θ2n(θ2θ1)(θ2θ3)(θ2θ4)(θ2θ5)+C3θ3n(θ3θ1)(θ3θ2)(θ3θ4)(θ3θ5)+C4θ4n(θ4θ1)(θ4θ2)(θ4θ3)(θ4θ5)+C5θ5n(θ5θ1)(θ5θ2)(θ5θ3)(θ5θ4),
(16)
where C1=b4(θ2+θ3+θ4+θ5)b3+(θ2θ5+θ2θ3+θ5θ3+θ2θ4+θ5θ4+θ3θ4)b2(θ2θ5θ3+θ2θ5θ4+θ2θ3θ4+θ5θ3θ4)b1+(θ2θ5θ3θ4)b0,C2=b4(θ1+θ3+θ4+θ5)b3+(θ1θ5+θ1θ3+θ1θ4+θ5θ3+θ5θ4+θ3θ4)b2(θ1θ5θ3+θ1θ5θ4+θ1θ3θ4+θ5θ3θ4)b1+(θ1θ5θ3θ4)b0,C3=b4(θ1+θ2+θ4+θ5)b3+(θ1θ2+θ1θ5+θ2θ5+θ1θ4+θ2θ4+θ5θ4)b2(θ1θ2θ5+θ1θ2θ4+θ1θ5θ4+θ2θ5θ4)b1+(θ1θ2θ5θ4)b0,C4=b4(θ1+θ2+θ3+θ5)b3+(θ1θ2+θ1θ5+θ1θ3+θ2θ5+θ2θ3+θ5θ3)b2(θ1θ2θ5+θ1θ2θ3+θ1θ5θ3+θ2θ5θ3)b1+(θ1θ2θ5θ3)b0,C5=b4(θ1+θ2+θ3+θ4)b3+(θ1θ2+θ1θ3+θ1θ4+θ2θ3+θ2θ4+θ3θ4)b2(θ1θ2θ3+θ1θ2θ4+θ1θ3θ4+θ2θ3θ4)b1+(θ1θ2θ3θ4)b0. Here, θ1,θ2,θ3,θ4 and θ5 are the roots of the equation x56x4+13x314x2+7x3=(x3)(x43x3+4x22x+1)=0. Moreover, the approximate value of θ1,θ2,θ3,θ4 and θ5 are given by θ1=3,θ2=1.30901699437495+0.951056516295154i,θ3=1.309016994374950.951056516295154i,θ4=0.190983005625053+0.587785252292473i,θ5=0.1909830056250530.587785252292473i. Note that θ1+θ2+θ3+θ4+θ5=6,θ1θ2+θ1θ5+θ1θ3+θ2θ5+θ1θ4+θ2θ3+θ5θ3+θ2θ4+θ5θ4+θ3θ4=13,θ1θ2θ5+θ1θ2θ3+θ1θ5θ3+θ1θ2θ4+θ1θ5θ4+θ2θ5θ3+θ1θ3θ4+θ2θ5θ4+θ2θ3θ4+θ5θ3θ4=14,θ1θ2θ5θ3+θ1θ2θ5θ4+θ1θ2θ3θ4+θ1θ5θ3θ4+θ2θ5θ3θ4=7,θ1θ2θ3θ4θ5=3. For all integers n, (Binet’s formulas of) binomial transforms of fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers (using initial conditions in (16)) can be expressed using Binet’s formulas as J^n=(θ133θ12+2θ12)θ1n(θ1θ2)(θ1θ3)(θ1θ4)(θ1θ5)+(θ233θ22+2θ22)θ2n(θ2θ1)(θ2θ3)(θ2θ4)(θ2θ5)+(θ333θ32+2θ32)θ3n(θ3θ1)(θ3θ2)(θ3θ4)(θ3θ5)+(θ433θ42+2θ42)θ4n(θ4θ1)(θ4θ2)(θ4θ3)(θ4θ5)+(θ533θ52+2θ52)θ5n(θ5θ1)(θ5θ2)(θ5θ3)(θ5θ4), j^n=(2θ149θ13+17θ1213θ1+5)θ1n(θ1θ2)(θ1θ3)(θ1θ4)(θ1θ5)+(2θ249θ23+17θ2213θ2+5)θ2n(θ2θ1)(θ2θ3)(θ2θ4)(θ2θ5)+(2θ349θ33+17θ3213θ3+5)θ3n(θ3θ1)(θ3θ2)(θ3θ4)(θ3θ5)+(2θ449θ43+17θ4213θ4+5)θ4n(θ4θ1)(θ4θ2)(θ4θ3)(θ4θ5)+(2θ549θ53+17θ5213θ5+5)θ5n(θ5θ1)(θ5θ2)(θ5θ3)(θ5θ4), S^n=(θ11)3θ1n(θ1θ2)(θ1θ3)(θ1θ4)(θ1θ5)+(θ21)3θ2n(θ2θ1)(θ2θ3)(θ2θ4)(θ2θ5)+(θ31)3θ3n(θ3θ1)(θ3θ2)(θ3θ4)(θ3θ5)+(θ41)3θ4n(θ4θ1)(θ4θ2)(θ4θ3)(θ4θ5)+(θ51)3θ5n(θ5θ1)(θ5θ2)(θ5θ3)(θ5θ4),R^n=θ1n+θ2n+θ3n+θ4n+θ5n, respectively.

3. Generating functions and obtaining Binet formula of binomial transform from generating function

The generating function of the binomial transform of the generalized fifth order Jacobsthal sequence Vn is a power series centered at the origin whose coefficients are the binomial transform of the generalized fifth order Jacobsthal sequence.

Next, we give the ordinary generating function fbn(x)=n=0bnxn of the sequence bn.

Lemma 4. Suppose that fbn(x)=n=0bnxn is the ordinary generating function of the binomial transform of the generalized fifth order Jacobsthal sequence {Vn}n0. Then, fbn(x) is given by

fbn(x)=V0+(V15V0)x+(8V04V1+V2)x2+(4V16V03V2+V3)x3+(V02V1+V22V3+V4)x416x+13x214x3+7x43x5.
(17)

Proof. Using Lemma 1, we obtain fbn(x)=b0+(b16b0)x+(b26b1+13b0)x2+(b36b2+13b114b0)x3+(b46b3+13b214b1+7b0)x416x+13x214x3+7x43x5=V0+(V15V0)x+(8V04V1+V2)x2+(4V16V03V2+V3)x3+(V02V1+V22V3+V4)x416x+13x214x3+7x43x5, where b0=V0,b1=V0+V1,b2=V0+2V1+V2,b3=V0+3V1+3V2+V3,b4=V0+4V1+6V2+4V3+V4.

Note that P. Barry shows in [26] that if A(x) is the generating function of the sequence {an}, then S(x)=11xA(x1x) is the generating function of the sequence {bn} with bn=i=0n(ni)ai. In our case, since A(x)=V0+(V1V0)x+(V2V1V0)x2+(V3V2V1V0)x3+(V4V3V2V1V0)x41xx2x3x42x5, see Lemma 2, we obtain S(x)=11xA(x1x)=V0+(V15V0)x+(8V04V1+V2)x2+(4V16V03V2+V3)x3+(V02V1+V22V3+V4)x416x+13x214x3+7x43x5. The Lemma 4 gives the following results as particular examples.

Corollary 3. Generating functions of the binomial transform of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers are n=0J^nxn=x3x2+2x32x416x+13x214x3+7x43x5,n=0j^nxn=29x+17x213x3+5x416x+13x214x3+7x43x5,n=0S^nxn=x3x2+3x3x416x+13x214x3+7x43x5,n=0R^nxn=524x+39x228x3+7x416x+13x214x3+7x43x5. respectively.

4. Simson formulas

There is a well-known Simson Identity (formula) for Fibonacci sequence {Fn}, namely, Fn+1Fn1Fn2=(1)n, which was derived first by R. Simson in 1753 and it is now called as Cassini Identity (formula) as well. This can be written in the form |Fn+1FnFnFn1|=(1)n. The following theorem gives generalization of this result to the generalized Pentanacci sequence {Wn}.

Theorem 5.(Simson formula of generalized Pentanacci numbers) For all integers n, we have

|Wn+4Wn+3Wn+2Wn+1WnWn+3Wn+2Wn+1WnWn1Wn+2Wn+1WnWn1Wn2Wn+1WnWn1Wn2Wn3WnWn1Wn2Wn3Wn4|=vn|W4W3W2W1W0W3W2W1W0W1W2W1W0W1W2W1W0W1W2W3W0W1W2W3W4|.
(18)

Proof. Eq. (18) is given in [27], Theorem 3.1.

Taking {Wn}={bn} in the above theorem and considering bn+5=6bn+413bn+3+14bn+27bn+1+3bn, r=6,s=13,t=14,u=7,v=3, we have the following proposition.

Proposition 1. For all integers n, Simson formula of binomial transforms of generalized fifth order Jacobsthal numbers is given as |bn+4bn+3bn+2bn+1bnbn+3bn+2bn+1bnbn1bn+2bn+1bnbn1bn2bn+1bnbn1bn2bn3bnbn1bn2bn3bn4|=3n|b4b3b2b1b0b3b2b1b0b1b2b1b0b1b2b1b0b1b2b3b0b1b2b3b4|.

The Proposition 1 gives the following results as particular examples.

Corollary 4. For all integers n, Simson formula of binomial transforms of the fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers are given as |J^n+4J^n+3J^n+2J^n+1J^nJ^n+3J^n+2J^n+1J^nJ^n1J^n+2J^n+1J^nJ^n1J^n2J^n+1J^nJ^n1J^n2J^n3J^nJ^n1J^n2J^n3J^n4|=44×3n4,|j^n+4j^n+3j^n+2j^n+1j^nj^n+3j^n+2j^n+1j^nj^n1j^n+2j^n+1j^nj^n1j^n2j^n+1j^nj^n1j^n2j^n3j^nj^n1j^n2j^n3j^n4|=38×3n,|S^n+4S^n+3S^n+2S^n+1S^nS^n+3S^n+2S^n+1S^nS^n1S^n+2S^n+1S^nS^n1S^n2S^n+1S^nS^n1S^n2S^n3S^nS^n1S^n2S^n3S^n4|=8×3n4,|R^n+4R^n+3R^n+2R^n+1R^nR^n+3R^n+2R^n+1R^nR^n1R^n+2R^n+1R^nR^n1R^n2R^n+1R^nR^n1R^n2R^n3R^nR^n1R^n2R^n3R^n4|=120125×3n4, respectively.

5. Some identities

In this section, we obtain some identities of binomial transforms of fifth order Jacobsthal, fifth order Jacobsthal-Lucas, adjusted fifth order Jacobsthal and modified fifth order Jacobsthal-Lucas numbers. First, we present a few basic relations between {J^n} and {j^n}.

Lemma 5. The following equalities are true: 171J^n=22j^n+6+123j^n+5259j^n+4+329j^n+3193j^n+2,57J^n=3j^n+5+9j^n+4+7j^n+313j^n+222j^n+1,57J^n=9j^n+4+46j^n+355j^n+2j^n+19j^n,57J^n=8j^n+3+62j^n+2127j^n+1+54j^n27j^n1,57J^n=14j^n+223j^n+158j^n+29j^n124j^n2, and 198j^n=5J^n+6285J^n+5+1532J^n+42764J^n+3+2003J^n+2,66j^n=85J^n+5+489J^n+4898J^n+3+656J^n+2+5J^n+1,22j^n=7J^n+4+69J^n+3178J^n+2+200J^n+185J^n,22j^n=27J^n+387J^n+2+102J^n+136J^n21J^n1,22j^n=75J^n+2249J^n+1+342J^n210J^n1+81J^n2.

Proof. Writing J^n=a×j^n+6+b×j^n+5+c×j^n+4+d×j^n+3+e×j^n+2 and solving the system of equations J^0=a×j^6+b×j^5+c×j^4+d×j^3+e×j^2J^1=a×j^7+b×j^6+c×j^5+d×j^4+e×j^3J^2=a×j^8+b×j^7+c×j^6+d×j^5+e×j^4J^3=a×j^9+b×j^8+c×j^7+d×j^6+e×j^5J^4=a×j^10+b×j^9+c×j^8+d×j^7+e×j^6 we find that a=22171,b=4157,c=259171,d=329171,e=193171. The other equalities can be proved similarly.

Now, we give a few basic relations between {J^n} and {S^n}.

Lemma 6. The following equalities are true: 18J^n=7S^n+639S^n+5+82S^n+4104S^n+3+61S^n+2,6J^n=S^n+53S^n+42S^n+3+4S^n+2+7S^n+1,2J^n=S^n+45S^n+3+6S^n+2+S^n,2J^n=S^n+37S^n+2+14S^n+16S^n+3S^n1,2J^n=S^n+2+S^n+1+8S^n4S^n1+3S^n2, and 99S^n=J^n+6+42J^n+5248J^n+4+457J^n+3332J^n+2,33S^n=16J^n+587J^n+4+157J^n+3113J^n+2+J^n+1,11S^n=3J^n+417J^n+3+37J^n+237J^n+1+16J^n,11S^n=J^n+32J^n+2+5J^n+15J^n+9J^n1,11S^n=4J^n+28J^n+1+9J^n+2J^n1+3J^n2.

Next, we present a few basic relations between {J^n} and {R^n}.

Lemma 7. The following equalities are true: 43245J^n=164R^n+6+1182R^n+57993R^n+4+15017R^n+317692R^n+2,14415J^n=722R^n+53375R^n+4+5771R^n+36280R^n+2+164R^n+1,4805J^n=319R^n+41205R^n+3+1276R^n+21630R^n+1+722R^n,4805J^n=709R^n+32871R^n+2+2836R^n+11511R^n+957R^n1,4805J^n=1383R^n+26381R^n+1+8415R^n4006R^n1+2127R^n2, and 396R^n=1163J^n+67881J^n+5+19664J^n+422810J^n+3+10379J^n+2,132R^n=301J^n+5+1515J^n+42176J^n+3+746J^n+2+1163J^n+1,44R^n=97J^n+4+579J^n+31156J^n+2+1090J^n+1301J^n,44R^n=3J^n+3+105J^n+2268J^n+1+378J^n291J^n1,44R^n=87J^n+2229J^n+1+336J^n270J^n19J^n2.

Now, we give a few basic relations between {j^n} and {S^n}.

Lemma 8. The following equalities are true: 36j^n=83S^n+6+465S^n+5872S^n+4+706S^n+383S^n+2,12j^n=11S^n+5+69S^n+4152S^n+3+166S^n+283S^n+1,4j^n=S^n+43S^n+3+4S^n+22S^n+111S^n,4j^n=3S^n+39S^n+2+12S^n+118S^n+3S^n1,4j^n=9S^n+227S^n+1+24S^n18S^n1+9S^n2, and 171S^n=44j^n+6+246j^n+5461j^n+4+373j^n+344j^n+2,57S^n=6j^n+5+37j^n+481j^n+3+88j^n+244j^n+1,57S^n=j^n+43j^n+3+4j^n+22j^n+118j^n,57S^n=3j^n+39j^n+2+12j^n+125j^n+3j^n1,57S^n=9j^n+227j^n+1+17j^n18j^n1+9j^n2.

Next, we present a few basic relations between {j^n} and {R^n}.

Lemma 9. The following equalities are true: 43245j^n=11881R^n+671634R^n+5+151312R^n+4141779R^n+3+41176R^n+2,14415j^n=116R^n+51047R^n+4+8185R^n+313997R^n+2+11881R^n+1,4805j^n=581R^n+4+3231R^n+35207R^n+2+4231R^n+1116R^n,4805j^n=255R^n+3+2346R^n+23903R^n+1+3951R^n1743R^n1,4805j^n=816R^n+2588R^n+1+381R^n+42R^n1765R^n2, and 342R^n=457j^n+62397j^n+5+3994j^n+42396j^n+31025j^n+2,114R^n=115j^n+5649j^n+4+1334j^n+31408j^n+2+457j^n+1,114R^n=41j^n+4161j^n+3+202j^n+2348j^n+1+345j^n,114R^n=85j^n+3331j^n+2+226j^n+1+58j^n+123j^n1,114R^n=179j^n+2879j^n+1+1248j^n472j^n1+255j^n2.

Now, we give a few basic relations between {S^n} and {R^n}.

Lemma 10. The following equalities are true: 43245S^n=3857R^n+6+23568R^n+550024R^n+4+47053R^n+313622R^n+2,14415S^n=142R^n+5+39R^n+42315R^n+3+4459R^n+23857R^n+1,4805S^n=297R^n+41387R^n+3+2149R^n+21617R^n+1+142R^n,4805S^n=395R^n+31712R^n+2+2541R^n+11937R^n+891R^n1,4805S^n=658R^n+22594R^n+1+3593R^n1874R^n1+1185R^n2, and 72R^n=287S^n+6+1509S^n+52516S^n+4+1510S^n+3+649S^n+2,24R^n=71S^n+5+405S^n+4836S^n+3+886S^n+2287S^n+1,8R^n=7S^n+4+29S^n+336S^n+2+70S^n+171S^n,8R^n=13S^n+3+55S^n+228S^n+122S^n21S^n1,8R^n=23S^n+2+141S^n+1204S^n+70S^n139S^n2.

6. On the recurrence properties of binomial transform of the generalized fifth order Jacobsthal sequence

Taking r1=6,s1=13,t1=14,u1=7,v1=3 and Hn=R^n in Theorem 3, we obtain the following Proposition.

Proposition 2. For nZ, binomial Transform of the generalized fifth order Jacobsthal sequence have the following identity: bn=1243n(b0R^n44bnR^n3+3b0R^2n2+12R^n2b2n6b0R^n2R^2n6b0R^4n8bnR^3n12R^2nb2n24R^nb3n+24b4n+8b0R^nR^3n+12bnR^nR^2n)=3n(b4nR^nb3n+12(R^n2R^2n)b2n16(R^n3+2R^3n3R^2nR^n)bn+124(R^n4+3R^2n26R^n2R^2n6R^4n+8R^3nR^n)b0).

Using Proposition 2 (and Corollary 2), we obtain the following corollary which gives the connection between the special cases of binomial transform of generalized fifth order Jacobsthal sequence at the positive index and the negative index: for binomial transform of fifth order Jacobsthal, fifth order Jacobsthal-Lucas numbers: take bn=J^n with J^0=0,J^1=1,J^2=3,J^3=7,J^4=15, take bn=j^n with j^0=2,j^1=3,j^2=9,j^3=30,j^4=96, take bn=S^n with S^0=0,S^1=1,S^2=3,S^3=8,S^4=22, take bn=R^n with R^0=5,R^1=6,R^2=10,R^3=24,R^4=70, respectively. Note that in this case we have Hn=R^n. Note also that GnS^n.

Corollary 5. For nZ, we have the following recurrence relations:

  • (a) Recurrence relations of binomial transforms of fifth order Jacobsthal numbers (take bn=J^n in Proposition 2): J^n=3n(J^4nR^nJ^3n+12(R^n2R^2n)J^2n16(R^n3+2R^3n3R^2nR^n)J^n).
  • (b) Recurrence relations of binomial transforms of fifth order Jacobsthal-Lucas numbers (take bn=j^n in Proposition 2): j^n=3n(j^4nR^nj^3n+12(R^n2R^2n)j^2n16(R^n3+2R^3n3R^2nR^n)j^n+112(R^n4+3R^2n26R^n2R^2n6R^4n+8R^3nR^n)).
  • (c) Recurrence relations of binomial transforms of adjusted fifth order Jacobsthal numbers (take bn=S^n in Proposition 2): S^n=3n(S^4nR^nS^3n+12(R^n2R^2n)S^2n16(R^n3+2R^3n3R^2nR^n)S^n).
  • (d) Recurrence relations of binomial transforms of modified fifth order Jacobsthal-Lucas numbers (take bn=R^n in Proposition 2 or take Hn=R^n in Corollary 2): R^n=1243n(R^n4+3R^2n26R^n2R^2n6R^4n+8R^3nR^n).

7. Sum formulas

7.1. Sums of terms with positive subscripts

The following proposition presents some formulas of binomial transform of generalized fifth order Jacobsthal numbers with positive subscripts.

Proposition 3. If r=6,s=13,t=14,u=7,v=3, then for n0 we have the following formulas:

  • (a) k=0nbk=12(bn+55bn+4+8bn+36bn+2+bn+1b4+5b38b2+6b1b0).
  • (b) k=0nb2k=188(21b2n+2103b2n+1+244b2n98b2n1+69b2n221b4+103b3156b2+98b1+19b0).
  • (c) k=0nb2k+1=188(23b2n+229b2n+1+196b2n78b2n1+63b2n223b4+117b3196b2+166b163b0).

Proof. Take r=6,s=13,t=14,u=7,v=3, in Theorem 2.1 in [28].

From the last proposition, we have the following corollary which gives sum formulas of binomial transform of fifth order Jacobsthal numbers (take bn=J^n with J^0=0,J^1=1,J^2=3,J^3=7,J^4=15).

Corollary 6. For n0 we have the following formulas:

  • (a) k=0nJ^k=12(J^n+55J^n+4+8J^n+36J^n+2+J^n+1+2).
  • (b) k=0nJ^2k=188(21J^2n+2103J^2n+1+244J^2n98J^2n1+69J^2n2+36).
  • (c) k=0nJ^2k+1=188(23J^2n+229J^2n+1+196J^2n78J^2n1+63J^2n2+52).

Taking bn=j^n with j^0=2,j^1=3,j^2=9,j^3=30,j^4=96 in the last proposition, we have the following corollary which presents sum formulas of binomial transform of fifth order Jacobsthal-Lucas numbers.

Corollary 7. For n0 we have the following formulas:

  • (a) k=0nj^k=12(j^n+55j^n+4+8j^n+36j^n+2+j^n+12).
  • (b) k=0nj^2k=188(21j^2n+2103j^2n+1+244j^2n98j^2n1+69j^2n2+2).
  • (c) k=0nj^2k+1=188(23j^2n+229j^2n+1+196j^2n78j^2n1+63j^2n290).

From the last proposition, we have the following corollary which gives sum formulas of binomial transform of adjusted fifth order Jacobsthal numbers (take bn=S^n with S^0=0,S^1=1,S^2=3,S^3=8,S^4=22).

Corollary 8. For n0 we have the following formulas:

  • (a) k=0nS^k=12(S^n+55S^n+4+8S^n+36S^n+2+S^n+1).
  • (b) k=0nS^2k=188(21S^2n+2103S^2n+1+244S^2n98S^2n1+69S^2n28).
  • (c) k=0nS^2k+1=188(23S^2n+229S^2n+1+196S^2n78S^2n1+63S^2n2+8).

Taking bn=R^n with R^0=5,R^1=6,R^2=10,R^3=24,R^4=70 in the last proposition, we have the following corollary which presents sum formulas of binomial transform of modified fifth order Jacobsthal-Lucas numbers.

Corollary 9. For n0 we have the following formulas:

  • (a) k=0nR^k=12(R^n+55R^n+4+8R^n+36R^n+2+R^n+1+1).
  • (b) k=0nR^2k=188(21R^2n+2103R^2n+1+244R^2n98R^2n1+69R^2n2+125).
  • (c) k=0nR^2k+1=188(23R^2n+229R^2n+1+196R^2n78R^2n1+63R^2n281).

7.2. Sums of terms with negative subscripts

The following proposition presents some formulas of binomial transform of generalized fifth order Jacobsthal numbers with negative subscripts.

Proposition 4. If r=6,s=13,t=14,u=7,v=3, then for n1 we have the following formulas:

  • (a) k=1nbk=12(bn+4+5bn+38bn+2+6bn+1bn+b45b3+8b26b1+b0).
  • (b) k=1nb2k=188(23b2n+3+117b2n+2196b2n+1+166b2n63b2n1+21b4103b3+156b298b119b0).
  • (c) k=1nb2k+1=188(21b2n+3+103b2n+2156b2n+1+98b2n69b2n1+23b4117b3+196b2166b1+63b0).

Proof. Take r=6,s=13,t=14,u=7,v=3, in Theorem 3.1 in [28].

From the last proposition, we have the following corollary which gives sum formulas of binomial transform of fifth order Jacobsthal numbers (take bn=J^n with J^0=0,J^1=1,J^2=3,J^3=7,J^4=15).

Corollary 10. For n1, binomial transform of fifth order Jacobsthal numbers have the following properties.

  • (a) k=1nJ^k=12(J^n+4+5J^n+38J^n+2+6J^n+1J^n2).
  • (b) k=1nJ^2k=188(23J^2n+3+117J^2n+2196J^2n+1+166J^2n63J^2n136).
  • (c) k=1nJ^2k+1=188(21J^2n+3+103J^2n+2156J^2n+1+98J^2n69J^2n152).

Taking bn=j^n with j^0=2,j^1=3,j^2=9,j^3=30,j^4=96 in the last proposition, we have the following corollary which presents sum formulas of binomial transform of fifth order Jacobsthal-Lucas numbers.

Corollary 11. For n1, binomial transform of fifth order Jacobsthal-Lucas numbers have the following properties.

  • (a) k=1nj^k=12(j^n+4+5j^n+38j^n+2+6j^n+1j^n+2).
  • (b) k=1nj^2k=188(23j^2n+3+117j^2n+2196j^2n+1+166j^2n63j^2n12).
  • (c) k=1nj^2k+1=188(21j^2n+3+103j^2n+2156j^2n+1+98j^2n69j^2n1+90).

From the last proposition, we have the following corollary which gives sum formulas of binomial transform of adjusted fifth order Jacobsthal numbers (take bn=S^n with S^0=0,S^1=1,S^2=3,S^3=8,S^4=22).

Corollary 12. For n1, binomial transform of adjusted fifth order Jacobsthal numbers have the following properties.

  • (a) k=1nS^k=12(S^n+4+5S^n+38S^n+2+6S^n+1S^n).
  • (b) k=1nS^2k=188(23S^2n+3+117S^2n+2196S^2n+1+166S^2n63S^2n1+8).
  • (c) k=1nS^2k+1=188(21S^2n+3+103S^2n+2156S^2n+1+98S^2n69S^2n18).

Taking bn=R^n with R^0=5,R^1=6,R^2=10,R^3=24,R^4=70 in the last proposition, we have the following corollary which presents sum formulas of binomial transform of modified fifth order Jacobsthal-Lucas numbers.

Corollary 13. For n1, binomial transform of modified fifth order Jacobsthal-Lucas numbers have the following properties.

  • (a)  k=1nR^k=12(R^n+4+5R^n+38R^n+2+6R^n+1R^n1).
  • (b)  k=1nR^2k=188(23R^2n+3+117R^2n+2196R^2n+1+166R^2n63R^2n1125).
  • (c)  k=1nR^2k+1=188(21R^2n+3+103R^2n+2156R^2n+1+98R^2n69R^2n1+81).

8. Matrices related with binomial transform of generalized fifth order Jacobsthal numbers

We define the square matrix A of order 5 as: A=(613147310000010000010000010), such that detA=3. From (1) we have
(bn+4bn+3bn+2bn+1bn)=(613147310000010000010000010)(bn+3bn+2bn+1bnbn1),
(19)
and from (6) (or using (19) and induction) we have (bn+4bn+3bn+2bn+1bn)=(613147310000010000010000010)n(b4b3b2b1b0). If we take bn=S^n in (19) we have
(S^n+4S^n+3S^n+2S^n+1S^n)=(613147310000010000010000010)(S^n+3S^n+2S^n+1S^nS^n1).
(20)
We also, for n0, define Bn=(k=0n+1l=kn+1p=ln+1S^kE1E6E113k=0nl=knp=lnS^kk=0nl=knp=lnS^kE2E7E123k=0n1l=kn1p=ln1S^kk=0n1l=kn1p=ln1S^kE3E8E133k=0n2l=kn2p=ln2S^kk=0n2l=kn2p=ln2S^kE4E9E143k=0n3l=kn3p=ln3S^kk=0n3l=kn3p=ln3S^kE5E10E153k=0n4l=kn4p=ln4S^k), and Cn=(bn+113bn+14bn17bn2+3bn314bn7bn1+3bn27bn+3bn13bnbn13bn1+14bn27bn3+3bn414bn17bn2+3bn37bn1+3bn23bn1bn113bn2+14bn37bn4+3bn514bn27bn3+3bn47bn2+3bn33bn2bn213bn3+14bn47bn5+3bn614bn37bn4+3bn57bn3+3bn43bn3bn313bn4+14bn57bn6+3bn714bn47bn5+3bn67bn4+3bn53bn4), where {E1E2E3E4E5E6E7E8E9E10E11E12E13E14E15}={13k=0nl=knp=lnS^k+14k=0n1l=kn1p=ln1S^k7k=0n2l=kn2p=ln2S^k+3k=0n3l=kn3p=ln3S^k13k=0n1l=kn1p=ln1S^k+14k=0n2l=kn2p=ln2S^k7k=0n3l=kn3p=ln3S^k+3k=0n4l=kn4p=ln4S^k13k=0n2l=kn2p=ln2S^k+14k=0n3l=kn3p=ln3S^k7k=0n4l=kn4p=ln4S^k+3k=0n5l=kn5p=ln5S^k13k=0n3l=kn3p=ln3S^k+14k=0n4l=kn4p=ln4S^k7k=0n5l=kn5p=ln5S^k+3k=0n6l=kn6p=ln6S^k13k=0n4l=kn4p=ln4S^k+14k=0n5l=kn5p=ln5S^k7k=0n6l=kn6p=ln6S^k+3k=0n7l=kn7p=ln7S^k14k=0nl=knp=lnS^k7k=0n1l=kn1p=ln1S^k+3k=0n2l=kn2p=ln2S^k14k=0n1l=kn1p=ln1S^k7k=0n2l=kn2p=ln2S^k+3k=0n3l=kn3p=ln3S^k14k=0n2l=kn2p=ln2S^k7k=0n3l=kn3p=ln3S^k+3k=0n4l=kn4p=ln4S^k14k=0n3l=kn3p=ln3S^k7k=0n4l=kn4p=ln4S^k+3k=0n5l=kn5p=ln5S^k14k=0n4l=kn4p=ln4S^k7k=0n5l=kn5p=ln5S^k+3k=0n6l=kn6p=ln6S^k7k=0nl=knp=lnS^k+3k=0n1l=kn1p=ln1S^k7k=0n1l=kn1p=ln1S^k+3k=0n2l=kn2p=ln2S^k7k=0n2l=kn2p=ln2S^k+3k=0n3l=kn3p=ln3S^k7k=0n3l=kn3p=ln3S^k+3k=0n4l=kn4p=ln4S^k7k=0n4l=kn4p=ln4S^k+3k=0n5l=kn5p=ln5S^k}. By convention, we assume that k=00l=k0p=l0S^k=0,k=01l=k1p=l1S^k=0,k=02l=k2p=l2S^k=0,k=03l=k3p=l3S^k=0,k=04l=k4p=l4S^k=13,k=05l=k5p=l5S^k=79,k=06l=k6p=l6S^k=727,k=07l=k7p=l7S^k=12881.

Theorem 6. For all integers m,n0, we have

  • (a) Bn=An.
  • (b) C1An=AnC1.
  • (c) Cn+m=CnBm=BmCn.

Proof.

  • (a) Proof can be done by mathematical induction on n.
  • (b) After matrix multiplication, (b) follows.
  • (c) We have Cn=ACn1. From the last equation, using induction, we obtain Cn=An1C1. Now Cn+m=An+m1C1=An1AmC1=An1C1Am=CnBm and similarly Cn+m=BmCn.

Theorem 7. For m,n0, we have bn+m=bnk=0m+1l=km+1p=lm+1S^k+bn1(13k=0ml=kmp=lmS^k+14k=0m1l=km1p=lm1S^k7k=0m2l=km2p=lm2S^k+3k=0m3l=km3p=lm3S^k)+bn2(14k=0ml=kmp=lmS^k7k=0m1l=km1p=lm1S^k+3k=0m2l=km2p=lm2S^k)+bn3(7k=0ml=kmp=lmS^k+3k=0m1l=km1p=lm1S^k)+3bn4k=0ml=kmp=lmS^k.

Proof. From the equation Cn+m=CnBm=BmCn, we see that an element of Cn+m is the product of row Cn and a column Bm. From the last equation, we say that an element of Cn+m is the product of a row Cn and column Bm. We just compare the linear combination of the 2nd row and 1st column entries of the matrices Cn+m and CnBm. This completes the proof.

Corollary 14. For m,n0, we have J^n+m=J^nk=0m+1l=km+1p=lm+1S^k+J^n1(13k=0ml=kmp=lmS^k+14k=0m1l=km1p=lm1S^k7k=0m2l=km2p=lm2S^k+3k=0m3l=km3p=lm3S^k)+J^n2(14k=0ml=kmp=lmS^k7k=0m1l=km1p=lm1S^k+3k=0m2l=km2p=lm2S^k)+J^n3(7k=0ml=kmp=lmS^k+3k=0m1l=km1p=lm1S^k)+3J^n4k=0ml=kmp=lmS^k, j^n+m=j^nk=0m+1l=km+1p=lm+1S^k+j^n1(13k=0ml=kmp=lmS^k+14k=0m1l=km1p=lm1S^k7k=0m2l=km2p=lm2S^k+3k=0m3l=km3p=lm3S^k)+j^n2(14k=0ml=kmp=lmS^k7k=0m1l=km1p=lm1S^k+3k=0m2l=km2p=lm2S^k)+j^n3(7k=0ml=kmp=lmS^k+3k=0m1l=km1p=lm1S^k)+3j^n4k=0ml=kmp=lmS^k, S^n+m=S^nk=0m+1l=km+1p=lm+1S^k+S^n1(13k=0ml=kmp=lmS^k+14k=0m1l=km1p=lm1S^k7k=0m2l=km2p=lm2S^k+3k=0m3l=km3p=lm3S^k)+S^n2(14k=0ml=kmp=lmS^k7k=0m1l=km1p=lm1S^k+3k=0m2l=km2p=lm2S^k)+S^n3(7k=0ml=kmp=lmS^k+3k=0m1l=km1p=lm1S^k)+3S^n4k=0ml=kmp=lmS^k, and R^n+m=R^nk=0m+1l=km+1p=lm+1S^k+R^n1(13k=0ml=kmp=lmS^k+14k=0m1l=km1p=lm1S^k7k=0m2l=km2p=lm2S^k+3k=0m3l=km3p=lm3S^k)+R^n2(14k=0ml=kmp=lmS^k7k=0m1l=km1p=lm1S^k+3k=0m2l=km2p=lm2S^k)+R^n3(7k=0ml=kmp=lmS^k+3k=0m1l=km1p=lm1S^k)+3R^n4k=0ml=kmp=lmS^k.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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