On the Binomial Interpolated Triangles

Transcription

1 Journal of Integer Sequences, Vol. 20 (2017), Artcle On the Bnomal Interpolated Trangles László Németh Insttute of Mathematcs Unversty of Sopron Bajcsy Zs. u. 4 H9400 Sopron Hungary nemeth.laszlo@un-sopron.hu Abstract The bnomal nterpolated transform of a sequence s a generalzaton of the wellknown bnomal transform. We examne a Pascal-lke trangle, on whch a bnomal nterpolated transform works between the left and rght dagonals, focusng on bnary recurrences. We gve the sums of the elements n rows and n rsng dagonals, and we defne two specal classes of these arthmetcal trangles. 1 Introducton Let us defne the sequence b = {b n } n=0 R as the bnomal transform of the gven sequence a = {a n } n=0 R by b n = n n ( ) a. Ths transformaton s nvertble wth formula a n = n n ( ) ( 1) n b. Several researchers [1, 2, 3, 5, 7] examned the propertes and the generalzatons of the bnomal transformaton. One of ts generalzatons s the so-called bnomal nterpolated transform [1] gven by b n = n ( ) n u v n a (1) for any non-zero u,v R. Bhadoura at al. [2] and Falcon and Plaza [3] showed for some (fallng and rsng) k-bnomal transform cases, when u = 1, v = k; u = k, v = 1 or u = v = k, that there are specal nfnte (Pascal-lke) trangles, whose left dagonal (left leg) contans the terms of a and the rght one (rght leg) the terms of b. 1

2 ... In our paper, we examne the nner part of ths type of trangle n a generalzed form focusng exclusvely on the bnary recurrence sequences a. We determne the sums and alternatng sums of rows, rsng dagonals and central term sequence and gve explct forms for entres of the trangle. Fnally, we defne and examne two specal classes of our arthmetcal trangles. 2 Bnomal nterpolated trangle Let an arthmetcal trangle (called derangement trangle [2]) be defned by terms a n,k (k-th entry n row n see Fgure 1) from the sequence {a n,0 } n=0, where a n,k = ua n,k 1 +va n 1,k 1 (1 k n), (2) u,v R and uv 0. We shall show that the rght dagonal sequence s a bnomal nterpolated transform of the left dagonal sequence wth parameters u and v. Moreover, we shall prove that the converse also holds, wth the parameters v/u and 1/u. a n,k 1 a n 1,k 1 u v a n,k a 0,0 a 1,0 a 1,1 a n 1,k 1 v u 1 u a n,k 1 a n,k a 2,0 a 2,1 a 2,2 a 3,0 a 3,1 a 3,2 a 3,3 a 4,0 a 4,1 a 4,2 a 4,3 a 4, Fgure 1: Bnomal nterpolated trangle Theorem 1. For all nteger r, k such that k 0 k n, where k 0 0 s a fxed nteger, we have k k 0 ( ) k k0 a n,k = u v k k0 a n k+k0 +,k 0. (3) Proof. We prove t by nducton on n and k. If k = k 0, then the formula (3) trvally holds for any n. We suppose that (3) s true up to k 0 k 1. Let 1 k = k k 0. Then 2

3 a n,k = ua n,k 1 +va n 1,k 1 k 1 ( ) k 1 k 1 ( ) k 1 = u u v k 1 a n k++1,k0 +v u v k 1 a n k+,k0 = = k 1 ( ) k 1 k 1 ( ) k 1 u +1 v k 1 a n k++1,k0 + u v k a n k+,k0 k =1 ( ) k 1 k 1 u v k a 1 n k+,k0 + =1 ( k 1 ) u v k a n k+,k0 k 1 ( ) k 1 k 1 = u k a n,k0 + u v k a 1 n k+,k0 +v k a n k,k0 + k 1 (( ) k 1 = v k a n k,k = k =1 ( ) k u v k a n k+,k0. ( k 1 =1 ( k 1 )) u v k a n k+,k0 +u k a n,k0 ) u v k a n k+,k0 Consderng the substtutons k 0 = 0 and k = n, or consderng a fxed term a n0,k 0 leads us to the followng corollary. Corollary 2. The rght dagonal sequence s the bnomal nterpolated transform of the left dagonal sequence, so n ( ) n b n = a n,n = u v n a,0. (4) Furthermore, let us fx k 0 and n 0, so that 0 k 0 n 0. Then the terms a,j = a n,k form a bnomal nterpolated sub-trangle (see Fgure 2), where = n n 0, j = k k 0 n 0 n, k 0 k n (n 0 k 0 ). The sub-trangle s rght dagonal sequence s the bnomal nterpolated transform of ts left dagonal sequence. We now express the terms a n,k by the rght dagonal sequence {a n,n } n=0. Theorem 3. For any 0 k n n k ( n k a k n = )( 1 u ) ( v ) n k a k+ k+ u. (5) 3

4 ... a 0,0 k 0 a 1,0 a 1,1... n a 0,0. a 1,0. a 1, Fgure 2: Sub-trangle of the bnomal nterpolated trangle Proof. We suppose that the sequence {b n = a n,n } wth {b n } n=0 = {a n,n } n=0 s gven. Let U = 1/u and V = v/u, then the rght equalty s a n,k 1 = Ua n,k +Va n 1,k 1, accordng to (2). Based on ths connecton, we can wrte the entres of the bnomal nterpolated trangle (from rght to left) by b n,r, where b n,0 = b n, b n,r = Ub n,r 1 +Vb n 1,r 1 (1 r n). Usng relaton (3), proved n Theorem 1, we obtan b n,r = r ( ) r U V r b n r+,0. Snce b n,r = a n,n r, consderng the substtuton k = n r, the thess follows. If k = 0, as a drect consequence of Theorem 3, the nverse transformaton of the bnomal nterpolated transform (1) s a n,0 = n ( n )( 1 u ) ( v u) n a,. (6) 3 Bnary bnomal nterpolated trangle Let a n,0 = a n, where {a n } n=0 s a bnary recursve sequence defned by a n = αa n 1 +βa n 2 (2 n, α,β R, αβ 0), (7) wth ntal values a 0,a 1 R ( a 0 + a 1 0). Then we call the bnomal nterpolated trangle bnary bnomal nterpolated trangle and we let BIT (a 0,a 1,α,β;u,v) (n short BIT ) denote t. Bhadoura et al. [2] gave three specal examples, where the trangles are generated by the 4-Lucas sequence. They are BIT (2,4,4,1;1,1), BIT (2,4,4,1;4,1) and BIT (2,4,4,1;4,4). 4

5 From now on, we wll use two mportant varables A = uα+2v and B = u 2 β uvα v 2. Frst of all, we gve some recursve formulas for terms a n,k. The results of the next techncal lemma wll be useful durng the proofs of the further theorems. Lemma 4. The followng recurrence relatons hold a n,k = αa n 1,k +βa n 2,k (2 k +2 n), (8) a k n = uβ v a n 1,k B v a n 1,k 1 (2 k +1 n), (9) a n,k = uα+v a n 1,k + B u u a n 2,k 1 (2 k +1 n), (10) a n,k = (uα+v)a n 1,k 1 +uβa n 2,k 1 (2 k +1 n), (11) a n,k = u2 β +v 2 a n 1,k 1 ub v v a n 1,k 2 (2 k n), (12) a n,k = Aa n 1,k 1 +Ba n 2,k 2 (2 k n), (13) a n,k = 2uβ vα a n,k 1 B β β a n,k 2 (2 k n). (14) Proof. Frst we prove relaton (8) by nducton. For k = 0 t corresponds to defnton (7). Let us suppose that ths formula s true up to k 1. So, f x = a n 3,k 1 and y = a n 2,k 1, then a n 1,k 1 = βx+αy and a n,k 1 = αβx+(β +α 2 )y hold. Fgure 3, whch s a sutable part of Fgure 1, depcts t. Now a n 2,k = vx + uy, a n 1,k = uβx + (uα + v)y and a n,k = (uαβ+vβ)x+(uα 2 +vα+uβ)y, whch gves (8). Moreover, we proved wth t, that Fgure 3 could be any part of Fgure 1. In order to prove the further equatons we can also use the relevant part of Fgure 3. For example, n the case (11) a n,k = (uα+v)y +uβx = uβx+(uα+v)y. The relaton (13) gves that the rght dagonal sequence b (and the -th dagonals parallel to t) satsfes the bnary recurrence relaton n 2 b n = Ab n 1 +Bb n 2 (15) wth ntal values b 0 = a 0, b 1 = ua 1 +va 0 (generally, b 0 = a,0, b 1 = a +1,1, 0). Moreover, the system of equatons A = 0, B = 0 gves v = uα/2 and u 2 (α 2 + 4β) = 0. Thus, as the condton uv 0 holds, f v uα/2 or α 2 + 4β 0, then A + B = 0, otherwse A = B = 0. 5

6 x y vx+uy βx+αy αβx+(α 2 +β)y uβx+(uα+v)y (uαβ +vβ)x + (uα 2 +vα+uβ)y (u 2 β +v 2 )x + (u 2 α+2uv)y (u 2 αβ +2uvβ)x + (u 2 α 2 +u 2 β +2uvα+v 2 )y Fgure 3: A part of the growng 3.1 Sums and alternatng sums of rows Let s = {s n } n=0 be the sequence whose elements are the sums of the values belongng to the n-th row of a bnomal nterpolated trangle. We have s n = n a n,k = k=0 n k=0 k ( ) k u v k a n k+,0. (16) We obtan a lnear recurrence for s, whose coeffcents depend only on α, β, A and B,.e., the coeffcents of the bnary recurrences related to sequences a and b. Theorem 5. The sequence s satsfes the followng fourth order lnear homogeneous recurrence n 4 s n = (α+a)s n 1 +(β αa+b)s n 2 (αb +βa)s n 3 βbs n 4. (17) Proof. The sum of row n can be gven by the prevous two ones n form s n = α(s n 1 a n 1,n 1 )+βs n 2 +a n,n 1 +a n,n. (18) Snce ua n,n 1 = a n,n va n 1,n 1 then from (18) we obtan us n = uαs n 1 +uβs n 2 +(u+1)a n,n (αu+v)a n 1,n 1. (19) Applyng (13) and as A αu v = v we have us n = uαs n 1 +uβs n 2 +(Au+v)a n 1,n 1 +B(u+1)a n 2,n 2. (20) Now we transform (19) nto us n 1 = uαs n 2 +uβs n 3 +(u+1)a n 1,n 1 (αu+v)a n 2,n 2. (21) 6

7 Transformng agan (19) nto form us n 2 = uαs n and usng (13) we gan ubs n 2 = ub(αs n 3 +βs n 4 ) (αu+v)a n 1,n 1 +(B(u+1)+(αu+v)A)a n 2,n 2. (22) Fnally, let us express a n 1,n 1 and a n 2,n 2 from (20) and (21), respectvely. We substtute them nto (22) and obtan (17). Remark 6. It should be notced that the recurrence (18) s the mnmal order recurrence of s when B(u+1) 2 +(αu+v)(au+v) 0. Indeed, n ths case the determnant of coeffcents of the system (from (20) and (21)) { (Au+v)a n 1,n 1 +B(u+1)a n 2,n 2 = us n uαs n 1 uβs n 2 (u+1)a n 1,n 1 (αu+v)a n 2,n 2 = us n 1 uαs n 2 uβs n 3 s dfferent from 0, so the system has a unque soluton, otherwse when B(u+1) 2 +(αu+v)(au+v) = 0 or, equvalently, wth a lttle bt of calculatons, when B(u + 1) 2 + (α u)(v + uα) = 0 the mnmal order of the lnear recurrence of s should be less then four. For example, the soluton u = 1 and v = α mples that A = α and B = β. Then the relaton (20) becomes n 2 s n = αs n 1 +βs n 2. (23) Obvously, summng relatons s n = αs n 1 + βs n 2, αs n 1 = α 2 s n 2 αβs n 3 and βs n 2 = αβs n 3 β 2 s n 4 we fnd that the recurrence (17) also holds for sequence s, but t s not the mnmal order one. Let the sequence s = { s n } n=0 be the alternatng sum sequence, where the terms are the values of rows of a bnomal nterpolated trangle, so that n n k ( ) k s n = ( 1) k a n,k = ( 1) k u v k a n k+,0. k=0 k=0 Theorem 7. The sequence s satsfes the followng fourth order lnear homogeneous recurrence n 4 s n = (α A) s n 1 +(β +αa+b) s n 2 (αb βa) s n 3 βb s n 4. (24) Proof. Rowbyrowthesgnsoftheentresnthealternatngsumsdonotchangendrectons parallel to the left dagonal (sgn(( 1) k a n,k ) = sgn(( 1) k a n 1,k )), but parallel to the rght dagonal they change (sgn(( 1) k a n,k ) sgn(( 1) k 1 a n 1,k 1 )). Hence we only have to change the sgn of A n the summaton relaton (17). Remark 8. If u = 1 and v = α, then the relaton (24) becomes more smple, n 3 s n = (α 2 +2β) s n 2 β 2 s n 4. (25) 7

8 3.2 Rsng dagonal sum sequence The sequence d = {d n } n=0 of sums of elements n rsng dagonals has the followng defnton n n 2 n 0 d n = a k,n k = a n k,k. (26) k= n 2 k=0 When we consder the Pascal arthmetcal trangle t s well-known that these sums provde the Fbonacc sequence. We gve the recurrence relaton for the sequence of sums of elements belongng to rsng dagonal (.e., shallow dagonal as they are defned n Wolfram Math World [8]) of our bnomal nterpolated trangles. Theorem 9. The rsng dagonals sums sequence d of a BIT satsfes the sxth order lnear recurrence relaton n 6 D n = AD n 2 +BD n 4, (27) where D n = d n +αd n 1 +βd n 2. Moreover, for even n (n = 2k, n 2), D n = b k also holds. We pont out that wth the equaton (27) s a concse expresson for the followng sxth order recurrence relaton d n = αd n 1 +(β +A)d n 2 αad n 3 +( βa+b)d n 4 αbd n 5 βbd n 6. (28) Proof. Wthout loss of generalty, let n = 2k. The values of d n+2 and d n+3 can be gven by the prevous two ones n forms d n+2 = αd n+1 +βd n +a k+1,k+1, d n+3 = α(d n+2 a k+1,k+1 )+βd n+1 +a k+2,k+1, generally for even and odd ndexes we have d n+2 = αd n+2 1 +βd n+2 2 +a k+,k+, d n+2+1 = α(d n+2 a k+,k+ )+βd n+2 1 +a k++1,k+. When we consder D n, wth n even, we obtan and accordng to (13) 1 a k+,k+ = d n+2 αd n+2 1 βd n+2 2 = D n+2, D n+6 AD n+4 BD n+2 = a k+3,k+3 +Aa k+2,k+2 +Ba k+1,k+1 = 0. 8

9 If we deal wth the case D n, wth n odd, then, usng relatons (10) and (2), we fnd Thus d n+3 = αd n+2 +βd n+1 + v u a k+1,k+1 + B u a k,k, d n+5 = αd n+4 +βd n+3 + va+b a k+1,k+1 + vb u u a k,k, d n+7 = αd n+6 +βd n+5 + va2 +vb +AB a k+1,k+1 + u vab +B2 a k,k. u u(d n+7 AD n+5 BD n+3 ) = ( (va 2 +vb +AB)+A(vA+B)+vB)a k+1,k+1 +( (vab +B 2 )+vab +B 2 )a k,k = Central elements The central elements of the bnomal trangles are the terms a 2k,k wth k 0. In ths subsecton we gve the relaton between them, moreover t turns out that the before mentoned recurrence relaton holds for all the columns (whch are parallel to the vertcal axs of the trangle). These sequences are defned by {a 2k+l,k } k=k 0 wth l Z and { 0, f l 0; k 0 = (29) l, f l < 0. Moreover, f l = 0, then t s the sequence of the central elements. Theorem 10. All the sequences {a 2k+l,k } k=k 0 of the bnomal nterpolated trangle wth l Z and k 0 defned by (29) satsfy the same bnary homogeneous recurrence relaton where c k = a 2k+l,k. Proof. We have to prove that c k+2 = (α 2 u+αv +2βu)c k+1 βbc k, (30) a 2(k+2)+l,k+2 = (α 2 u+αv +2βu)a 2(k+1)+l,k+1 β(u 2 β uvα v 2 )a 2k+l,k. (31) Frst, supposng that uα+v 0 and recallng (11) and (8) we obtan a 2k+3+l,k+1 = (uα+v)a 2k+2+l,k +uβa 2k+1+l,k = (uα+v)(αa 2k+1+l,k +βa 2k+l,k )+uβa 2k+1+l,k = (uα 2 +vα+uβ) a 2k+2+l,k+1 uβa 2k+l,k uα+v +(uα+v)βa 2k+l,k, 9

10 furthermore a 2k+4+l,k+2 = (uα+v)a 2k+3+l,k+1 +uβa 2k+2+l,k+1 = (uα 2 +vα+uβ)(a 2k+2+l,k+1 uβa 2k+l,k )+(uα+v) 2 βa 2k+l,k +uβa 2k+2+l,k+1 = (uα 2 +vα+2uβ)a 2k+2+l,k+1 +β( u 2 β +uvα+v 2 )a 2k+l,k. Second, when v = uα the equalty (30) becomes or explctly c k+2 = 2uβc k+1 u 2 β 2 c k, a 2(k+2)+l,k+2 = 2uβa 2(k+1)+l,k+1 u 2 β 2 ca 2k+l,k. (32) Thus, snce the sequence s a geometrc progresson we have a 2(k+2)+l,k+2 = uβa 2(k+1)+l,k+1 = u 2 β 2 a 2k+l,k, and clearly (32) holds because t corresponds to the dentty uβa 2(k+1)+l,k+1 = 2uβa 2(k+1)+l,k+1 uβa 2(k+1)+l,k Explct formula Theorem 11. Let D = α 2 +4β. If D 0, then the explct formula of a n,k s a n,k = (vd+αv 2βu)a n,0 +2βa n,1 2vD where x 1 = (α+d)/2, x 2 = (α D)/2 and a n,0 = (D α)a 0,0 +2a 1,0 2D a n,1 = (D α)a 0,0 +2a 1,0 x n 1 2D ( ) k βu vx2 + β If D = 0 and A 0, then the explct formula s a n,k = (vd αv +2βu)a n,0 2βa n,1 2vD ( ) k βu vx1, (33) x n 1 + (D+α)a 0,0 2a 1,0 x n 2D 2, (34) 1 (ux 1 +v)+ (D+α)a 0,0 2a 1,0 2D ( ( αan,1 )) ( ) k a n,0 +k A a A n,0, α 10 β x n 1 2 (ux 2 +v). (35)

11 where ( a n,0 = a 0,0 +n 2a ) 1,0 αa (α ) n 0,0 (36) α 2 ( a n,1 = ua n,0 +v a 0,0 +(n 1) 2a ) 1,0 αa (α ) 0,0 n 1. (37) α 2 If D = 0 and A = 0, then ( a n,0 = a 0,0 +n 2a ) 1,0 αa (α ) n 0,0 (38) α 2 a n,1 = u 2a 1,0 αa ( 0,0 α ) n. (39) α 2 and a n,k = 0, f k 2. Proof. We suppose that α 2 + 4β 0. Frstly, we gve the explct form of the elements n case of the man path (k = 0). As the recurson s bnary wth coeffcent α and β, then the characterstc equaton of (7) s x 2 αx β = 0 wth roots x 1 x 2. Consequently, for all n 0 the term a n,0 s a lnear combnaton of the powers x n 1, x n 2. In order to determne the coeffcents of ths lnear combnaton, we need to solve the system of equatons px 1 +qx 2 = a,0, = 0,1. From ths system we fnd p = a 1,0 a 0,0 x 2 x 1 x 2 and q = a 1,0x 1 a 1,0. Thus observng that x 1 = a+d 2,x 2 = a D,x 1 x 2 = D, 2 x 1 x 2 where D = α 2 +4β, we easly fnd equalty (34). Moreover a n,1 = ua n,0 + va n 1,0 yelds (35). Secondly, we gve smlarly the explct formula of the sequence {a n,k } n k=0 wth ntal condtons a n,0 and a n,1. Because of (14), ts characterstc equaton s and the roots are 2βu αv +vd y 1 = 2β y 2 2βu αv y αuv βu2 +v 2 β β = u+ v(d α) 2β = βu vx 2 β = 0 and y 2 = βu vx 1. β When we consder the case D 2 = α 2 +4β = 0 and A 0 (or, equvalently, β = α 2 /4 and v = uα/2) we use the same method adopted before. Takng n account that the characterstc equaton x 2 αx β = 0 has the unque root x 0 = α/2, we have to solve the system (p+q)x 0 = a,0, = 0,1, whch provdes the coeffcents p, q for a n,0 = (p+qn)x n 0 n equalty (36). In ths case the characterstc equaton of sequence {a k n} n k=0 also has just one root y 0 = A/α 0 and the equatons ( p+ q)y 0 = a n, = 0,1 and k yeld the fnal formula. Fnally, we examne the case D 2 = α 2 +4β = 0 and A = 0. Now the equvalent condtons β = α 2 /4 and v = uα/2 mply that B = 0 and the equalty (37) s smplfed nto (39). Moreover f k 2, then the relaton (14) becomes a n,k = 0a n,k 1 +0a n,k 2 = 0. 11

12 4 Specal types of bnary bnomal nterpolated trangles The classcal Pascal trangle has vertcal symmetry and ts nner elements satsfy the wellknown rule of addton, namely every elements s the sum of the two terms drectly above t. In ths secton we gve the classes of our trangles whch have the same propertes, and we answer the queston, Is Pascal s trangle a bnomal nterpolated trangle? A bnomal nterpolated trangle s (vertcally) symmetrcal f a n,k = a n,n k (0 k n). Theorem 12. Let a 0,0 0 and a 1,0 be gven. A bnary bnomal nterpolated trangle s symmetrcal f and only f or where u 1 or u = 1, v = α = 2a 1,0 a 0,0 α = 2a 1,0, β = α2 a 0,0 4, v = α(u 1), 2 u = 1, v = α = 2a 1,0 a 0,0, β = α2 4. (See [1] for the bnomal transform n a specal case). Proof. Indeed clearly the condton a n,k = a n,n k mples the followng system of equatons α = A = uα+2v β = B = u 2 β uvα v 2. b 1 = a 1,0 = ua 1,0 +va 0,0 From the frst equaton we fnd v = α(u 1)/2 and substtutng n the second equaton we obtan, wth some calculatons, (α 2 +4β)(u+1)(u 1) = 0. Now, we have the three possbltes α 2 + 4β = 0, u = 1, u = 1. Obvously the case u = 1 mples v = 0, a contradcton. The case u = 1 gves v = α and α = 2a 1,0 /a 0,0. Consderng the case α 2 +4β = 0 we obtan the second part of the statement of the theorem. Fnally, the equaltes α 2 +4β = 0 and u = 1 yeld the thrd last case. Corollary 13. Let λ = a 1,0 /a 0,0 and a 0,0 0, u 1. Then the form of a symmetrcal bnomal nterpolated trangle can be wrtten by BIT (a 0,λa 0,2λ,β; 1,2λ), (40) 12

13 or BIT (a 0,λa 0,2λ, λ 2 ;u,λ(1 u)), (41) BIT (a 0,λa 0,2λ, λ 2 ; 1,2λ). (42) Fgure 4 shows a symmetrcal bnomal nterpolated trangle, namely BIT (2, 1, 1, 1; 1, 1) generated by Lucas numbers (A n [6]) Fgure 4: Symmetrcal bnomal nterpolated trangle BIT (2, 1, 1, 1; 1, 1) Remark 14. Accordng to (23) the recurson of s n the symmetrcal bnomal nterpolated trangles case s s n = 2λs n 1 +βs n 2 or s n = 2λs n 1 λ 2 s n 2. Moreover, when we consder the trangle (42) for the sequence d we obtan d n = λd n 1 +λd n 2 λ 2 d n 3. Theorem 15. Pascal s trangle s not a bnomal nterpolated trangle. Proof. The sum of above terms condton and relaton (9) should mply aβ/v = B/v = 1 and the vertcal symmetry mples β = B. Thus we fnd u = 1 and v = β. Now the frst set of solutons comng from Theorem 12 gves v = β = α = 2, snce n Pascal s trangle we have a 0,0 = a 1,0 = 1. The recurrence (7) becomes a n,0 = 2a n 1,0 2a n 2,0 whch s clearly not satsfed n Pascal s trangle by the frst elements (all equal to 1) of any row. From the second set of solutons we fnd α = 2, β = 1, v = 1 u, but, snce u = 1, from these solutons we must have v = 2 and on the other hand snce v = β = 1 we fnd a contradcton. Now we gve the condtons of u and v, so that an nner entry of BIT could be the sum of values not only left but also drectly above t by coeffcents u and v. Usng (9), we can gan t n two dfferent ways (see Fgure 5). Our cases are case 1: u = uβ v, v = B v, case 2: v = uβ v, u = B v. 13

14 a n 1,k 1 v a n,k a n 1,k u a n 1,k 1 a n 1,k u v a n,k Fgure 5: Coeffcents of summng downwards a Case 1 a 1 βa 0 +αa 1 βa 0 +αa 1 αβa 0 + (α 2 +β)a 1 (α 2 β +β 2 )a 0 + (α 3 +2αβ)a 1 αβa 0 + (α 2 +β)a 1 (α 2 β +β 2 )a 0 + (α 3 +2αβ)a 1 (α 3 β +2αβ 2 )a 0 + (α 4 +3α 2 β +β 2 )a 1 (α 4 β +3α 2 β 2 +β 3 )a 0 + (α 5 +4α 3 β +3αβ 2 )a 1 Fgure 6: BIT (a 0,a 1,α,β;α,β) Solvng the system { u = uβ v v = B v we can easly obtan that u = α and v = β, thus the trangle s, BIT (a 0,a 1,α,β;α,β). (43) We derve some nterestng propertes of ths trangle. From Fgure 6, whch shows the frst four rows of trangle BIT (a 0,a 1,α,β;α,β) (see also Fgure 7), we notce that the rows and the left dagonal satsfy the same recurrence. Indeed, usng the equaltes u = α, v = β, the recurrence relaton (14) becomes a n,k = αa n,k 1 + βa n,k 2. Moreover, the terms along the rsng dagonals are the same. The next theorem and corollary wll provde t precsely. β a n+1,k 1 a n,k 1 a n 1,k 1 α β a n,k α β α a n 1,k α a n,k+1 Fgure 7: Coeffcents of summng n case 1 14

15 Theorem 16. The entres n trangle BIT (a 0,a 1,α,β;α,β) can be wrtten (n 1) by a n,k = a 0 n+k 2 2 ( n+k 2 ) α n+k 2 2 β +1 +a 1 n+k 1 2 ( n+k 1 ) α n+k 2 1 β. Proof. We consder the results of Theorem 11. Indeed, usng the equaltes u = α, v = β and x 1 +x 2 = α, αx 1 +β = x 2 1, αx 2 +β = x 2 2, D +α = 2x 1, D α = 2x 2 the equatons (33) (35) become a n,k = a n,0x 2 +a n,1 D where D = α 2 +4β = x 1 x 2 0 and a n,0 = a 0x 2 +a 1 D a n,1 = a 0x 2 +a 1 x n+1 D A lttle calculaton shows that ( x n+k 1 2 a n,k = a 0 β x 1 x 2 Usng the Grard-Warng formula [4] 1 x n+k 1 x k 1 + a n,0x 1 a n,1 x k D 2, x n 1 + a 0x 1 a 1 x n D 2, 1 + a 0x 1 a 1 D )+a 1 ( x n+k x n x2 n+k x 1 x 2 X N+1 Y N+1 N 2 ( ) N = ( 1) (X +Y) N 2 (XY), X Y ). (44) where, n our case, X = x 1, Y = x 2, X + Y = α, XY = β and N = n + k 2 or N = n+k 1, the thess follows. The Grard-Warng formula also holds n the case x 1 = x 2,.e., D = 0, takng the lmt x 1 x 2 on both members. We menton that D = 0 mples A 0. Corollary 17. In case of the trangle BIT (a 0,a 1,α,β;α,β) the rsng dagonal sequence {a n k,k } n 2 k=0 s a constant sequence for any k. Proof. We apply the formula proved n Theorem 16 to a n k,k and obtan an expresson that does not depend on the ndex k. Fgure 8 shows BIT (a 0,a 1,1,1;1,1) as an example for case 1. Usng the result of Theorem 16, n the case α = 1 and β = 1, the bnomal coeffcents are the coeffcents of the elementsnthersngdagonals. Thusa n,k = a 0 f n+k 1 +a 1 f n+k, wheref n sthen th Fbonacc number (A000045), so that, f 0 = 0,f 1 = f 1 = 1. (We also fnd from the above relaton (44) the connecton wth Fbonacc numbers.) The specal Fbonacc bnomal nterpolated trangle s BIT (1,1,1,1;1,1) (A199512). Falcon and Plaza [3] provded an other example n Table 4 for ths case, namely BIT (0,1,3,1;3,1), whch s generated by the 3-Fbonacc sequence (A006190). 15

16 a Case 2 a 1 p+a 1 a 0 +a 1 a 0 +2a 1 2a 0 +3a 1 a 0 +2a 1 2a 0 +3a 1 3a 0 +5a 1 5a 0 +8a 1 2a 0 +3a 1 3a 0 +5a 1 5a 0 +8a 1 8a 0 +13a 1 13a 0 +21a 1 From the equatons system we obtan Fgure 8: BIT (a 0,a 1,1,1;1,1) v 1 = α 1+ (α 1) 2 +4β, u 1 = v2 1 2 β, v 2 = α 1 (α 1) 2 +4β, u 2 = v2 2 2 β. If (α 1) 2 +4β = 0,.e., β = (1/4)(α 1) 2, then, replacng the correspondng values for u and v, after smplfcaton we have BIT (a 0,a 1,α, (α 1)2 ; 1, α 1 ). (45) 4 2 If (α 1) 2 +4β 0, then the trangles are BIT (a 0,a 1,α,β;u 1,v 1 ), (46) BIT (a 0,a 1,α,β;u 2,v 2 ). (47) Fnally, we compare the trangles (43) (47) wth the symmetrcal ones (40) (42). In all the cases for the symmetrcal bnary nterpolated bnomal trangle wth some calculatons we gan the followng corollary. Corollary 18. The symmetrcal bnary nterpolated bnomal trangles whose an nner entry could be the sum of values not only left but also drectly above t by coeffcents u and v are whch has only terms a 0 and ±a 0 /2 and whch has only terms a 0. BIT (a 0, a 0 2, 1, 1; 1, 1), BIT (a 0,a 0,2, 1;2, 1), 16

17 Remark 19. When both parameters a 0,0, a 1,0 are zero we fnd the trval trangle exclusvely composed of null entres. The trangle wth all entres equal to 1 correspond to more bnomal nterpolated trangles, for example, BIT (1,1,2, 1; 1,2) or BIT (1,1,2, 1; 1,2). The trangle whose left dagonal and rows are the sequences of natural numbers (A000027) s BIT (1,2,2, 1;2, 1) (A094727). 5 Acknowledgment The author would lke to thank the anonymous referee for carefully readng the manuscrpt and for hs/her useful suggestons and mprovements. References [1] S. Barbero, U. Cerrut and N. Murru, A generalzaton of the bnomal nterpolated operator and ts acton on lnear recurrent sequences, J. Integer Sequences 13 (2010), Artcle [2] P. Bhadoura, D. Jhala and B. Sngh, Bnomal transforms of the k-lucas sequences and ts propertes, J. Math. Computer Sc. 8 (2014), [3] S. Falcon and A. Plaza, Bnomal transform of k-fbonacc sequence, Int. J. Nonln. Sc. Num. 10 (2009), [4] H. W. Gould, The Grard-Warng power sum formulas for symmetrc functons and Fbonacc sequences, Fbonacc Quart. 37 (1999) [5] J. Pan, Some propertes of the multple bnomal transform and the Hankel transform of shfted sequences, J. Integer Sequences 14 (2011), Artcle [6] N. J. A. Sloane, The On-Lne Encyclopeda of Integer Sequences, [7] M. Z. Spvey and L. L. Stel, The k-bnomal transforms and the Hankel transform, J. Integer Sequences 9 (2006), Artcle [8] Wolfram Math World, Pascal s trangle, Mathematcs Subject Classfcaton: Prmary 11B37; Secondary 11B39, 11B65, 11B75. Keywords: bnomal operator, bnomal transform, bnomal nterpolated transform, recurrence sequence. 17

18 (Concerned wth sequences A000027, A000032, A000045, A006190, A094727, and A ) Receved Aprl ; revsed verson receved July Publshed n Journal of Integer Sequences, July Return to Journal of Integer Sequences home page. 18