Power sums in hyperbolic Pascal triangles

Transcription

1 Power sums n hyperbolc Pascal trangles László Németh, László Szalay arxv: v1 [mathco] 15 Mar 2017 Abstract In ths paper, we descrbe a method to determne the power sum of the elements of the rows n the hyperbolc Pascal trangles correspondng to {4, q} wth q 5 The method s based on the theory of lnear recurrences, and the results are demonstrated by evaluatng the k th power sum n the range 2 k 11 Key Words: Pascal trangle, Hyperbolc Pascal trangle, Bnomal coeffcents, Power sum MSC code: 05A10, 11B65, 11B99 The fnal publcaton s avalable at Analele Stntfce ale Unverstat Ovdus Constanta, (wwwanstuocmathro) 1 Introducton Bnomal coeffcents, Pascal s trangle and ts generalzatons have been wdely studed One of the examned propertes s the power sum S k (n) = n =0 ( ) k n, n 0, k {0, 1, 2, } Apart from S 0 (n) = n + 1, S 1 (n) = 2 n, and S 2 (n) = ( ) 2n n, there s no general soluton for S k (n), but recurrences are gven for the cases 3 k 10 For example, Franel [5, 6] obtaned the recurrence S 3 (n + 1) = 7n2 + 7n + 2 S (n + 1) 2 3 (n) + 8n2 (n + 1) S 3(n 1) 2 for k = 3 The dffculty of the problem s also ndcated by the result, that there s no closed form, only asymptotc formula for S k (n) n the cases 3 k 9 For more detals and further references of power sums of bnomal coeffcent see [1, 4, 11] In ths artcle, we nvestgate the analogous queston, and gve a general method to determne the power sums n hyperbolc Pascal trangles, whch are a recently dscovered geometrcal generalzatons of Pascal s classcal trangle The method s llustrated by evaluatng the k th power sum n the range 2 k 11, and we calculate the correspondng Unversty of Sopron, Insttute of Mathematcs, Hungary nemethlaszlo@un-sopronhu Unversty J Selye, Department of Mathematcs and Informatcs, Slovaka; and Unversty of Sopron, Insttute of Mathematcs, Hungary szalaylaszlo@un-sopronhu 1

2 2 L Németh, L Szalay recurrences up to k = 11 A hyperbolc Pascal trangle contans two types of elements (say A and B) n the rows, therefore ts structure s more complcated than the Pascal s trangle s, but at the same tme the exstence of elements havng type B facltates the determnaton of power sums snce they behave as separatng tems between elements havng type A A short summary on hyperbolc Pascal trangles can be found n the next subsecton 11 Hyperbolc Pascal trangles In the hyperbolc plane there are nfnte types of regular mosacs, they are denoted by Schläfl s symbol {p, q}, where the postve ntegers p and q has the property (p 2)(q 2) > 4 Each regular mosac nduces a so-called hyperbolc Pascal trangle (see [2]), followng and generalzng the connecton between the classcal Pascal s trangle and the Eucldean regular square mosac {4, 4} For more detals see [2, 8, 9], but here we also collect some necessary nformaton There are several approaches to generalze Pascal s arthmetc trangle (see, for nstance [3]) The hyperbolc Pascal trangle based on the mosac {p, q} can be fgured as a dgraph, where the vertces and the edges are the vertces and the edges of a well defned part of the lattce {p, q}, respectvely, further each vertex possesses the value whch gves the number of dfferent shortest paths from the base vertex Fgure 1 llustrates the hyperbolc Pascal trangle HPT p,q, when {p, q} = {4, 6} Fgure 1: Hyperbolc Pascal trangle lnked to {4, 6} up to row 5 Generally, for {4, q} the base vertex has two edges, the leftmost and the rghtmost vertces have three, the others have q edges The square shaped cells surrounded by approprate edges are correspondng to the squares n the regular mosac Apart from the wnger elements, certan vertces (called Type A for convenence) have two ascendants and q 2 descendants, the others ( Type B ) have one ascendant and q 1 descendants In the fgures, we denote the vertces type A by red crcles and the vertces type B by cyan damonds, further the wngers by whte damonds The vertces whch are n-edge-long far from the base vertex are n row n The general method of drawng s the followng Gong along the vertces of the j th row, accordng to type of the elements (wnger, A, B), we draw approprate number of edges downwards (2, q 2, q 1, respectvely) Neghbour edges of two neghbour vertces of the

3 Power sums n hyperbolc Pascal trangles 3 j th row meet n the (j + 1) th row, constructng a vertex wth type A The other descendants of row j n row j + 1 have type B In the sequel, ) n ( denotes the k kth element n row n, whch s ether the sum of the values of ts two ascendants or the value of ts unque ascendant We note, that the hyperbolc Pascal trangle has the property of vertcal symmetry In the remanng part of the paper we consder the wnger nodes (havng value 1) as elements wth type B Denote by a n and b n the number of vertces of type A and B n row n, respectvely, further let s n = a n + b n, whch gves the total number of the vertces of row n 0 Then the ternary homogeneous recurrence relaton s n = (q 1)s n 1 (q 1)s n 2 + s n 3 (n 4), holds wth ntal values s 1 = 2, s 2 = 3, s 3 = q (see [2]) Moreover, let â n, ˆb n and ŝ n denote the sum of type A, type B and all elements of the n th row, respectvely It was justfed n [2] that the three sequences {â n }, {ˆb n } and {ŝ n } can be descrbed by the same ternary homogenous recurrence relaton ther ntal values are x n = qx n 1 (q + 1)x n 2 + 2x n 3 (n 4), â 1 = 0, â 2 = 2, â 3 = 6; ˆb1 = 2, ˆb 2 = 2, ˆb 3 = 2q 6; ŝ 1 = 2, ŝ 2 = 4, ŝ 3 = 2q 12 New results In ths artcle, we descrbe a method to determne the k th -power sum (s k ) n = s n 1 We wll llustrate the technque by the cases 2 k 11 The results are gven by dfferent lnear recursons Note that s n = (s 0 ) n, further ŝ n = (s 1 ) n Hence we nterested n the problem f k 2 Here we also ntroduce the notaton (a k ) n and (b k ) n for the sum of the k th power of elements of type A and B n row n, respectvely Clearly, (s k ) n = (a k ) n + (b k ) n 2 The method Let us start wth a further techncal note Take two consecutve elements ) n l ( and ) n l+1 ( of HPT 4,q, and let denote X and Y the not necessarly dstnct types of them, respectvely Let the th and j th power-product of them s denoted by x y j = ) n l ( n ) l+1 (j, and the sum of all such products (where the frst term has type X, the second does Y ) of row n by (x y j ) n For example, n case of HPT 4,6 we fnd (a 2 b 3 ) 3 = = 81 and (bb 3 ) 3 = = 16 (see Fgure 1) Clearly, by the vertcal symmetry of HPT 4,q, (a b j ) n = (b j a ) n holds An other easy but mportant observaton s that (b 1 b) n = (b 2 b 2 ) n = = (bb 1 ) n, snce two consecutve elements havng type B are equal =0 ) n (k

4 4 L Németh, L Szalay 21 Case k = 2 Before the general descrpton, we restrct ourselves to the sum of squares to facltate the justfcaton of the method Hence fx k = 2, and take (s 2 ) n = s n 1 =0 ) n (2 = (a 2 ) n + (b 2 ) n Accordng to the the structure of the trangle HPT 4,q and the expanson of (x + y) 2, we wll also use the sums (ab) n, (ba) n and (bb) n Frst consder the sum (a 2 ) n+1 Observe that an element of type A n row n + 1 s a sum of two elements ether of a type A and B or of a type B from row n Snce, apart from the wngers, each element of row n takes part twce n constructng the elements of type A n row n + 1, then (a 2 ) n+1 = 2(a 2 ) n + ( ) 2 (ab) n + 1 ( ) 2 (ba) n + 1 ( ) 2 (bb) n + 2(b 2 ) n 2 1 = 2(a 2 ) n + 2(ab) n + 2(ba) n + 2(bb) n + 2(b 2 ) n 2 Clearly, an element of type B n row n + 1 concdes an element of type ether A ((q 4)- tmes) or B ((q 3)-tmes, apart from the wngers) from row n, thus we see (b 2 ) n+1 = (q 4)(a 2 ) n + (q 3)(b 2 ) n 2(q 4), where 2(q 4) s agan the correcton caused by the wngers To explan (ab) n+1 = (a 2 ) n + (ab) n + (ba) n + (bb) n + (b 2 ) n 1, we put together the constructon rule of elements type A and B, respectvely, and the fact we need to consder the neghbour pars A, B (n type) Snce the left wnger 1 does not appear n (ab) n+1, we have the correcton 1 By the vertcal symmetry, the relaton (ab) n+1 = (ba) n+1 (1) holds Fnally, (bb) n+1 = (q 5)(a 2 ) n + (q 4)(b 2 ) n 2(q 4) holds snce two neghbours of type B are comng from an element of type ether A or B n drect manner As a summary, usng (1) we have the system (a 2 ) n+1 = 2(a 2 ) n + 4(ab) n + 2(b 2 ) n + 2(bb) n 2, (ab) n+1 = (a 2 ) n + 2(ab) n + (b 2 ) n + (bb) n 1, (b 2 ) n+1 = (q 4)(a 2 ) n + (q 3)(b 2 ) n 2(q 4), (bb) n+1 = (q 5)(a 2 ) n + (q 4)(b 2 ) n 2(q 4) (2)

5 Power sums n hyperbolc Pascal trangles 5 22 General case Recall that (b k 1 b) n = (b k 2 b 2 ) n = = (bb k 1 ) n In the sequel, we wll denote them by u n We use analoguously the consderatons and the explanatons from the prevous subsecton (where the case k = 2 was handled), and together wth the bnomal theorem we gan the system of recurrence equatons (a k ) n+1 = (a k j b j ) n+1 = k ( ) k k ( ) ( k ( ) ) k k (a k b ) n + (b k a ) n + 2 u n 2, =0 =0 =0 k j ( ) k j k j ( ) k j (a k j b j+ ) n + (b k j a j+ ) n =0 =0 ( k j ( ) ) k j + 1 u n 1, (3) =0 (b k ) n+1 = (q 4)(a k ) n + (q 3)(b k ) n 2(q 4), u n+1 = (q 5)(a k ) n + (q 4)(b k ) n 2(q 4) wth k + 2 sequences, where j = 1,, k 1 To be more cut-clear, we equvalently have k 1 ( ) k (a k ) n+1 = 2(a k ) n + 2 (a k b ) n + 2(b k ) n + (2 k 2)u n 2, =1 k j 1 ( k j 1 k j (a k j b j ) n+1 = (a k ) n + =0 ) (a k j b j+ ) n + =0 ( k j ) (a j+ b k j ) n +(b k ) n + (2 k j 1)u n 1, (4) (b k ) n+1 = (q 4)(a k ) n + (q 3)(b k ) n 2(q 4), u n+1 = (q 5)(a k ) n + (q 4)(b k ) n 2(q 4) Now we clam to elmnate the sequences (a k ) n and (b k ) n from (4) 23 Background to elmnate (a k ) n and (b k ) n from the system (4) Our purpose s to elmnate the sequences (a k ) n and (b k ) n by gvng a recursve formula to descrbe them In the next part, we develope a tool for handlng such problems n general System (4) can be nterpreted as a vector recurson of the form ḡ t+1 = Mḡ t + h gven by a sutable matrx M and vector h, when we consder (a k ) n, (a k 1 b) n,, (b k ) n, u n as coordnate sequences of ḡ n Frst we concentrate on the homogenous case, e when h s the zero vector 0 Let ν 2 be an nteger, further let ḡ 0,, ḡ ν 1 R ν denote lnearly ndependent ntal vectors Wth the coeffcents α 0,, α ν 1 R we can set up the homogenous lnear vector recurson ḡ t = α ν 1 ḡ t α 0 ḡ t ν (t ν) (5)

6 6 L Németh, L Szalay Lemma 21 There exsts unquely a matrx M R ν ν such that ḡ t+1 = Mḡ t (6) holds for any t N Moreover, f (6) holds for a vector sequence ḡ t wth a gven (not necessarly regular) matrx M R ν ν, then ḡ t satsfes (5), where the coeffcents concde the negatve of the coeffcents of the characterstc polynomal of M Proof Obvously, by (5) one can obtan ḡ ν R ν Put G = [ḡ 0,, ḡ ν 1 ] R ν ν and G = [ḡ 1,, ḡ ν ] R ν ν, where the matrx G s clearly regular Frst we show that the only possblty s M = G G 1 Assume that M satsfes (6) f t = 0,, ν 1 The system of vector equatons ḡ 1 = Mḡ 0 ḡ ν = Mḡ ν 1 s equvalent to the matrx equaton G = MG But G s regular, therefore M = G G 1 really exsts Now we justfy that M = G G 1 s sutable for arbtrary subscrpt t, e (6) holds for any t N The defnton of the matrx M proves the statement for t = 0, 1,, ν 1 In order to show for arbtrary t, we use the technque of nducton Hence suppose that (6) holds for t = 0, 1,, τ Then ḡ τ+1 = α ν 1 ḡ τ + + α 0 ḡ τ ν+1 = α ν 1 G G 1 ḡ τ α 0 G G 1 ḡ τ ν = G G 1 (α ν 1 ḡ τ α 0 ḡ τ ν ) = G G 1 ḡ τ Fnally, we prove that f M satsfes (6), then we arrve at (5) wth the gven condtons Let k(x) = x ν α ν 1 x ν 1 α 0 denote the characterstc polynomal of M By the Cayley Hamlton theorem, k(m) s zero a matrx, consequently k(m)ḡ 0 = 0, e M ν ḡ 0 α ν 1 M ν 1 ḡ 0 α 0 ḡ 0 = 0 (7) Snce for any natural number t the equalty ḡ t+1 = Mḡ t mples ḡ t+1 = M t+1 ḡ 0, (7) can be rewrtten as ḡ ν = α ν 1 ḡ ν α 0 ḡ 0 Further and smlarly holds ḡ ν+1 = Mḡ ν = M(α ν 1 ḡ ν α 0 ḡ 0 ) = α ν 1 ḡ ν + + α 0 ḡ 1, ḡ t+1 = Mḡ t = α ν 1 ḡ t + + α 0 ḡ t ν (t > ν) Remark 1 In ths lemma, we proved a bt more than we need to nvestgate the power sums n hyyperbolc Pascal trangles Indeed, we wll use only the drecton whch provdes the coeffcents α from the characterstc polynomal of M A statement of [10] also proves the second part of Lemma 21

7 Power sums n hyperbolc Pascal trangles 7 24 Connecton between the homogenous and nhomogenous vector recurrences Snce system (4) s nhomogenous, and Lemma 21 s able to handle only homogenous vector recursons, therefore we must clarfy the transt between them Assume that the vector h s fxed, and we have ḡ t+1 = Mḡ t + h (t N), further suppose that the characterstc polynomal of M s k(x) = x ν α ν 1 x ν 1 α 0 Obvously, ḡ t+2 = Mḡ t+1 + h, and ḡ t+2 ḡ t+1 = M(ḡ t+1 ḡ t ) Put d t = ḡ t+1 ḡ t Thus d t+1 = M d t mples d t = α ν 1 dt α 0 dt ν, and then ḡ t+1 = (α ν 1 + 1)ḡ t + (α ν 2 α ν 1 )ḡ t (α 0 α 1 )ḡ t+1 ν α 0 ḡ t ν Fnally, observe that (x 1)k(x) = x ν+1 (α ν 1 + 1)x ν (α ν 2 α ν 1 )x ν 1 (α 0 α 1 )x + α 0, where k(x) s the characterstc polynomal of M 25 Determnaton of the characterstc polynomal Consder now the matrx M = [m,j ] N (k+2) (k+2) generated by the system (4) If we ntroduce bnomal coeffcents wth negatve lower ndex, or when the lower ndex s greater then the upper ndex, and take such a value zero (ths s one of the conventonal approaches), the left upper k (k+1) mnor matrx can smply be gven Indeed, f 0 k 1, 0 j k, by (4), ( ) ( ) ( ) ( ) k k k k m,j = + = + k j j j k j Further m 0,k+1 = 2 k 2, and f 1 k 1 and j = k + 1, then m,j = 2 k 1 Moreover, m k,0 = q 4, m k,k = q 3, m k+1,0 = q 5, m k+1,k = q 4 The entres have not been lsted up here (m,j wth k k + 1, 1 j k + 1, j k) are zero Obvously (see (4)), n our case the vector h = [ (q 4) 2(q 4) ]

8 8 L Németh, L Szalay By vsualzng the nformatons we gan M = 2 k 2 2 k 1 1 ( k ) ( j + k k j) q q 3 0 q q 4 0 As usual, we wll use the formula k(x) = det(xi M) for determnng the characterstc polynomal of the matrx M, where I s the unt matrx We carry out the evaluaton n two steps, fnally we obtan a qute smple-lookng determnant Frst we deal wth M tself Let R u denote the row u of the matrx M The followng lemma descrbes row equvalent transformatons of M, whch keeps the value of the determnant Before applyng the lemma, frst we make a preparaton step (agan a row equvalent transformaton) for modfyng M, namely Rk new = R k R k+1 to result the penultmate row [ ] Lemma 22 For u = 0, 1,, k 1 apply successvely the row equvalent operaton R new u = ( ) δ ( 1) t R u+t, (8) t where δ = k u 1 Then the elements of the left upper (k + 1) (k + 1) mnor of M new are zero, except the man dagonal, and the antdagonal elements whch are 1 If the man dagonal and antdagonal meet, the common element s 2 Further n the last column we have m new,k+1 Proof Let j be arbtrarly fxed Then m new u,j = = = = 1 f 1 k 1, and mnew 0,k+1 = 0 ( ) (( ) ( )) δ k u t k u t ( 1) t + t j k j ( )( ) δ k u t ( )( ) δ k u t ( 1) t + ( 1) t t j t k j ( ) ( ) k u δ k u δ +, j δ k j δ where the last equalty s a consequence of the more general dentty ( )( ) δ z t ( 1) t = t r ( ) z δ r δ

9 Power sums n hyperbolc Pascal trangles 9 (see, for nstance, page 28 n [7]) Clearly, δ = k u mples ( ) k u δ + j δ ( ) k u δ = k j δ 0, f j k u and j u; 1, f ether j = k u or j = u; 2, f j = k u and j = u Note that the last case can be occurred only f k = 2u Contnung wth the last column, for 1 u < k we easly conclude m new u,k+1 = ( δ (2 ( 1) t) t k u t 1 ) = ( δ = 2 k u ( 1) t t ) ( 1 2 ( ) δ ( 1) t 2 k u t t ) t 0 = 2 k u ( 1 1 2) δ = 1 ( ) δ ( 1) t t Clearly, the same argument leads to 0 n case of the last element of the 0th row Now let M = M xi, and we wll carry out the same row equvalent operatons on M what we dd on M Only one modfcaton s, that after the preparaton step we nsert an addtonal operaton Namely, Rk new = R k R k+1 s followed by Rk+1 new = R k+1 (q 5)R k Hence keepng tabs on the last two rows of M, we see [ q (q 3) x 0 q q 4 x ] = [ x x (q 5)x (q 4)x Snce M = M xi effects the elements only n the dagonal of M (by x), accordng to (8), for 0 u < j k we fnd ( ) δ m new u,j = ( 1) j u+1 x j u Fnally, the last column of M new s gven by the vector [ ( 1) k x 1 + ( 1) k 1 x 1 + x 1 x x (q 4)x ] ] Hence, we conclude that M new s the sum of x ( ( k 1) x k ) ( ) ( 2 x ( 1) k k k 1 x ( 1) k+1 k k) x ( 1) k x 0 x ( ) ( ) ( k 1 1 x ( 1) k 1 k 1 k 2 x ( 1) k k 1 k 1) x ( 1) k 1 x 0 0 x ( 1) k 2( ) ( k 2 k 3 x ( 1) k 1 k 2 k 2) x ( 1) k 2 x ( x 1 ) 1 x x x x (q 5)x (q 4)x

10 10 L Németh, L Szalay and At the end, the characterstc polynomal k(x) = ( 1) k det( M new ), and by the prevous subsecton we can return the nhomogenous case wth the polynomal (x 1)k(x) = (x 1)( 1) k det( M new ) In the range 1 k 11 the appendx wll provde the results, snce knowng the characterstc polynomal, one can obtan drectly the correspondng recursve formula 26 Example: return to the case k = 2 System (2) provdes the matrx M, and then M = M xi = 2 x x 1 1 q 4 0 q 3 x 0 q 5 0 q 4 x The consecutve row equvalent transformatons R 2 = R 2 R 3, R 3 (q 5)R 2, and then R 0 = R 0 ( ) 2 R1 + ( ) 2 R2, R = R 1 ( 1 1) R2 return wth x 2x x x M new = 0 x x x 0 0 x x (q 5)x (q 4)x Thus the determnant and then we obtan k 1 (x) = det( M new ) = x 4 (q + 1)x 3 + 6x 2 2x, k(x) = (x 1)( 1) 2 k 1 (x) = x 5 (q + 2)x 4 + (q + 7)x 3 8x 2 + 2x, whch provde the recursve rule (s 2 ) n = (q + 2)(s 2 ) n 1 (q + 7)(s 2 ) n 2 + 8(s 2 ) n 3 2(s 2 ) n 4 The ntal values can be gven from the frst 4 rows of HPT 4,q : (s 2 ) 1 = 2, (s 2 ) 2 = 6, (s 2 ) 3 = 4q + 4, (s 3 ) 4 = 4q 2 + 6q 20

11 Power sums n hyperbolc Pascal trangles 11 3 Appendx and conjectures The method works for arbtrary postve nteger k In the next table we collect the coeffcents of (s k ) n = c j (q)(s k ) n j j=1 k c 1 (q) c 2 (q) c 3 (q) c 4 (q) c 5 (q) 0 q 1 q q q q + 2 q q + 4 q 19 2q q + 7 6q 41 7q q q 71 9q 17 10q q q 99 17q q q q q q q q q q q q q q q q q q q q q q q q q q q k c 6 (q) c 7 (q) c 8 (q) q q Partally from the results we have the followng Conjecture 1 Let k be gven Then the lnear recurrence correspondng to k has order k/2 + 3 Further the coeffcents c j (q) are lnear polynomals n q Our conjecture s based not only on the frst few cases but also on the followng approach We are able to decrease the sze of (3) whch entals the reducton of the dmenson of matrx M (but we loose ts smplcty) The new system of [k/2] + 3 recurrent sequences s the followng

12 12 L Németh, L Szalay Put l = (k 1)/2 and m = (k 1)/2 Obvously, f k s odd, then m = l, otherwse m = l + 1 Let (c j ) n = (a k j b j ) n + (a j b k j ) n, where j = 1,, l Moreover, f k s even, then let (c l+1 ) n = (a l+1 b l+1 ) n Now the new system of recurrences admts m ( ) k (a k ) n+1 = 2(a k ) n + 2 (c ) n + 2(b k ) n + (2 k 2)u n 2, =1 (b k ) n+1 = (q 4)(a k ) n + (q 3)(b k ) n 2(q 4), m ( ) k j m (( ) k j (c j ) n+1 = (a k ) n + (c ) n + + j =j =1 +(b k ) n + (2 k j 1)u n 1, u n+1 = (q 5)(a k ) n + (q 4)(b k ) n 2(q 4) where j = 1,, l When k s even, thenthere s an addton sequence References l+1 ( ) k l 1 (c l+1 ) n+1 = (a k ) n + (c l+1 ) n + (c ) n =1 +(b k ) n + (2 k l 1 1)u n 1 ( )) j (c ) n [1] Askar Dzhumadl daev, Damr Yelusszov, Power Sums of Bnomal Coeffcents, J Integer Seq, 16, (2013), Artcle 1314 [2] Belbachr, H, Németh, L, Szalay, L, Hyperbolc Pascal trangles, Appl Math Comp, 273 (2016), [3] Belbachr, H and Szalay, L, On the arthmetc trangles, Saula Math Sem, 9 (17) (2014), [4] Cusck,T W, Recurrences for sums of powers of bnomal coeffcents, Journal of Combnatoral Theory, Seres A, 52 (1), (1989), [5] Franel, J, On a queston of Lasant, L ntermédare des mathématcens, 1 (1894), [6] Franel, J, On a queston of J Franel, L ntermédare des mathématcens, 2 (1895), [7] Gould, H W, Combnatoral denttes, Morgantown, W Va, 1972 [8] Németh, L Szalay, L, Alternatng sums n hyperbolc Pascal trangles, Mskolc Mathematcal Notes, 17 (2), (2016), [9] Németh, L Szalay, L, Recurrence sequences n the hyperbolc Pascal trangle correspondng to the regular mosac {4, 5}, Annales Mathematcae et Informatcae, 46 (2016)

13 Power sums n hyperbolc Pascal trangles 13 [10] Németh, L On the hyperbolc Pascal pyramd, Betr Algebra Geom 57, (2016) [11] Yuan Jn, Zh-Juan Lu, Asmus L Schmdt, On recurrences for sums of powers of bnomal coeffcents, J Number Theory, 128 (10),