Notes on Numbe Theoy and Discete Mathematics Pint ISSN 1310 513, Online ISSN 367 875 Vol, 018, No 1, 16 DOI: 10756/nntdm018116- On a genealization of Euleian numbes Claudio Pita-Ruiz Facultad de Ingenieía, Univesidad Panameicana Augusto Rodin 98, México, Ciudad de México, 0390, México e-mail: cpita@upedumx Received: 7 July 016 Accepted: 31 Januay 018 s Abstact: We conside the sequence an+b pp n, poduct of the p-th degee n-polynomial an+b p, whee a, b C, a 0,, p N, and the l s sp s -th degee n-polynomial P n l αsn+βs ps, s whee αs, β s C, s, p s N, s,, l In the expansion of the polynomial an+b pp n+p+ l n in tems of the binomials s sps i p+ l, i 0, 1,, p + l s sps s sp s, the esulting coefficients A αs,βs,s,ps a,b, p, i ae the genealized Euleian numbes we conside in this wok the case P n 1, a 1, b 0, 1 coesponds to the standad Euleian numbes We obtain esults on symmeties, ecuences, ow sums, and altenating ow sums, that genealize the coesponding well-known esults fo the standad Euleian numbes The main tool we use to obtain ou esults thoughout the wok, is the Z-tansfom of sequences Keywods: Genealized Euleian numbes 010 Mathematics Subject Classification: 11B83 1 Intoduction Euleian numbes have been mathematical combinatoial objects of inteest to mathematicians along the yeas: beginning with Eule s wok [10] in the eighteenth centuy, they ae still consideed as objects wothy to study [11, 1] We have nowadays seveal genealizations of these objects [17, 18, 19, 0, 5] Within the woks of L Calitz besides the expositoy wok [3], we find also some genealizations of Euleian numbes [6, 7], including q-genealizations of them [,, 5] among othe elated woks [8] Euleian numbes appea as coefficients of the sequence of polynomials 1, z + 1, z + z + 1, z 3 +11z +11z+1, called Euleian polynomials, consideed by L Eule [10], pp 85, 86 16
We will use the notation A p, i fo the coesponding Euleian numbe in the p-th ow and i-th column of the so-called Euleian numbes tiangle, with p 1,, ows, and i 0, 1,, columns, namely p i 0 1 3 1 0 1 0 1 1 3 0 1 1 0 1 11 11 1 Some shifted vesions of this tiangle appea in the liteatue also as Euleian numbes tiangle Some impotant known facts about Euleian numbes ae the following: Explicit fomula: A p, i i j0 p + 1 1 j i j p 1 j Symmety: Recuence: Row sums: A p, i A p, p + 1 i A p, i ia p 1, i + p + 1 i A p 1, i 1 3 p A p, i p! Altenating ow sums: p whee B p+1 is the p + 1-th Benoulli numbe 1 i A p, i p+1 1 p+1 B p+1, 5 p + 1 Besides the mentioned popeties, a emakable fact is that Euleian numbes ae the coefficients appeaing when we wite n p as a linea combination of the binomials n p, n+1 p,, n+p p This is the Wopitsky identity []: n p p n + p i A p, i 6 p We will wok with the Z-tansfom of sequences, which is a map Z that takes complex sequences a n a 0, a 1,, a n, into complex functions Z a n z o simply Z a n given by the Lauent seies Z a n a n n0 z n defined fo z > R, whee R > 0 is the adius of convegence of the Taylo seies n0 a nz n the geneating function of the sequence a n 17
If Z a n A z, we also say that the sequence a n is the invese Z-tansfom of the complex function A z, and we wite a n Z 1 A z We will ecall now some basic facts about the Z tansfom, that we will use thoughout the wok fo futhe eading, see [1, 3] The sequence λ n whee λ is a given non-zeo complex numbe, has Z-tansfom Z λ n n0 λ n z n 1 1 λ z z z λ, 7 defined fo z > λ In paticula, the Z-tansfom of the constant sequence 1 is Z 1 z z 1 8 Fou impotant popeties of the Z-tansfom which fomal poofs ae easy execises left to the eade, ae the following: 1 Z is linea and injective We will be using this fact without futhe comments Advance-shifting popety If Z a n A z, and k N is given, then k 1 Z a n+k z A k a j z 9 z j 3 Multiplication by the sequence n If Z a n A z, then j0 Z na n z d A z 10 dz Fomula 10 implies Z n a n z d dz A z + z d A z 11 dz Fom 8 and 10, we see that the Z-tansfom of the sequence n is Similaly, we have that Z n z d dz z z 1 z z 1 1 Z n z d z z z + 1 dz z 1 z 1 3, 13 Z n 3 z d z z + 1 dz z 1 3 z z + z + 1 z 1, 1 and so on The Z-tansfom of the sequence n, whee N is given, is n z Z +1 15 z 1 18
The poof is an easy induction on, left to the eade Accoding to the advance-shifting popety 9, togethe with 15, we see that fo 0 k, we have n + k z k+1 Z +1 16 z 1 Obseve that, accoding to Wopitsky identity 6 and fomula 16, the Z-tansfom of the sequence n p p N given is Z n p p A p, i zp i+1 z 1 p+1, 17 It is a known fact that the p + 1 binomials n+p i p, i 0, 1,, p fom a basis of the vecto space of polynomials of degee at most p see [], Sec 3: Wopitsky identity 6 gives us an explicit fomula expessing the p-th degee polynomial n p as a linea combination of the elements of the mentioned basis, and Euleian numbes ae the coesponding coodinates Thus, fo any polynomial f with deg f p, we can wite f n p n + p i c i, 18 p with uniquely detemined constants c i depending only on f As Stanley [] suggests, a good name fo these constants is f-euleian numbes Fo example, Shanks [1] consides the p-th degee polynomial f n n p, and shows that f n p c n+p i i p fo some constants c i that do not depend on n He gives an explicit expession fo c i in tems of a deteminant obtained fom Came s ule not eally fiendly fo algebaic manipulation Fist noticed by Calitz [1] see also [16], the constants c i A f p, i in the expansion 18 of the polynomial f n in tems of binomials n+p i p, i 0, 1,, p, ae c i i j0 p + 1 1 j f i j 19 j These ae then the f-euleian numbes associated to polynomial f of degee at most p Then, the expansion 18 can be witten as f n p i j0 p + 1 1 j f i j j n + p i p 0 In the pesent wok we conside f-euleian numbes fo some geneal polynomials f including the Shanks ones, and study some of thei popeties In paticula, we conside polynomials which ae powes of binomial coefficients f n an+b p,which is a polynomial of degee p, whee a, b ae given complex numbes, a 0, and N is also given Moe geneally, we conside polynomials of the type p p pl an + b α n + β αl n + β l f n, 1 19 l
whee α s, β s C, and s and p s non-negative integes, ae given, s,, l In this case f is a p + l s sp s -th degee polynomial Fo each p N, the polynomial 1 can be witten as a linea combination of the p + l s sp s + 1 binomials n+p+ l s sps i p+ l, s sps i 0, 1,, p + l s sp s To study the coefficients of this linea combination is the main goal of this wok In Section we intoduce ou genealized Euleian numbes, give some special values of them, and in Subsection 1 we wite explicitly the fist few ows of some tiangles of these numbes Sections 3,, 5 and 6 ae devoted to establish genealizations of the popeties of symmety, ecuence 3, ow sums, and altenating ow sums 5, espectively Finally we comment that ou main inteest thoughout the wok is to have good genealizations of the well-known algebaic natue and popeties of standad Euleian numbes We do not conside the combinatoial pat of the life of Euleian numbes: we think that suely thee ae inteesting combinatoial intepetations fo the genealized Euleian numbes consideed in this wok This should give mateial fo the futue The genealized Euleian numbes Note: We will be witing σ fo l s sp s Having in mind the pincipal p-th degee polynomial an+b p, we will conside the poduct 1, which is a p + σ-th degee polynomial, and wite the expansion 0 p l ps an + b αs n + β s The coefficients A as,bs,s,ps a,b, p, i s j0 s p+σ A as,bs,s,ps a,b, p, i n + p + σ i p + σ i p l ps p + σ + 1 a i j + b 1 j αs i j + β s, j 3 whee, i 0, 1,, p+ σ, ae the genealized Euleian numbes we want to study in this wok The notation A as,bs,s,ps a,b, p, i suggests that the pincipal polynomial is an+b p, and that this is accompanied in a poduct by polynomials α sn+β s ps, a s s,, l We have A s,b s, s,p s a,b, p, i 0 fo i < 0 o i > p + σ When p p l 0 o l 0, we have the expansion of the p-th degee polynomial an+b p, namely p an + b p n + p i A a,b, p, i, p which coefficients ae the genealized Euleian numbes s s A a,b, p, i i p p + 1 a i j + b 1 j, 5 j j0 0
and in tun, when in these numbes 5 we have a 1, b 0, 1, the esulting numbes A 1,0,1 p, i ae the standad Euleian numbes 1 Accoding to 16, and 3, the Z-tansfom of the sequence an+b p l αsn+βs ps s is an p l ps + b αs n + β s Z z p+σ s s A as,bs,s,ps a,b, s p, i z p+σ i z 1 p+σ+1 6 In paticula we have p an + b Z z p A a,b, p, i z p i z 1 p+1 7 The fist two elements of the p-th ow in the genealized Euleian numbes A as,bs,s,ps a,b, p, i tiangle ae fom 3 with i 0, 1 A αs,βs,s,ps a,b, p, 0 A αs,βs,s,ps a,b, p, 1 p l ps b βs, 8 s s p l ps p l ps a + b αs + β s b βs p + σ + 1 9 s s In paticula, if 0 b <, we have A αs,βs,s,ps a,b, p, 0 0 This happens, fo example, when b 0 see Remak afte fomula 35 and tiangle GENT, Subsection 1 Similaly, the last element of the p-th ow coesponding to i p + σ is s s A αs,βs,s,ps a,b, p, p + σ p+σ p l ps p + σ + 1 a p + σ j + b 1 j αs p + σ j + β s 30 j j0 s s We claim that 30 can be witten as In fact, we have A αs,βs,s,ps a,b, p, p + σ A αs,βs,s,ps p l ps a b + 1 αs β s + s 1 31 a,b, p, p + σ + 1 3 p l ps p + σ + 1 a p + σ + 1 j + b 1 j αs p + σ + 1 j + β s j j0 s s p l ps p+σ+1 1 p+σ+1 b 0 p+σ + j0 s 1 j p + σ + 1 j βs s s p l a p + σ j + b + a αs p + σ j + β s + α s 1 s s s ps
o That is, fom 3 we see that p+σ p l ps p + σ + 1 a p + σ j + b + a 1 j αs p + σ j + β s + α s j j0 s s p l ps b 1 p+σ βs 33 s s In 33 eplace b by b a and β s by β s α s to get p+σ p l ps p + σ + 1 a p + σ j + b 1 j αs p + σ j + β s j j0 s s p l ps b a 1 p+σ βs α s 3 s s Thus, accoding to 30, expession 3 is p l ps A αs,βs,s,ps b a a,b, p, p + σ 1 p+σ βs α s, A αs,βs,s,ps a,b, p, p + σ s p l ps a b + 1 αs β s + s 1, which poves ou claim 31 In paticula, if 1 a b < 1, we have A αs,βs,s,ps a,b, p, p + σ 0 this happens, fo example, when a b Let us conside as an example the genealized Euleian numbes A 1,7,,1 3, 5, p, i: the fist two elements accoding to 8 and 9, and the last element accoding to 31, of the second ow of the coesponding tiangle, ae the following 1 A 1,7,,1 5 7 3, 5,, 0 10900, A 1,7,,1 3 5 1 + 7 5 7 3, 5,, 1 8 + + 1 113100, 1 A 1,7,,1 3 + 5 + 1 1 7 + 1 3, 5,, 10 1633500 In fact, the second ow of the tiangle of genealized Euleian numbes A 1,7,,1 3, 5, p, i is s s s 10900 113100 5651800 16939955 33908380 5136850 58773550 185575 13608790 79890 1633500 which means that the expansion of 3n 5 n+7 is n + 10 n + 9 n + 8 n + 7 10900 113100 + 5651800 16939955 10 10 10 10 n + 6 n + 5 n + n + 3 +33908380 5136850 + 5877550 185575 10 10 10 10 n + n + 1 n +13608790 79890 + 1633500 35 10 10 10
Remak Fom fomula 5, it is clea that if a 1 and b 0 we have A 1,0, p, i 0 fo i 0, 1,, 1 Thus we can wite as n p p i A 1,0, p, i n+p i p In geneal, the value of the non-negative index i befoe of which we have A a,b, p, i 0, depends on a, b and, as it is evident fom fomulas 8 and 9 Howeve, in all cases when a 1 o b 0, we will conside the index i beginning fom i 0 1 Some genealized Euleian numbes tiangles We show now some concete tiangles of genealized Euleian numbes We will be using them as examples of the popeties and esults we will discuss fo genealized Euleian numbes in the est of the aticle We will wite GENT# to mean genealized Euleian numbes tiangle numbe # GENT1: Genealized Euleian Numbes A 3,,3,, 3,1 p, i Explicit Fomula: A 3,,3,, 3,1 p, i i j0 1j p+7 j i j 3 p 3i j+ 3 Expansion: n 3 p 3n+ 3 p+6 A3,,3,, 3,1 p, i n+p+6 i p+6 p i 0 1 3 5 6 7 8 9 1 0 100 3936 53787 1088 1 760 5 0 100 36 001 111671 116076 396036 19375 5 3 0 100 136 69915 936370 56655 15 GENT: Genealized Euleian Numbes A 1,0,3 p, i Explicit Fomula: A 1,0,3 p, i i j0 1j 3p+1 i j p j 3 Expansion: n p 3 3p i3 A 1,0,3 p, i n+3p i 3p p i 3 5 6 7 8 9 10 11 1 1 1 1 9 9 1 3 1 5 05 760 05 5 1 1 3 6750 968 181 181 968 6750 3 1 3
GENT3: Genealized Euleian Numbes A 3,, p, i Explicit Fomula: A 3,, p, i i j0 1j p+1 3i j+ p j Expansion: 3n+ p p A 3,, p, i n+p i p p i 0 1 3 5 6 7 8 1 1 7 1 1 95 9 95 1 3 1 993 1973 33676 1973 993 1 1 9991 569 3978637 7507078 3978637 569 9991 1 GENT: Genealized Euleian Numbes A 1,,3 p, i Explicit Fomula: A 1,,3 p, i i j0 1j 3p+1 i j+ p j 3 Expansion: n+ p 3 3p A 1,,3 p, i n+3p i 3p p i 0 1 3 5 6 7 8 9 1 6 1 16 1 36 35 1 7 1 3 6 360 880 195 306 09 10 5 10 1 The second ows of the genealized Euleian numbes A 1,0, p, i tiangles fo example GENT, contain known esults, as we show now The numbes A 1,0,, i ae A 1,0,, i i + 1 i j 1 j, 36 j j0 whee i, + 1,, By using the identity k + 1 k + j 1 j j j0, 37 k fomula 68 fom Gould s book [13], we see that the genealized Euleian numbes A 1,0,, i ae squaes of binomial coefficients Moe pecisely, fo k 0, 1,,, we have A 1,0,, k + k + 1 k + j 1 j j j0 38 k Thus, the coesponding expansion n i A 1,0,, i n+ i can be witten as n + k n k0 k which is identity 617 fom Gould s book [13], 39
3 Symmeties In this section we show some symmeties fo genealized Euleian numbes A αs,βs,s,ps a,b, p, i Poposition 1 a The genealized Euleian numbes A a,b, p, i have the symmety A a,b, p, i A a,a b+ 1, p, p i, 0 fo i 0, 1,, p b The genealized Euleian numbes A αs,βs,s,ps a,b, p, i, whee s β s + 1 α s, s, 3,, l, have the symmety fo i 0, 1,, p + σ A αs,βs,s,ps a,b, p, i A αs,βs,s,ps a,a b+ 1, p, p + σ i, 1 Poof We conside the expansion p l ps an + b αs n + β s s s p+σ A αs,βs,s,ps a,b, p, i p+σ A αs,βs,s,ps n + p + σ i p + σ n + i a,b, p, p + σ i p + σ 3 In eplace n by 1 n, to obtain p l ps a 1 n + b αs 1 n + β s o s s p+σ p l ps an + a b + 1 αs n + α s β s + s 1 s Expession 3 says that p l ps an + a b + 1 αs n + β s s s s p+σ A αs,βs,s,ps a,b, p, i p+σ 1 n + p + σ i p + σ A αs,βs,s,ps n + i a,b, p, i p + σ A αs,βs,s,ps n + i a,a b+ 1, p, p + σ i p + σ 5 a Set s 0 in and 5 to conclude that, fo i 0, 1,, p, we have that A a,b, p, i is equal to A a,a b+ 1, p, p i, as desied b If s β s + 1 α s, s, 3,, l, we can wite as p l ps an + a b + 1 αs n + β s s s p+σ and again, compaing 5 with 6, we obtain the desied conclusion 1 A αs,βs,s,ps n + i a,b, p, i, 6 p + σ, 5
What Poposition 1 says is that if we have p l ps an + b αs n + β s s s p+σ A αs,βs,s,ps a,b, p, i n + p + σ i p + σ, 7 then, in any of the cases: i s 0, s, 3,, l, o ii s β s + 1 α s, s, 3,, l, we also have p l ps an + a b + 1 αs n + β s 8 p+σ s A αs,βs,s,ps a,b, p, p + σ i s n + p + σ i p + σ Two examples fom Poposition 1 ae the following, we wite togethe expessions 7 and 8 n + n + 1 n 5n 1 + 13 + 36, n + n + 1 n 5n + 6 36 + 13 + 3n 1n + 1 n+3 3 n+5 5 + 16 n+ 5 + 850 n+3 5 + 3900 n+ 5 + 711 n+1 5 + n 5, 3n + n + 1 n+3 3 n+5 5 + 711 n+ 5 + 3900 n+3 5 + 850 n+ 5 + 16 n+1 5 n 5 Obseve that in the case 3n + n + 3 3 n + 5 n + n + 3 n + n + 1 + 3 + 535 + 535 + 3 + 5 5 5 5 5 n, 5 besides the hypothesis 3 β + 1 α 3 + 1, we have also b + 1 a + 1 3 Thus, expessions 7 and 8 ae identical That is, in this case we have the symmety A,3,3,1 3,, 1, i A,3,3,1 3,, 1, 5 i, i 0, 1,, 5 Obseve also that if b + 1 a, expession 0 looks as A a,b, p, i A a,b, p, p i, i 0, 1,, p This is the case of genealized Euleian numbes A 3,, p, i see GENT3 in Subsection 1 Yet anothe paticula case fom 0 is A,1, p, i A,, p, p i, i 0, 1,, p That is, we have p n + 1 p n + p i p n + i A,1, p, i A,, p, i p p p n + p n + p i p n + i A,, p, i A,1, p, i p p Fo example n + 1 6 n + 6 6 n + 3 n + n + 1 n 7 + 35 + 1 +, 6 6 6 6 n + 6 n + 5 n + n + 3 + 1 + 35 + 7 6 6 6 6 6, 9 50
As a final example, we show the following cuious situation 3n n + 3 n + n + 1 n 9 + 180 + 61 + 36 3n + 1 n + 3 n + n + 1 n 36 + 61 + 180 + 9 3n + 3 n + n + 3 n + n + 1 9 + 180 + 61 + 36 3n + n + n + 3 n + n + 1 36 + 61 + 180 + 9, 51, 5, 53 5 The symmety 0 says that A 3,b,, i A 3, b,, i, 0 i Thus with b 0 we have A 3,0,, i A 3,,, i, which explains the elation between 51 and 5 Also with b 1 we have A 3,1,, i A 3,3,, i, which explains the elation between 5 and 53 The elations 51 5 and 53 5, ae pat of the cuiosity of this example In fact, by eplacing n by n in 51 we obtain 5 If in 53 we eplace n by n, we obtain 3n n 1 n n + 1 n + 9 + 180 + 61 + 36 55 Now eplace n by n + in 55 to obtain 5 In the case a 1, b 0, the symmety 0 says that A 1,0, p, p i A 1,, p, i But plainly we have, fo i 0, 1,, p, that Thus, the symmety 0 can be witten in this case as A 1,, p, i A 1,0, p, + i, 56 A 1,0, p, p + 1 i A 1,0, p, i 57 See Lemma in [1] The tiangle GENT in Subsection 1 shows the symmety 57 Of couse, the known case of the standad Euleian numbes A 1,0,1 p, i is also included in 57 It is not difficult to see that A 1,, p, i 0 fo i p 1 + 1,, p Then 56 is valid fo i 0, 1,, p 1 Fo example, we have n 5 5 n + 5 i c i 10 and whee c 0 c 5 1, c 1 c 5 and c c 3 100 Next we conside the case a and b 1 n + 5 5 5 n + 10 i c i, 10 Poposition Fo odd N, the genealized Euleian numbes A,1, p, i satisfy the symmety A,1, p, i + 1 A,1, p, p i, 58 whee i 0, 1,, p 1 7
Poof The genealized Euleian numbes fom 6, with a, b 1 A,1, p, i i j0 p p + 1 i j + 1 1 j, j o that is ae such that A,1, p, i 0 fo i 0, 1,, 1 1 Thus, we can wite p n + 1 p n + p i A,1, p, i, p p n + 1 p 1 p 1 i 1 In 59 eplace n by 1 1 n to obtain p n p 1 p n + 1 A,1, p, i + 1 n + p i 1 p 59 n + i A,1, p, p i 60 p A,1, p, i + 1 p 1 A,1, p, i + 1 Compaing 60 with 61 we obtain the desied conclusion n 1 + p i p n + i p, 61 In paticula the case 1 of 58, we have that A,1,1 p, i A,1,1 p, p i which can also be concluded fom 9 and 50 An example of this paticula case with p is the expansion n + n + 3 n + n + 1 n n + 1 + 76 + 30 + 76 + 6 An example of 58, with 3 and p, is n + 1 n + 5 n + n + 3 n + n + 1 + 93 + 56 + 56 + 93 3 6 6 6 6 6 + n 63 6 Finally, we comment that fo the genealized Euleian numbes A 1,0,1,q,1,3 p, i, whee q is any non-negative intege given, we have the symmety A 1,0,1,q,1,3 p, i + 1 A 1,0,1,q,1,3 p, 3p + q i, 6 whee i 0, 1,, 3p + q The poof of this fact is simila to the poof of Poposition, and we leave it to the eade An example with q and p 1 is n + 1 n + n + 3 n + n + 1 n n + 3 + 90 + 3 + 3 5 5 5 5 5 8
Recuence Let us see the case 1 We want to elate the genealized Euleian numbe A αs,βs,s,ps a,b,1 p, i in the p-th ow and i-th column i 0, 1,, p + σ of the coesponding tiangle of genealized Euleian numbes, with some of the genealized Euleian numbes A αs,βs,s,ps a,b,1 p 1, fom the p 1-th ow Ou stategy will be to obtain the Z-tansfom of the sequence an + b p l s namely Z an + b p l ps αs n + β s s s z p+σ A αs,βs,s,ps a,b,1 p, i z p+σ i αsn+βs ps, s z 1 p+σ+1, 65 see 6 and 7, and elate it with the Z-tansfom of the sequence an+b p 1 l αsn+βs ps s s The beginning is easy: we use 10 and 65 with p eplaced by p 1, to wite 65 as l ps Z an + b p αs n + β s az az d dz s n an + b p 1 s l ps αs n + β s + bz an + b p 1 s z p 1+σ A αs,βs,s,ps a,b,1 s p 1, i z p 1+σ i z 1 p 1+σ+1 + b l ps αs n + β s s z p 1+σ A αs,βs,s,ps a,b,1 Afte pefoming the deivative and some simplifications we get s p 1, i z p 1+σ i z 1 p 1+σ+1 Z z p+σ an + b p l ps αs n + β s s s a p + σ i + 1 b A αs,βs,s,ps a,b,1 p 1, i 1 + ai + b A α,β, 1,p 1 z p+σ i a,b,1 p 1, i z 1 p+σ+1 66 Fom 65 and 66, we obtain the ecuence A αs,βs,s,ps a,b,1 p, i ai + b A αs,βs,s,ps a,b,1 p 1, i + a p + σ i + 1 b A αs,βs,s,ps a,b,1 p 1, i 1 Fomula 67 contains the ecuence A a,b,1 p, i ai + b A a,b,1 p 1, i + a p i + 1 b A a,b,1 p 1, i 1, 68 see fomula 0 in [5], and in tun fomula 68 contains the case of the well-known ecuence 3 fo the standad Euleian numbes A 1,0,1 p, i 67 9
Fo example, fo the genealized Euleian numbes A 3,,3,, 3,1 p, i we have the ecuence accoding to 67 A 3,,3,, 3,1 p, i i 3 A 3,,3,, 3,1 p 1, i + p i + 17 A 3,,3,, 3,1 p 1, i 1 69 If p 3 and i 5 we have see GENT1 in Subsection 1 A 3,,3,, 3,1 3, 5 7A 3,,3,, 3,1, 5+13A 3,,3,, 3,1, 7 116076+13 111671 56655 We can mimic the pevious pocedue in ode to obtain the following ecuence in the case we wite σ fo σ A αs,βs,s,ps a,b, p, i 70 1 ai + b ai + b 1 Aαs,βs,s,ps a,b, p 1, i + 1 + 1 a i + ai a b + a p + σ + 1 +b a + ab a p + σ + ab p + σ a b a p + σ Aαs,βs,s,ps a,b, p 1, i 1 ai p σ 1 + bai p σ 1 + b 1Aαs,βs,s,ps a,b, p 1, i Fo example, the ecuence fo the genealized Euleian numbes A 3,, p, i is 3i + A 3,, p, i A 3,, p 1, i + 9i + 18ip A 3,, p 1, i 1 71 3 i p 1 + A 3,, p 1, i If p and i 5 we have see GENT3 in Subsection 1 A 3,,, 5 17 10 A 3,, 3, 5 + 133A 3,, 3, + A 3,, 3, 3 136 993 + 133 1973 + 55 33676 3978637 Recuence 70 does not offe any motivation to continue tying the next paticula cases 3,, seeking a conjectue fo the geneal case The good news come fom the fact that if a 1, we have a nice geneal ecuence Poposition 3 The ecuence fo the genealized Euleian numbes A αs,βs,s,ps 1,b, p, i is given by A αs,βs,s,ps 1,b, p, i i k + b p + σ + k i b A αs,βs,s,ps 1,b, p 1, i k, 7 k k k0 whee i 0, 1,, p + σ 30
Poof Fom and 3 we can wite p+σ A αs,βs,s,ps n + p + σ i 1,b, p, i p + σ p 1 l ps n + b n + b αs n + β s 73 n + b p 1+σ s Thus, to pove 7 we have to pove that p+σ i k+b k k0 p 1+σ n + b s A αs,βs,s,ps a,b, p 1, i p+σ+k i b k A αs,βs,s,ps 1,b, p 1, i n + p 1 + σ i p 1 + σ A αs,βs,s,ps n+p+σ i 1,b, p 1, i k p + σ n + p 1 + σ i p 1 + σ 7 In the expession of the left-hand side of 7, we intoduce the new index I i k, which uns fom I 0 up to I p 1 + σ Then we wite 7 as p+σ k0 p 1+σ I0 i k+b k k0 p+σ+k i b A αs,βs,s,ps n+p+σ i 1,b, p 1, i k k p + σ I +b p+σ I b k k The poof ends if we show that n+p+σ I k p + σ A αs,βs,s,ps 1,b, p 1, I I + b p + σ I b n + p + σ I k k k p + σ k0 n + b n + p 1 + σ I 75 p 1 + σ If in identity 63 of Gould s book [13], namely β γ α + κ α α γ + δ, κ δ κ β + γ β + γ δ δ κ we set α n + I 1, β p + σ I b, γ I + b, δ and κ k, we obtain 75 Then the poof is complete In paticula we have fom 7 with, thata αs,βs,s,ps 1,b, p, 0 b α A s,β s, s,p s 1,b, p 1, 0 see 8 If we set i 1 in 7 and use 8 and 9, we obtain the following fomula with flavo of Pascal s tiangle ecuence b b b + 1 p + σ + 1 + p + σ b p 1 + σ + 1 1 31
Let us see a concete numeical example fom the ecuence 7 The genealized Euleian numbes A 1,0,3 p, i have the ecuence i i 1 3p + 1 i A 1,0,3 p, i A 1,0,3 p 1, i + A 1,0,3 p 1, i 1 76 3 1 i 3p + i + A 1,0,3 p 1, i 1 3p + 3 i + A 1,0,3 p 1, i 3 3 If in 76 we set p, i 7 see GENT in Subsection 1, then we get A 1,0,3, 7 7 6 6 5 7 A 1,0,3 3, 7 + A 1,0,3 3, 6 + A 1,0,3 3, 5 + 3 1 1 7 6 6 5 7 8 05 + 760 + 05 + 5 3 1 1 3 181 8 A 1,0,3 3, 3 Finally, obseve that the genealized Euleian numbes A 1,0, 3, i fom the thid ow of the coesponding tiangle, can be witten in tems of squaes of binomial coefficients in tems of the genealized Euleian numbes A 1,0,, i see 38 In this case the ecuence 7 gives us the explicit fomula A 1,0, 3, i k0 i k i 3 + k i k + k i 77 That is, fo i,, 3, we have the identity i 3 3 + 1 i j i k 3 + k i 1 j 78 j i k + k i j0 k0 5 Row sums Let us conside the genealized Euleian numbes A αs,βs,s,ps a,b, p, i involved in the expansion We want an expession fo the sum of the elements of the p-th ow of the coesponding tiangle of genealized Euleian numbes, namely p+σ The esult is given in the following Poposition A αs,βs,s,ps a,b, p, i 79 Poposition The sum of the genealized Euleian numbes A αs,βs,s,ps a,b, p, i, i 0, 1,, p+ σ, is given by p+σ A αs,βs,s,ps a,b, p, i p + σ! ap! p Obseve that this esult does not depend on b no β s 3 l s α sp s s ps 80 s!
Poof The coefficient of n p+σ in an+b p l αsn+βs ps s the left-hand side of, is s The coefficient of n p+σ in p+σ a p! p A αs,βs,s,ps a,b, l s α sp s s ps, 81 s! p, i n+p+σ i p+σ the ight-hand side of is p+σ 1 A αs,βs,s,ps a,b, p, i 8 p + σ! The conclusion 80 comes fom equating 81 and 8 In paticula, the sum of genealized Euleian numbes A a,b, 1, i of the fist ow of the coesponding genealized Euleian numbes tiangle, is equal to a One moe example: the sum 9 A 1,0,3 3, i is 1 + 5 + 05 + 760 + 05 + 5 + 1 1680 see GENT in Subsection 1 Accoding to 80, we have 9 A 1 1,0,3 3, i 3 3! 3! 3 1680 In fact, as Poposition says, the sum 9 A 1,b,3 3, i is equal to 1680, fo any b C Fo example, if b, we have the sum 9 A 1,,3 3, i 6 + 360 + 880 + 195 + 306 09 + 10 5 + 10 1 1680 see GENT in Subsection 1 A final example: let us conside the expansion fo 3n+b, namely 3n + b b n + + b + 5b 3 + 8b + 15b + 9 n + 3 83 3 + b 1b 3 + 3 n + b + 90b + 180 + b + 11b 3 8b 63b + 61 n + 1 b 3 n + Even though the genealized Euleian numbes A 3,b,, i, i 0, 1,, 3,, do depend on b, the sum A 3,b,, i does not In fact, we have accoding to 80 A 3,b,, i! 3 86! Since the sum 80 does not depend on b no β s, we have in paticula 80 with a α s 1 p+σ Then, we can wite 80 as p+σ p+σ A 1,βs,s,ps 1,b, p, i A αs,βs,s,ps a,b, p, i a p l s Moeove, fom 8 we see that p A 1,0, p, i p!! p, and A 1,0,s,ps 1,0, p, i 33 p+σ α sp s s sp s p + σ!! p l s 8 ps s! A 1,0,s,ps 1,0, p, i 85 A 1,0,s p s, i sp s! ps 86 s!
Then, by using 8 and 86, we can wite 85 as p l sp s A 1,0, p, i A 1,0,s p s, i p! l! p s s s p s! s! ps p! l s sp s! p + σ! p + σ p, p,, s p s whee p+σ p, p,, sp s is the multinomial coefficient that p+σ A 1,0,s,ps 1,0, p, i 1 p+σ A 1,0,s,ps 1,0, p, i, 87 p+σ! p! p! l p l Finally, we have fom 87! p+σ A 1,0,s,ps 1,0, p, i 88 l a p α sp s p + σ p l sp s s A 1,0, p, i A 1,0,s p s, i p, s p,, s p s s Thus, in ode to know the sum of genealized Euleian numbes p+σ A αs,βs,s,ps a,b, p, i involved in the expansion, it is sufficient to know the sums of genealized Euleian numbes involved in the simple expansions n p p ps n + p i n sp s n + s p s i A 1,0, p, i and A 1,0,s p s, i, p s s p s 89 s,, l, espectively Fo example, let us conside the expansion 35 In ode to know the sum of coefficients, that is 10900 113100 + 5651800 16939955 + 33908380 5136850 +5877550 185575 + 13608790 79890 + 1633500 0667150, it is sufficient to know the simple expansion n n + n + 3 n + n + 1 n + 16 + 36 + 16 +, 8 8 8 8 8 and use 88 to see that the equied sum is 3 8 8+ 8 1 + 16 + 36 + 16 + 1 0667150 Let us conside now the case of the genealized Euleian numbes A a,0, p, i, with p Fom fomulas 85 and 86 we see that the sum of the numbes A a,0,, i, i 0, 1,,, is given by A a,0,, i a!, which is a times the well-known fomula fo the sum of! squaes of binomial coefficients see 38, namely k0 k Moeove, if p ps, s,, l, fomula 88 looks as l s1 s A 1,βs,s, a,b, 1, i a 1+ + + l 1 + + + l 1 3 1,,, l 1 l l 90
Fo example, fom the expansion 63 we see that the coesponding sum of coefficients is 1 + 93 + 56 180 As fomula 90 says, this sum is 6 6 3 180 6 Altenating ow sums Fo the standad Euleian numbes A 1,0,1 p, i, the known esult on altenating ow sums is 5 In paticula, we have p i1 1i A 1,0,1 p, i 0 if p is even That is, we know that p i1 1i A 1,0, p, i 0 when p is even and 1 In the following Poposition we will show that the altenating sum p i1 1i A 1,0, p, i is equal to 0 when p is even, and is any positive odd numbe Poposition 5 If p is even and is odd, we have a p 1 i A 1,0, p, i 0 91 b i p 1 1 i A 1,, p, i 0 9 Poof a We use 57: fo i, + 1,, p, we have A 1,0, p, p + 1 i A 1,0, p, i Thus p i 1 i A 1,0, p, i p+1 1 i p+1 1 i p+1 1 0, i 1 i A 1,0, p, i + 1 i A 1,0, p, i + p i p+1+1 p+1 1 i 1 i A 1,0, p, i 1 p+1 i A 1,0, p, p + 1 i 1 i A 1,0, p, i A 1,0, p, p + 1 i as desied b We have A 1,, p, i A 1,0, p, + i fo i 0, 1,, p 1 Then we have that accoding to 91, fo p even and odd, 0 p i p 1 1 i A 1,0, p, i 1 p 1 1 i A 1,0, p, + i 1 1 i A 1,, p, i, fom whee 9 follows See fo example GENT in Subsection 1, whee the altenating ow sums is 0 fo p and p 35
Poposition 6 If is odd, we have p 1 1 i A,1, p, i + 1 0, 93 in each of the following cases: a p even and 3mod b p odd and 1mod Poof We will use 58 Poposition : the genealized Euleian numbes A,1, p, i have the symmety A,1, p, i + 1 A,1, p, p i, fo i 0,, p 1 Obseve that in any of the cases a o b, the numbe p 1 is odd Then 0, p 1 1 p 1 1 1 p 1 1 1 + 1 1 i A,1, p, i + 1 1 i A,1, p, i + 1 + p 1 1 1 i A,1, p, i + 1 p 1 + p 1 1 p 1 1 i 1 p 1 +1 1 1 1 i A,1, p, i + 1 i A,1, p, i + 1 1 p 1 i A,1, p, p i as desied The expansion 63 p and 3 is an example of the case a The expansion n+1 5 n+3 5 + 15 n+ 5 + 15 n+1 5 + n 5,p 1 and 5 is an example of the case b In the geneal case of the expansion, we have the following esult Poposition 7 The altenating sum of genealized Euleian numbes A a,b, p, j, 0 j p, is given by p j0 p+1! p 1 j A a,b, p, j 9 p i 0 p p b j + 1 p i j a 1 1+ j1 i j B 1+ j1 i j1 i j j i j 1 + j1 i j i 1 0 j1 Poof We expand the binomial an+b p in ode to wite it as a polynomial in n to get p an + b 1! p p i 0 p p p a i 1+ +i b p i 1 b 1 p i b + 1 p i n i 1+ +i 95 i 1 0 i 1 i 36
Then we have fom 6, 95 and 17 that z p A a,b, p, j z p j j0 1! p z 1 p+1 p p p p i 0 z i 1+ +i k0 i 1 0 i 1 A 1,0,1 i 1 + + i, k z i 1+ +i k i a i 1+ +i b p i 1 b 1 p i b + 1 p i z 1 i 1+ +i +1, 96 fom whee we get p j0 A a,b, p, j z p j 1! p i 0 j1 i j k0 p p p b j + 1 p i j a j1 i j 97 i 1 0 A 1,0,1 j1 j1 i j, k i j z k+ j1 i j z 1 p j1 i j By setting z 1 in 97 we obtain that p j0 p! p 1 j A a,b, p, j p i 0 p p b j + 1 p i j i 1 0 j1 i j a j1 i j j1 i j and finally we use 5 to obtain fom 98 the desied conclusion 9 k0 1 k A 1,0,1 j1 Let us see some paticula cases of 9 In the case 1 fomula 9 looks as p 1 j A a,b,1 p, j p+1 j0 which can be witten as p 1 j A a,b,1 p, j p+1 p b p B 1 + j0 With p 1, fomula 100 says that j0 p i j, k 98 p b p i a i 1 1+i B 1+i, 99 i 1 + i p i + 1 b p i 1 a i+1 1 +i B +i + i 1 1 j A a,b,1 1, j bb 1 + a 1 B b a, 100 In fact, the expansion is an+b 1 b n+1 1 +a b n 1, and the altenating sum of coefficients b a b b a 37
With p fomula 100 says that 1 j A a,b,1, j 8 b B 1 + ba 1 B b ab j0 In fact, the expansion is an+b 1 b n+ + a + ab b n+1 altenating sums of coefficients is b a + ab b + a b b ab In the case, fomula 9 looks as p j0 1 j A a,b, p, j p+1 p i 0 i 1 0 If p 1 fomula 101 says that j0 p p p i 1 i + a b n, and the a i 1+i b p i 1 b 1 p i 1 i 1+i +1 B i1 +i +1 i 1 + i + 1 1 j A a,b, 1, j b b 1 B 1 + a b 1 1 B In fact, the expansion is an+b altenating sum of coefficients is b If p fomula 101 says that 8 101 a b ab + b b n+ + a+b 3 b n+1 + a b+1 n, and the a+b 3 b + a b+1 a b ab + b 1 j A a,b,, j 10 j0 b b 1 B 1 + ab b 1 b 1 1 B + a 3 b 1 1 B b a 3 ab b 3 + b + 6ab ab 3 + a 3 b In paticula, we have j0 1j A 3,b,, j b 16b 3 + 0b + 8b 7 If we set b, we obtain j0 1j A 3,,, j 106, which is the value of the altenating sum 1 95 + 9 95 + 1 see GENT3 in Subsection 1 If b 1, fomula 10 is j0 1 j A a,1,, j 16a 3 1 B a 3 In fact, the expansion of an+1 is an + 1 1 n + 3 a a + 1 + 1 a 11a + 6a 1 n + + 1 a 11a 6a 1 n + 1 + 1 n a a 1, and the altenating sum of coefficients is 1 a a + 1 + 1 a 11a + 6a 1 1 a 11a 6a 1 + 1 a a 1 a 3 38
We conside now two paticula cases of the altenating sum p j0 1j A a,b,1 p, j Poposition 8 a If a and b 1 we have p 1 j A,1,1 p, j p E p 103 j0 In paticula we have p j0 1j A,1,1 p, j 0 if p is odd see case b of Poposition 6 b If a b 1 p j0 1 j A 1,1,1 p, j p+1 p + 1 p+1 1 B p+1 10 This is essentially a shifted vesion of 5 In paticula we have p j0 1j A 1,1,1 p, j 0 if p is even Poof We will use that and p j0 p + 1 j+1 1 j+1 1 B j+1 p + 1 p E p p + 1 E p, 105 j + 1 p j0 p + 1 1 j+1 B j+1 p + 1 j + 1 E p 1 p+1 1 B p+1, 106 whee E p ae the Eule polynomials, and E p ae the Eule numbes These fomulas can be obtained, fo example, fom, 5, 9, 10 and 1, of [9] a Accoding to 99 we have p 1 j A,1,1 p, j j0 Now use 105 to obtain 103 b Accoding to 99 we have p j0 Now use 106 to obtain 10 p p + 1 1 j A 1,1,1 p, j p+1 p + 1 p p + 1 i+1 1 i+1 B i+1 i + 1 p j0 p + 1 1 j+1 B j+1 j + 1 Fo example, with b 1 and p fomula 103 says that j0 1j A,1,1, j 16E 80 In fact, fom 6 we see that the altenating sum of coefficients is 1 76 + 30 76 + 1 80 An inteesting paticula case of fomula 101 is when b 1, consideed in the following Poposition 39
Poposition 9 We have the following fomula fo the altenating sum of genealized Euleian numbes A a, 1, p, j, p 1 j A a, 1, p, j p 107 Poof Fom 101 we obtain p j0 1 j A a, 1, p, j p+1 j0 p i 0 i 1 0 p p p Set i 1 + i j in the sums of the ight-hand side of 108 to wite p j0 p+1 p 1 j A a, 1, p, j p i 0 j0 p + p+1 p i 1 i 1 p i a i 1+i 1 i 1+i +1 B i1 +i +1 i 1 + i + 1 p p 1 p i a j 1 j+1 B j+1 j i i j + 1 p + p+1 1 p p Now use that p 1 p i i j1 p j1 a j 1 j B j j p a j 1 j B j j 1 i j p p 1 i p i j 1 i p p 1 i p 0, i j 1 i identity 33 of Gould s book [13] to obtain the conclusion 107 Fo example, the expansion of an+ 1 is an + 1 1 n + + a a 1 n + 3 + 88a + a + 3 n + 6 16 3 + a + a 1 n + 1 + 16a 8a + 1 n, 16 6 and the altenating sum of coefficients is 1 6 a a 1 + 88a + a + 3 a + a 1 + 16a 8a + 1 16 3 16 6 108 1 Refeences [1] Calitz, L 195 Note on a Pape of Shanks, Ame Math Monthly, 59,, 39 1 [] Calitz, L 1978 A Note on q-euleian Numbes, J Combin Theoy Se A, 5, 90 9 0
[3] Calitz, L 1959 Euleian numbes and polynomials, Math Mag, 3, 7 60 [] Calitz, L 195 q-benoulli and Euleian numbes, Tans Ame Math Soc, 76, 33 350 [5] Calitz, L 1975 A combinatoial popety of q-euleian numbes, Ame Math Monthly, 8, 51 5 [6] Calitz, L 1960 Euleian numbes and polynomials of highe ode, Duke Math J, 7, 01 3 [7] Calitz, L 196 Extended Benoulli and Euleian numbes, Duke Math J, 31, 667 689 [8] Calitz, L, & Riodan, J 1953 Conguences fo Euleian Numbes, Duke Math J, 0, 339 33 [9] Dilche, K Benoulli and Eule Polynomials, Digital Libay of Mathematical Functions, Chapte Available at: http://dlmfnistgov/ [10] Eule, L 1755 Institutiones Calculi Diffeentialis, Academiae Impeialis Scientiaum Petopolitanae [11] Foata, D 010 Euleian Polynomials: fom Eule s Time to the Pesent, in The Legacy of Alladi Ramakishnan in the Mathematical Sciences, Spinge, 53 73 [1] Foata, D, & Schützenbege, M P 1970 Théoie géométique des polynômes Euléiens, Lectue Notes in Mathematics, Vol 138, Spinge [13] Gould, H W 197 Combinatoial Identities, Mogantown, W Va [1] Gaf, U 00 Applied Laplace Tansfoms and z-tansfoms fo Scientists and Enginees: A Computational Appoach using a Mathematica Package, Bikhäuse [15] Hsu, L C, & Shiue, P J S 1999 On cetain summation poblems and genealizations of Euleian polynomials and numbes, Discete Math, 0, 37 7 [16] Koutas, M V 199 Euleian Numbes Associated with Sequences of Polynomials, Fibonacci Quat, 3, 57 [17] Lehme, D H 198 Genealized Euleian Numbes, J Combin Theoy Se A, 3, 195 15 [18] Lin, Z 013 On some genealized q-euleian polynomials, DMTCS Poc AS, 39 50 [19] Petesen, T K 015 Euleian Numbes, Bikhäuse [0] Savage, C D, & Viswanathan, G, The 1/k-Euleian Polynomials, Available online at: http://wwwncsuedu/ savage/papers/the_1_ove_k_euleian_ Polynomialspdf 1
[1] Shanks, E B 1951 Iteated Sums of Powes of the Binomial Coefficients, Ame Math Monthly, 58, 0 07 [] Stanley, R P 1997 Enumeative Combinatoics, Vol 1, Cambidge Studies in Advanced Mathematics, 9, Cambidge Univesity Pess [3] Vilch, R 1987 Z Tansfom Theoy and Applications, D Reidel Publishing Company [] Wopitzky, J 1883 Studien übe die Benoullischen und Euleschen Zahlen, J Reine Angew Math, 9, 03 3 [5] Xiong, T, Tsao, H P, & Hall, J I 013 Geneal Euleian Numbes and Euleian Polynomials, Jounal of Mathematics, Vol 013, ID 6913