TAGAWA, Hiroyuki

English

Wakayama University Academic Repository

??????????????????On convolutions of extended Catalan numbers?? ??∗Hiroyuki TAGAWA(?????????)????, ???????????????????????????????????????????, ????, ????????????????????????????????????????????????1 ??????Cn ?, ?????? (0, 0)?? (n, n)????? y = x???????????? Step (1, 0),i.e. ?, ???? Step (0, 1), i.e. ?, ???????????????????????, n = 3 ?????,????  ,  ,  ,  , ? 5??????, C3 = 5 ??????, n = 0, 1, 2, 3, 4, 5 ????????? Cn ?C0 = 0, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42???, ???, Cn = 1n+1(2nn)????????????????,(mn)?, ????, i.e.(mn)= m!n!(m−n)!??????????, ??????????????????, n+ 1 ?????????????, n+ 2????????????, ??????????????, ??????????? (cf. [4], [7])??? {an}, {bn} ????, cn =∑nk=0 akbn−k ???????? {cn} ?, {an} ? {bn} ???????????? (convolution)???????????, ??????????? R[x] ?? f(x) =∑mk≥0 akxk? g(x) =∑mk≥0 bkxk ?? f(x)g(x) ?∑n≥0(∑nk=0 akbn−k)xn ?????, xn ???? {an} ? {bn} ?????????????, ??????????????????????????????, ????,??????, ????????????????, ??????????????, r ≥ 1, n ≥ 0 ????∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1Cik =r(2n+ r − 1)!n!(n+ r)!(1)???????????? (cf. [1], [6])???, a ∈ C, n ∈ Z (n ≥ 0) ????, ???????? (a)n ?(a)n =1 if n = 0,∏n−1i=0 (a+ i) if n ≥ 1???, r ≥ 2, a1, a2, . . . , ar, b1, b2, . . . , br−1 ∈ C ????, ????? rFr−1(a1, a2, . . . , arb1, b2, . . . , br−1; z)?rFr−1(a1, a2, . . . , arb1, b2, . . . , br−1; z)=∑k≥0zk∏ri=1(ai)kk!∏r−1i=1 (bi)k∗???????, JSPS ??? JP16K05060 ???????????.− 11−拡張されたカタラン数の合成積について2019 年 10 月 2 日受理??????????????????On convolutions of extended Catalan numbers∗Hiroyuki TAGAWA(?????????), ???????????????????????????????????????????, ????, ????????????????????????????????????????????????1??Cn ?, ?????? (0, 0)?? (n, n)????? y = x???????????? Step (1, 0),i.e. ?, ???? Step (0, 1), i.e. ?, ???????????????????????, n = 3 ?????,????  ,  ,  ,  , ? 5??????, C3 = 5 ??????, n = 0, 1, 2, 3, 4, 5 ????????? Cn ?C0 = 0, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, ???, Cn = 1n+1(2nn)????????????? ,(mn)?, , i.e.(mn)= m!n!(m−n)!??????????, ??????????????????, n+ 1 ?????????????, n+ 2????????????, ??????????????, ??????????? (cf. [4], [7])??? {an}, {bn} ????, cn =∑nk=0 akbn−k ??????? {cn} , {an} {bn} ?? ???(convolution)? , ??????????? R[x] f(x) =∑mk≥0 akxkg(x) =∑mk≥0 bkxk ?? f(x)g(x)∑n≥0(∑nk=0 akbn−k)xn , xn ??? {an} ? {bn}????????????, ??????????????????????????????, ????,??????, ????????????????, ??????????????, r ≥ 1, n ≥ 0 ????∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1Cik =r(2n+ r − 1)!n!(n+ r)!(1)???????????? (cf. [1], [6])???, a ∈ C, n ∈ Z (n ≥ 0) ????, ???????? (a)n ?(a)n =1 if n = 0,∏n−1i=0 (a+ i) if n ≥ 1???, r ≥ 2, a1, a2, . . . , ar, b1, b2, . . . , br−1 ∈ C ????, ????? rFr−1(a1, a2, . . . , arb1, b2, . . . , br−1; z)?rFr−1(a1, a2, . . . , arb1, b2, . . . , br−1; z)=∑k≥0zk∏ri=1(ai)kk!∏r−1i=1 (bi)k∗???????, JSPS ??? JP16K05060 ???????????.和歌山大学教育学部紀要　自然科学　第70集　（2020）− 12−?????????, C ?????????, Z ?????????????????????, a ∈ C, m ≥ 1 ?????? { (a)(m+1)nn!(a+1)mn } ?????????????????????, ????, ?????????????? Cn(m, s), ????????????? C2n ???????????????????2 ????????????, m ≥ 1, n, s ≥ 0 ????, ?????? (0, 0) ?? (mn+ s, n) ???? y = xm ???????????? Step (1, 0) ???? Step (0, 1) ???????????? Cn(m, s) ???????, C4(3, 5) ?,?? A ?? B ????????????AB??, Cn(1, 0) = Cn ???, Cn(m, s) ?????, ??????????Lemma 2.1. m ≥ 1, n, s ≥ 0 ????Cn(m, s) =(s+ 1)((m+ 1)n+ s)!n!(mn+ s+ 1)!(=(s+ 1)(m+1)nn!(s+ 2)mn). (2)Proof. (2) ???? C˜n(m, s) ???, Cn(m, s) = C˜n(m, s) ? n ??????????? s ?????????????? 2?????????????, Cn(m, s) ?????C0(m, s) = 1, (3)Cn(m, 0) = Cn−1(m,m) (n ≥ 1), (4)Cn(m, s) = Cn(m, s− 1) + Cn−1(m,m+ s) (n, s ≥ 1) (5)????n = 0 ?????, C0(m, s) = 1 = C˜0(m, s) ??????????n− 1 ???????????(n ≥ 1)?n ≥ 1 ??????, s = 0 ?????, (4) ? n ?????????????Cn(m, 0) = Cn−1(m,m) = C˜n−1(m,m) =(m+ 1)((m+ 1)(n− 1) +m)!(n− 1)!(m(n− 1) +m+ 1)! =((m+ 1)n)!n!(mn+ 1)!= C˜n(m, 0)??????????s− 1 ??????????? (s ≥ 1)?????, (5) ?, n ???????????,?? s ?????????????Cn(m, s) = Cn(m, s− 1) + Cn−1(m,m+ s) = C˜n(m, s− 1) + C˜n−1(m,m+ s) (6)=s((m+ 1)n+ s− 1)!n!(mn+ s)!+(m+ s+ 1)((m+ 1)(n− 1) +m+ s)!(n− 1)!(m(n− 1) +m+ s+ 1)! = C˜n(m, s) (7)???????????????, (2) ? 2????????????????, ????, Cn(m, 0), Cm(p− 1, r− 1) ? Fuss-Catalan number, Raney number ?????????????? ([7], [9])???, (2) ? s = 0 ????????, ??????????????????????? (cf. [8])????????? Cn(m, s) ????????????????Proposition 2.2. r,m ≥ 1, n ≥ 0, a ∈ C ????∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(a)(m+1)ikik!(a+ 1)mik=(ar)(m+1)nn!(ar + 1)mn. (8)拡張されたカタラン数の合成積について− 13−?????????, C ?????????, Z ?????????????????????, a ∈ C, m ≥ 1 ?????? { (a)(m+1)nn!(a+1)mn } ?????????????????????, ????, ?????????????? Cn(m, s), ????????????? C2n ???????????????????2 ????????????, m ≥ 1, n, s ≥ 0 ????, ?????? (0, 0) ?? (mn+ s, n) ???? y = xm ???????????? Step (1, 0) ???? Step (0, 1) ???????????? Cn(m, s) ???????, C4(3, 5) ?,?? A ?? B ????????????AB??, Cn(1, 0) = Cn ???, Cn(m, s) ?????, ??????????Lemma 2.1. m ≥ 1, n, s ≥ 0 ????Cn(m, s) =(s+ 1)((m+ 1)n+ s)!n!(mn+ s+ 1)!(=(s+ 1)(m+1)nn!(s+ 2)mn). (2)Proof. (2) ???? C˜n(m, s) ???, Cn(m, s) = C˜n(m, s) ? n ??????????? s ?????????????? 2?????????????, Cn(m, s) ?????C0(m, s) = 1, (3)Cn(m, 0) = Cn−1(m,m) (n ≥ 1), (4)Cn(m, s) = Cn(m, s− 1) + Cn−1(m,m+ s) (n, s ≥ 1) (5)????n = 0 ?????, C0(m, s) = 1 = C˜0(m, s) ??????????n− 1 ???????????(n ≥ 1)?n ≥ 1 ??????, s = 0 ?????, (4) ? n ?????????????Cn(m, 0) = Cn−1(m,m) = C˜n−1(m,m) =(m+ 1)((m+ 1)(n− 1) +m)!(n− 1)!(m(n− 1) +m+ 1)! =((m+ 1)n)!n!(mn+ 1)!= C˜n(m, 0)??????????s− 1 ??????????? (s ≥ 1)?????, (5) ?, n ???????????,?? s ?????????????Cn(m, s) = Cn(m, s− 1) + Cn−1(m,m+ s) = C˜n(m, s− 1) + C˜n−1(m,m+ s) (6)=s((m+ 1)n+ s− 1)!n!(mn+ s)!+(m+ s+ 1)((m+ 1)(n− 1) +m+ s)!(n− 1)!(m(n− 1) +m+ s+ 1)! = C˜n(m, s) (7)???????????????, (2) ? 2????????????????, ????, Cn(m, 0), Cm(p− 1, r− 1) ? Fuss-Catalan number, Raney number ?????????????? ([7], [9])???, (2) ? s = 0 ????????, ??????????????????????? (cf. [8])????????? Cn(m, s) ????????????????Proposition 2.2. r,m ≥ 1, n ≥ 0, a ∈ C ????∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(a)(m+1)ikik!(a+ 1)mik=(ar)(m+1)nn!(ar + 1)mn. (8)??, a = s+ 1 (s ≥ 0) ????∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1Cik(m, s) = Cn(m, r(s+ 1)− 1). (9)??, ???????Lemma 2.3. a, b ∈ C, m ≥ 1, n ≥ 0 ????n∑k=0(−n)k(a)(m+1)k(b)mkk!(a+ 1)mk(b− n+ 1)(m+1)k =(b− a+mn)(a− b+ 1)n−1(b+mn)(−b+ 1)n−1 . (10)??, (10) ????????????2m+2F2m+1(−n, am+1 , a+1m+1 , . . . , a+mm+1 , bm , b+1m , . . . , b+m−1ma+1m ,a+2m , . . . ,a+mm ,b−n+1m+1 ,b−n+2m+1 , . . . ,b−n+m+1m+1; 1)=(b− a+mn)(a− b+ 1)n−1(b+mn)(−b+ 1)n−1 . (11)Proof. ??, ??????????? (cf. [2], [3])?2m+2F2m+1(−n, am+1 , a+1m+1 , . . . , a+mm+1 , bm , b+1m , . . . , b+m−1ma+1m ,a+2m , . . . ,a+mm ,b−nm+1 ,b−n+1m+1 , . . . ,b−n+mm+1; 1)=(a− b+ 1)n(1− b)n , (12)2m+2F2m+1(−n, am+1 , a+1m+1 , . . . , a+mm+1 , bm , b+1m , . . . , b+m−1mam ,a+1m , . . . ,a+m−1m ,b−n+1m+1 ,b−n+2m+1 , . . . ,b−n+m+1m+1; 1)= − (a− b)n(b+mn)(1− b)n−1 . (13)??, ?????????? 3???????? (cf. [5] ? (2.8)?).r+2Fr+1(a0, a1, . . . , ar+1b, c, d1, d2, . . . , dr−1; z)=c− 1c− b · r+2Fr+1(a0, a1, . . . , ar+1b, c− 1, d1, d2, . . . , dr−1; z)− b− 1c− b · r+2Fr+1(a0, a2, . . . , ar+1b− 1, c, d1, d2, . . . , dr−1; z). (14)?????, (14)????, a0 = −n, (a1, a2, . . . , ar+1) = ( am+1 , a+1m+1 , . . . , a+mm+1 , bm , b+1m , . . . , b+m−1m ), (b, c) =(a+mm ,b−n+m+1m+1 ), (d1, d2, . . . , dr−1) = (a+1m ,a+2m , . . . ,a+m−1m ,b−n+1m+1 ,b−n+2m+1 , . . . ,b−n+mm+1 ), z = 1 ??????, (12), (13) ?????? (11) ??????Remark 2.4. ???????????, ?? (10) ?????????, ???Fn(a, b,m; z) =n∑k=0(−n)k(a)(m+1)k(b)mkzkk!(a+ 1)mk(b− n+ 1)(m+1)k???, n ≥ 1 ?????????????????, n ????????????????????Fn(a, b,m; z) = −a(m+ 1)(a− b+ n− 2)(a+m)(b− n+ 1) Fn−1(a− 1, b,m; z)+(a− 1)(a(m+ 1)− (b− n+ 1)m)(a+m)(b− n+ 1) Fn−1(a, b,m; z)− az(a+ (m+ 1)n− 1)(b)m(a+m)(b− n+ 1)m+1 Fn−1(a+m, b+m,m; z). (15)??, (15) ?????, ??? zk (0 ≤ k ≤ n) ???????(a+m)(a+ (m+ 1)k − 1)(b− n+ 1 + (m+ 1)k)n= (a− 1)(a+mk)(a− b+ n− 2)(k − n)(m+ 1)− (a− 1)(a+ (m+ 1)k − 1)(a+ (a− b+ n− 1)m)(k − n)+ (a+m)(a+ (m+ 1)n− 1)(b− n+ 1 + (m+ 1)k)k????????, ????????????和歌山大学教育学部紀要　自然科学　第70集　（2020）− 14−???, (8) ????????????????, (8) ???? Gn(m,a, r), i.e.Gn(m,a, r) =(ar)(m+1)nn!(ar + 1)mn????Lemma 2.5. r ≥ 2, 1 ≤ u ≤ r − 1, n ≥ 0 ????n∑k=0Gk(m,a, u)Gn−k(m,a, r − u) = Gn(m, a, r). (16)Proof. (16) ?????????, (10) ????, (a, b) ? (au,−a(r − u)−mn) ??????????????n∑i=0Gk(m,a, u)Gn−k(m, a, r − u)=(a(r − u))(m+1)nn!(a(r − u) + 1)mnn∑k=0(−n)k(au)(m+1)k(−a(r − u)−mn)mkk!(au+ 1)mk(−a(r − u)− (m+ 1)n+ 1)(m+1)k=(a(r − u))(m+1)nn!(a(r − u) + 1)mn ×r(ar +mn+ 1)n−1(r − u)(a(r − u) +mn+ 1)n−1=(ar)(m+1)nn!(ar + 1)mn= Gn(m,a, r)????????????????, (8) ????Proof of Proposition 2.2. r ???????????? r = 1 ?????????? Cn(m, s) ??????????r − 1 ??????????? (r ≥ 2)?????, ??????? (16) ??∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(a)(m+1)ikik!(a+ 1)mik=n∑i=0(a)(m+1)ii!(a+ 1)mi∑i1+i2+···+ir−1=n−ii1,i2,...,ir−1≥0r−1∏k=1(a)(m+1)ikik!(a+ 1)mik=n∑i=0Gi(m,a, 1)Gn−i(m, a, r − 1) = Gn(m,a, r)????????Corollary 2.6. m ≥ 1, r ≥ 1, a ∈ C ????m+1Fm(am+1 ,a+1m+1 , . . . ,a+mm+1a+1m ,a+2m , . . . ,a+mm; z)r= m+1Fm(arm+1 ,ar+1m+1 , . . . ,ar+mm+1ar+1m ,ar+2m , . . . ,ar+mm; z). (17)Proof. (17) ? z ? (m+1)m+1zmm ????????? zn (n ≥ 0) ???????, (8) ??????????????, ???? (17) ????????, (8) ? (a,m) ?????????????, ??????????????????????Corollary 2.7. n ≥ 0, r ≥ 1 ????(i) ∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(2ikik)=4n( r2 )nn!. (18)(ii) ∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1C2ik =22n−1r( r2 + n+ 1)n−1n!. (19)拡張されたカタラン数の合成積について− 15−???, (8) ????????????????, (8) ???? Gn(m,a, r), i.e.Gn(m,a, r) =(ar)(m+1)nn!(ar + 1)mn????Lemma 2.5. r ≥ 2, 1 ≤ u ≤ r − 1, n ≥ 0 ????n∑k=0Gk(m,a, u)Gn−k(m,a, r − u) = Gn(m, a, r). (16)Proof. (16) ?????????, (10) ????, (a, b) ? (au,−a(r − u)−mn) ??????????????n∑i=0Gk(m,a, u)Gn−k(m, a, r − u)=(a(r − u))(m+1)nn!(a(r − u) + 1)mnn∑k=0(−n)k(au)(m+1)k(−a(r − u)−mn)mkk!(au+ 1)mk(−a(r − u)− (m+ 1)n+ 1)(m+1)k=(a(r − u))(m+1)nn!(a(r − u) + 1)mn ×r(ar +mn+ 1)n−1(r − u)(a(r − u) +mn+ 1)n−1=(ar)(m+1)nn!(ar + 1)mn= Gn(m,a, r)????????????????, (8) ????Proof of Proposition 2.2. r ???????????? r = 1 ?????????? Cn(m, s) ??????????r − 1 ??????????? (r ≥ 2)?????, ??????? (16) ??∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(a)(m+1)ikik!(a+ 1)mik=n∑i=0(a)(m+1)ii!(a+ 1)mi∑i1+i2+···+ir−1=n−ii1,i2,...,ir−1≥0r−1∏k=1(a)(m+1)ikik!(a+ 1)mik=n∑i=0Gi(m,a, 1)Gn−i(m, a, r − 1) = Gn(m,a, r)????????Corollary 2.6. m ≥ 1, r ≥ 1, a ∈ C ????m+1Fm(am+1 ,a+1m+1 , . . . ,a+mm+1a+1m ,a+2m , . . . ,a+mm; z)r= m+1Fm(arm+1 ,ar+1m+1 , . . . ,ar+mm+1ar+1m ,ar+2m , . . . ,ar+mm; z). (17)Proof. (17) ? z ? (m+1)m+1zmm ????????? zn (n ≥ 0) ???????, (8) ??????????????, ???? (17) ????????, (8) ? (a,m) ?????????????, ??????????????????????Corollary 2.7. n ≥ 0, r ≥ 1 ????(i) ∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1(2ikik)=4n( r2 )nn!. (18)(ii) ∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1C2ik =22n−1r( r2 + n+ 1)n−1n!. (19)(iii) (=(1)) ∑i1+i2+···+ir=ni1,i2,...,ir≥0r∏k=1Cik =r(2n+ r − 1)!n!(n+ r)!= Cn(1, r − 1). (20)Proof. n ≥ 0 ????Gn(0,12, 1) =14n(2nn), Gn(1,12, 1) =C2n4n, Gn(1, 1, 1) = Cn??????????????????, Gn(m,a, 1) =(a)(m+1)nn!(a+1)mn???????, (8) ????, (m,a) ?(0, 12 ), (1,12 ), (1, 1) ??????????????????????Remark 2.8. (19) ????, r = 2 ????? [7] ? Addition Problem A33 (a) ??????n∑i=0C2iC2n−2i = 4nCn (21)??????????[1] E. Catalan, “Sur les nombres de Segner”, Rend. Circ. Mat. Palermo, 1 (1887), 190-201.[2] http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/03/01/05/0017/[3] http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/03/01/05/0018/[4] T. Koshy, Catalan Numbers with Applications, Oxford University Press, 2009.[5] C. Krattenthaler, “A Systematic List of Two- and Three-term Contiguous Relations for Basic Hyper-geometric Series”, available at http://www.mat.univie.ac.at/∼kratt/artikel/contrel.html[6] A. Regev, “A proof of Ctalan’s convolution formula”, Integers, Electronic Journal of CombinatorialNumbers Theory, 12 (2012), #A29.[7] R. P. Stanley, Catalan numbers, Cambridge Univ. Press, 2015.[8] ????, “?????????????”, ?????????? 46? (1990), 19-34.[9] https://en.wikipedia.org/wiki/Fuss%E2%80%93Catalan number

On convolutions of extended Catalan numbers

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On convolutions of extended Catalan numbers

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