THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 8 0 ALEXANDERBERKOVICH 0 2 Abstract. IrevisitanautomatedproofofAndrews’pentagonalnumberthe- n orem found by Riese. I uncover a simple polynomial identity hidden behind a his proof. I explain how to use this identity to prove Andrews’ result along J withavarietyofnewformulasofsimilartype. Irevealaninterestingrelation 0 betweenthetri-pentagonaltheoremanditems(19),(20),(94),(98)onthecel- 2 ebrated Slater list. Finally,Iestablishanew infinitefamilyofmultipleseries identities. ] T N . 1. Introduction h t a The Gaussian or q-binomial coefficients are polynomials in q defined by m [ n+m := ((qq))nn(+q)mm, if n,m∈N, 1 (cid:20) n (cid:21)q (0, otherwise. v Here (q) = n (1 qj). We shall require the more general q-shifted factorials 8 n j=1 − defined by 0 Q 0 1, if n=0, 3 1. (a)n =(a;q)n := jn=−01(1−aqj) if n>0, 0 Q−j=n1 1−a1q−j if n<0. 8 0 We note that Q 1 : =0, if n<0, v (q) n i X and that L 1 r lim = . a L→∞(cid:20)j(cid:21)q (q)j Here and hereafter q <1. We shall also use the following notations | | k (a ,a ,...,a ) =(a ,a ,...,a ;q) := (a ) , 1 2 k n 1 2 k n i n i=1 Y (a) =(a;q) := lim (a) , ∞ ∞ n n→∞ q [z;q] :=(z, ) . ∞ ∞ z 2000 Mathematics Subject Classification. 33D15,11B65. Key words and phrases. multidimensional q-binomial identities, quintuple product identity, Baileylemmas,rationalfunctionidentities,recursionrelations. ResearchwassupportedinpartbyNSAgrantH98230-07-01-0011. 1 2 ALEXANDERBERKOVICH The literature on q-series abounds with numerous identities of the type ∞ 2L A(i,q) =B(L,q). (1.1) L i i=−∞ (cid:20) − (cid:21)q X For example, ∞ (−1)jq(2j) L2Lj =δL,0, (1.2) j=−∞ (cid:20) − (cid:21)q X ∞ ( 1)jq(j2) 2L =( 1)LqL2(q;q2)L, (1.3) − L j − j=−∞ (cid:20) − (cid:21)q2 X ∞ ( 1)jq2(2j) 2L =qL(q;q2)L. (1.4) − L j j=−∞ (cid:20) − (cid:21)q X Here δ is the Kronecker delta and i := i(i−1). i,j 2 2 The observant reader might have recognized (1.2) as a special case of (cid:0) (cid:1) ∞ (−1)jzjq(j+21) 2LL+ja =(1z)L+a(qz)L, (1.5) j=−∞ (cid:20) − (cid:21)q X where a N. The above formula is due to Cauchy. It is a finite form of the ∈ celebrated Jacobi triple product identity ∞ ( 1)jzjq(j+21) =(q)∞[qz;q]∞. (1.6) − j=−∞ X As for the formulas (1.3) and (1.4), they are, essentially, items G(4) and E(3), respectively, in Slater’s table [9]. Severalyearsago,Andrews[1]revisitedtheumbralmethodsusedbyL.J.Rogers. In [1], he discussed multidimensional identities of the form 2L A(L,q) =F(L,q), L+i i (cid:20) (cid:21)q X where L=(L ,L ,...,L ), i=(i ,i ,...,i ) and 1 2 d 1 2 d d 2L 2L k = . L+i L +i (cid:20) (cid:21)q k=1(cid:20) k k(cid:21)q Y In particular, he proved that ( 1)i+j+kq(i+j2+k) 2L 2M 2N − L i M j N k iX,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q (1.7) (q) (q) (q) = 2L 2M 2N , (q) (q) (q) L+M−N L+N−M M+N−L and that ( 1)iq(i+2j) 2L 2M =( 1)L(q;q2)L(q2;q2)M(−1)M−L. (1.8) − L i M j − (q) i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q M−L X THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 3 We note that (1.7) is a three-dimensional generalization of (1.2). Indeed, if we let N =0, we obtain that (−1)i+jq(i+2j) L2Li M2Mj =(q)2LδL,M. (1.9) i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q X If we now set M = 0 we end up with (1.2), as claimed. Also, (1.8) with M = 0 reduces to (1.2). On the other hand, (1.8) with L = 0 becomes (1.5) with z = 1, a=0. − In what follows, we will use a small variant of (1.9) (−1)i+jq(i+j2+1) 2LL+i1 2MM +j1 =−(q)2L+1δL,M. (1.10) i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q X To verify that (1.10) holds for M =0 (or L=0) we use (1.5) twice as follows ( 1)i+jq(i+j2+1) 2L+1 1 = − L i j i,j (cid:20) − (cid:21)q(cid:20)− (cid:21)q X ( 1)iq(i+21) 2L+1 ( 1)iq(2i) 2L+1 = − L i − − L i i (cid:20) − (cid:21)q i (cid:20) − (cid:21)q X X (1) (q) (q) (1) =0 (q) δ = (q) δ . L+1 L L+1 L 1 L,0 2L+1 L,0 − − − In [8], Riese used his qMultiSum package to provide a simple recurrence proof of (1.7). In the next section, I will rederive and generalize Riese’s recurrences. As a bonus, I will get a uniform proof of (1.7) – (1.10). Moreover, I will show that the same proof can be employed to establish four new identities: ( 1)iq(i+j)2 2L 2M − L i M j Xi,j (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2 (1.11) (q;q2) =( 1)M L−M (q2;q4) (q2;q4) , − ( q;q2) L M L−M − ( 1)i+jq(i+2j) 2L 2M =( 1)L+Mq(L−M)2 (q)2M , (1.12) − L i M j − (q) i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q2 M−L X ( 1)i+jq2(i+2j) 2L 2M − L i M j i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q2 (1.13) X =qL−M(q;q2) ( q;q2) (q2(L+1−M);q2) , L 2M M − and ( 1)i+j+kq(i+j2+k) 2L 2M 2N − L i M j N k i,j,k (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2(cid:20) − (cid:21)q2 (1.14) X =(q;q2) (q;q2) (q;q2) . L+M−N L+N−M M+N−L We remark that (1.14) is a perfect quadratic analogue of (1.7). Setting L = 0 in (1.11) yields (1.5) with q q2 and a = 0, z = 1. Setting M = 0 there yields → −q (1.5) with q q2, a=0, z = 1. Analogously, we can verify that (1.12) reduces to → q (1.2)and(1.3)andthat(1.13)reducesto(1.4)and(1.5)withq q2, a=0, z =1. → 4 ALEXANDERBERKOVICH Finally,(1.14)isathree-dimensionalgeneralizationof(1.3). Indeed,ifweletN =0 in (1.14) we get (−1)i+jq(i+2j) L2Li M2Mj =(q;q2)L+M(−1)M+Lq(L−M)2. (1.15) i,j (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2 X Letting M =0 in (1.15) yields (1.3). The remainder of this manuscript is organized as follows. In the next section, I willshowhowtouseasimplepolynomialidentitytoprove(1.7)–(1.15). InSection 3,Iwillemploysomewell-knownq-binomialtransformationstoderivenewtwoand three-dimensional identities. In Section 4, I will prove, among other results, that for k 0 ≥ qN12+N22+···+Nk2+1 (q) (q) (q) (q2;q2) (q;q2) n1,...X,nk+1≥0 n1 n2··· nk nk+1 nk+nk+1 (1.16) (q4+6k;q4+6k) [q2+2k;q4+6k] ∞ ∞ = , (q) [q1+k;q4+6k] ∞ ∞ where N :=n +n + +n , 1 i k+1. i i i+1 k+1 ··· ≤ ≤ Finally, in Section 5, I will provide fresh insights into the Tri-Pentagonal Theorem 1. (Andrews) 1 ( 1)i+j+kq(i+j2+k)+i2+j2+k2 = qi2+j2+k2 . (1.17) (q)3 − (q) (q) (q) ∞ i,j,k i,j,k≥0 i+j−k i+k−j j+k−i X X Recently, this theorem was given a very interesting partition theoretical inter- pretation in [3]. 2. Simple polynomial identity and its implications It is a fair statement that one does not need a computer to check that (1 x )(1 x q)+(1 x )(1 x q) 1 1 2 2 − − − − (1 x x q)(1 x x q2) 1 2 1 2 − − − (2.1) (1 x )(1 x q)(1 x )(1 x q) 1 1 2 2 − − − − − +x x (1+q)(1 x q)(1 x q)=0 1 2 1 2 − − holds true for any x ,x ,q R. 1 2 ∈ Next, we divide (2.1) by (1 x x q)(1 x x q2) and let x = qL+a+i, x = qL−i 1 2 1 2 1 2 − − to get (1 qL+a+i)(1 qL+a+i+1) − − (1 q2L+a+1)(1 q2L+a+2) − − (1 qL−i)(1 qL−i+1) + − − (1 q2L+a+1)(1 q2L+a+2) − − (2.2) (1 qL+a+i)(1 qL+a+i+1)(1 qL−i)(1 qL−i+1) − − − − − (1 q2L+a+1)(1 q2L+a+2) − − (1 qL+a+i+1)(1 qL−i+1) +q2L+a(1+q) − − =1. (1 q2L+a+1)(1 q2L+a+2) − − THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 5 2(L+1)+a If we multiply (2.2) by we deduce that (L+1) i (cid:20) − (cid:21)q 2L+a 2L+a 2L+a + +q2L+a(1+q) L (i 1) L (i+1) L i (cid:20) − − (cid:21)q (cid:20) − (cid:21)q (cid:20) − (cid:21)q (2.3) 2(L 1)+a 2(L+1)+a (1 q2L+a)(1 q2L+a−1) − = . − − − (L 1) i (L+1) i (cid:20) − − (cid:21)q (cid:20) − (cid:21)q We now define L,M L,M,q ,q 2L+a 2M +a f =f 1 2 := . a a i,j i,j L i M j (cid:18) (cid:19) (cid:18) (cid:19) (cid:20) − (cid:21)q1(cid:20) − (cid:21)q2 Obviously, (2.3) implies that L,M L,M L,M f +f +q2L+a(1+q )f a i 2,j a i,j 1 1 a i 1,j (cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19) (2.4) L 1,M L+1,M (1 q2L+a)(1 q2L+a−1)f − =f − − 1 − 1 a i 1,j a i 1,j (cid:18) − (cid:19) (cid:18) − (cid:19) and that L,M L,M L,M f +f +q2M+a(1+q )f a i,j 2 a i,j 2 2 a i,j 1 (cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19) (2.5) L,M 1 L,M +1 (1 q2M+a)(1 q2M+a−1)f − =f . − − 2 − 2 a i,j 1 a i,j 1 (cid:18) − (cid:19) (cid:18) − (cid:19) It is easy to combine (2.4) and (2.5) as follows L,M L,M f +q2L+a(1+q )f a i 2,j 1 1 a i 1,j (cid:18) − (cid:19) (cid:18) − (cid:19) L 1,M L+1,M (1 q2L+a)(1 q2L+a−1)f − f − − 1 − 1 a i 1,j − a i 1,j (cid:18) − (cid:19) (cid:18) − (cid:19) (2.6) L,M L,M =f +q2M+a(1+q )f a i,j 2 2 2 a i,j 1 (cid:18) − (cid:19) (cid:18) − (cid:19) L,M 1 L,M +1 (1 q2M+a)(1 q2M+a−1)f − f . − − 2 − 2 a i,j 1 − a i,j 1 (cid:18) − (cid:19) (cid:18) − (cid:19) To proceed, we require one more definition L,M,q ,q F (L,M)=F (L,M,x,y,q ,q ,q ):= xiyjqP(i+j)f 1 2 , (2.7) a a 0 1 2 0 a i,j i,j (cid:18) (cid:19) X where P(z) is some polynomial in z. Clearly, L,M L,M y2 xiyjqP(i+j−1)f =x2 xiyjqP(i+j−1)f . (2.8) 0 a i 2,j 0 a i,j 2 i,j (cid:18) − (cid:19) i,j (cid:18) − (cid:19) X X 6 ALEXANDERBERKOVICH Next, we multiply (2.6) by xiyjqP(i+j−1) and sum over i,j. Taking advantage of 0 (2.8) we derive that x q2L+a(1+q )F (L,M) (1 q2L+a)(1 q2L+a−1)F (L 1,M) 1 1 a − − 1 − 1 a − (cid:8) Fa(L+1,M) − (2.9) =y q22M+a(1+q2)Fa(L,M)−(1−q22M+(cid:9)a)(1−q22M+a−1)Fa(L,M −1) (cid:8) Fa(L,M +1) , − provided x = y . (cid:9) | | | | Observethatfora=0,1the recurrence(2.9)togetherwiththe boundaryvalues F (L,0), L 0 and F (0,M), M 0 specifies F (L,M) completely for L a a a ≥ ≥ ≥ 0, M 0. ≥ Itisplainthatthelefthandsidesin(1.8)–(1.13)and(1.15)areoftheform(2.7) with x = y = 1. It is also straightforward to check that the right hand sides | | | | there satisfy (2.9). This implies that (1.8)–(1.13) and (1.15) hold true if they hold when L 0, M = 0 and L = 0, M 0. But this is indeed the case as we saw in ≥ ≥ the Introduction. Fortunately, more is true. Define Ym(L,M,N):= ( 1)i+j+kq(i+j2+k) 2L 2M 2N . (2.10) − L+i M +j N +k i,j,k (cid:20) (cid:21)qm(cid:20) (cid:21)qm(cid:20) (cid:21)qm X Clearly,Y andY arethe lefthandsidesof(1.7)and(1.14),respectively. Multiply 1 2 (2.6) with q1 = q2 = qm by (−1)i+j+kq(i+j+2k−1) N2Nk and sum over i,j,k to (cid:20) − (cid:21)qm derive that q2Lm(1+qm)Y (L,M,N) (1 q2Lm)(1 q2Lm−m)Y (L 1,M,N) m m − − − − Y (L+1,M,N) m − =q2Mm(1+qm)Y (L,M,N) (1 q2Mm)(1 q2Mm−m)Y (L,M 1,N) m m − − − − Y (L,M +1,N). m − (2.11) We remark that (2.11) with m = 1 is, essentially, the recurrence derived by Riese in [8]. Once again, (2.11) together with the boundary values Y (0,M,N), M m ≥ 0, N 0 and Y (L,0,N), L 0, N 0 specify Y (L,M,N) completely for m m ≥ ≥ ≥ L,M,N 0. Next, we check that the right hand sides of (1.7) and (1.14) satisfy ≥ (2.11) with m = 1 and m = 2, respectively. Moreover, on the boundary these identities reduce to the two proven identities: (1.9) and (1.15). And so, (1.7) and (1.14) hold true, as claimed. The reader may wonder if the polynomial identity (2.1) can be extended to n variables : x ,x ,x ,...,x . This is indeed possible. The following generalization 1 2 3 n was suggested to me by Alain Lascoux: n n n n−1 (x ) + ( 1)t(qt−1x x x ) = (qx ) (qt ( 1)t)e , i 2 − i1 i2··· it 2 i 1 − − t+1 iY=1 Xt=1i1<iX2<···<it Yi=1 Xt=1 (2.12) where e ’s are the elementary symmetric functions in x ,x ,...,x . i 1 2 n THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 7 3. q-binomial transformations We begin by recalling some well-known formulas qr2 (q)2L 2r =qj2 2L , (3.1) (q) (q) r j L j r≥0 L−r 2r(cid:20) − (cid:21)q (cid:20) − (cid:21)q X 1 L 2r 2L qL−2r( ;q2) = , (3.2) −q L−2r 2r r j L 2j r≥0 (cid:20) (cid:21)q2(cid:20) − (cid:21)q4 (cid:20) − (cid:21)q X L 2r+a 2L q2r(r+a)( q) =q2j(j+a) , (3.3) − L−2r−a 2r+a r j L 2j a r≥0 (cid:20) (cid:21)q2(cid:20) − (cid:21)q2 (cid:20) − − (cid:21)q X where a=0,1. Weremarkthat(3.1)wasusedbyBressoud[5]togiveasimpleproofoftheRogers– Ramanujan identities. It can be recognized as a special case of the Bailey Lemma initsversionduetoAndrews[2]andPaule[7]. Thetransformations(3.2)and(3.3) were introduced by Berkovichand Warnaar in [4]. In [1], Andrews applied (3.1) to (1.7) three times to obtain the tri-pentagonal theorem (1.17). It is interesting that a single application of (3.1) to (1.7) yields a new three-dimensional identity ( 1)i+j+kq(i+j2+k)+i2 2L 2M 2N − L i M j N k i,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q X (3.4) =q(N−M)2(q) 2L . L+M+N L+N M (cid:20) − (cid:21)q Indeed, we have that ( 1)i+j+kq(i+j2+k)+i2 2L 2M 2N − L i M j N k i,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q X (3.5) qr2 (q) (q) (q) = 2L 2M 2N . (q) (q) (q) (q) L−r r+M−N r+N−M M+N−r r≥0 X The right hand side of (3.5) can be written in the form qn2(q) (q) (q) (qn−L,qn−M−N) RHS(3.5)= 2L 2M 2N θ(L n) rqr(L+M+N+1), (q) (q) (q) ≥ (q,q2n+1) 2n L−n M+N−n r r≥0 X (3.6) where n:= N M and | − | 1, if L n, θ(L n)= ≥ (3.7) ≥ (0, otherwise. The sum in (3.6) can be evaluated by the q-Chu-Vandermonde formula [[6], (II.7)] as (q) M+N+L . (q2n+1) (q) M+N−n n+L 8 ALEXANDERBERKOVICH And so, qn2(q) (q) (q) (q) RHS(3.5)= 2L 2M 2N M+N+L (q) (q) (q) (q) M+N+n M+N−n L−n L+n =q(N−M)2 2L (q) , L+N M L+M+N (cid:20) − (cid:21)q as claimed. Obviously,(3.2)and(3.3)canalsobeemployedtoproducenewtwo-dimensional identities. For example, we can replace q by q4 in (1.9) and apply (3.2) to obtain (−1)i+jq4(i+2j) L2Li M2M2j =qM−2L(−1q;q2)M−2L 2ML (q4;q4)2L. i,j (cid:20) − (cid:21)q4(cid:20) − (cid:21)q (cid:20) (cid:21)q2 X (3.8) Or we can replace q by q2 in (1.9) and apply (3.3) with a=0 twice to get ( 1)i+jq(i+j)(i+j+1)+2i2+2j2 2L 2M − L 2i M 2j i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q X = qr2 (q2;q2)L(q2;q2)M . (3.9) (q2;q2) (q) (q) r L−r M−r r≡0 mod2, X r≥0 Analogously, replacing q by q2 in (1.10) and using (3.3) with a=1 we obtain that 2L 2M q ( 1)i+jq(i+j)(i+j+1)+2i(i+1)+2j(j+1) − − L (2i+1) M (2j+1) i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q X = qr2 (q2;q2)L(q2;q2)M . (q2;q2) (q) (q) r L−r M−r r≡1 mod2, X r>0 (3.10) 4. Two infinite families of multiple series identities If we let L in (3.9) and (3.10) we end up with the following result →∞ qr2 (q2;q2)M = (−q)a (q2;q2) (q) (q2;q2) r M−r ∞ r≡a mod2, X r≥0 (4.1) 2M ( 1)i+jq(i+j)(i+j+1)+2i(i+a)+2j(j+a) , − M 2j a i,j (cid:20) − − (cid:21)q X where a=0,1. Remarkably, the double sum on the righthand side of (4.1) can be reducedtoasinglesum. Tothis end,weperformacleverchangeofthe summation variables j 3j+r a with r =0, 1 and i i j. This yields → − ± → − 1 ∞ RHS(4.1)= 1 ( 1)r ( 1)iq3i2+i(1+2r) (q2;q2) − − ∞ r=−1 i=−∞ X X (4.2) ∞ q24j2+2j(8r+1−6δa,1)+3r2+r+δa,1(1−4r) 2M . M 6j 2r+a j=−∞ (cid:20) − − (cid:21)q X THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 9 We now make use of a special case of (1.6) ∞ ( 1)iq3i2+i(1+2r) =(q2;q2) (1 δ ), r =0, 1 (4.3) ∞ r,1 − − ± i=−∞ X to simplify (4.2) further. This way we obtain qr2 (q2;q2)M = (q2;q2) (q) r M−r r≡a mod2, X r≥0 ∞ q24j2+2j(1−6δa,1)+δa,1 2M (4.4) M 6j+a − j=−∞ (cid:20) − (cid:21)q X ∞ q24j2−2j(7+6δa,1)+5δa,1 2M . M 6j+a+2 j=−∞ (cid:20) − (cid:21)q X Clearly, one could have arrived at (4.4) by taking a more direct route by applying (3.3) to the polynomial identity ∞ 2M +a 2M +a q(3j+1+a)2j =q2M(M+a), a=0,1, j=−∞ (cid:20)M −3j(cid:21)q2 −(cid:20)M −3j−1(cid:21)q2! X which is, essentially, A(5) and A(8) in [9]. However, I feel that the passage from (4.1) to (4.4) is a good warm-up exercise to prepare the reader for the development in the next section. We cannow follow a well trodden path [2] anditerate (3.1) to get for k 1 and ≥ a=0,1 qN12+···+Nk2+1(q)2M = (q) (q) (q) (q2;q2) (q;q2) n1,...,Xnk+1≥0, M−N1 n1··· nk nk+1 nk+nk+1 nk+1≡a mod2 ∞ q12(2+3k)j2+2j(1−δa,16(1+k))+(1+k)δa,1 2M (4.5) M 6j+a − j=−∞ (cid:20) − (cid:21)q X ∞ q12(2+3k)j2+2j(−7−12k−δa,16(1+k))+2+k(a+2)2+5δa,1 2M , M 6j+a+2 j=−∞ (cid:20) − (cid:21)q X whereN =n + +n , 1 i k+1. If wenowletM in(4.5)wearrive i i k+1 ··· ≤ ≤ →∞ at qN12+···+Nk2+1 = (q) (q) (q2;q2) (q;q2) n1,...,Xnk+1≥0, n1··· nk nk+1 nk+nk+1 nk+1≡a mod2 (4.6) q(1+k)δa,1 ∞ q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j), (q) a − a ∞ j=−∞ X 10 ALEXANDERBERKOVICH where a = 0,1 and za := q2+4a+4k(1+δa,1). At this stage we recall the quintuple product identity [[6], Ex. 5.6] ∞ (−1)nq3n2−1nz3n(1+zqn)=(q,−z,−zq)∞(qz2,zq2;q2)∞ n=X−∞ (4.7) [z2;q] ∞ =(q) . ∞ [z;q] ∞ This identity enables us to rewrite (4.5) as qN12+···+Nk2+1 = (q) (q) (q2;q2) (q;q2) nnk1+,1..≡.,Xank+m1≥od0,2 n1··· nk nk+1 nk+nk+1 (4.8) (q16+24k;q16+24k) q(1+k)δa,1 (q) ∞[za;q16+24k]∞[q16+24kza2;q32+48k]∞, ∞ where a=0,1. It remains to establish (1.16). To this end we add together (4.5) with a=0 and (4.5) with a = 1. This way we immediately obtain the correct left hand side of (1.16). Making use of (4.7) on the right we derive that ∞ 1 q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j)+ (q) 1 − 1 ∞ j=−∞ X q1+k ∞ q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j)= (q) 2 − 2 ∞ j=X−∞ (4.9) ∞ 1 ( 1)jq(2+3k)(3j−1)jq3(1+k)j(1+q1+kq(2+3k)2j)= (q) − ∞ j=−∞ X (q4+6k;q4+6k) [q2+2k;q4+6k] ∞ ∞ , (q) [q1+k;q4+6k] ∞ ∞ asdesired. Itisinstructivetocompare(1.16)withasomewhatsimilarformula[[2], (1.8)] qN˜12+···+N˜k2 = (q) (q) (q) (q) n1,.X..,nk≥0 n1 n2··· nk−1 2nk (4.10) (q4+6k;q4+6k) [q2+2k;q4+6k] ∞ ∞ , (q) [ q1+k;q4+6k] ∞ ∞ − where N˜ :=n + +n , 1 i k. i i k ··· ≤ ≤ 5. One-dimensional version of the tri-pentagonal theorem This paper arose from my attempt to ascertain if Andrews’ formula (1.17) was “genuinely” three-dimensional. In the course of this investigation I found that