We present some new relations between a continued fraction U(q) of order 12 (established by M. S. M. Naika et al.) and U(q n) for n = 7, 9, 11, 13.Наведено деякі нові співвідношення між ланцюговим дробом U(q) дванадцятого порядку (який описано М. С. М. Найка та іншими авторами) і U(qn) для n=7,9,11та13

Abdulrawf, A.A. Kahtan

Sathish Kumar, C.

Vasuki, K.R.

Ukrains’kyi Matematychnyi Zhurnal

English

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

UDC 511.72K. R. Vasuki, Abdulrawf A. A. Kahtan, G. Sharath, C. Sathish Kumar(Univ. Mysore, India)ON A CONTINUED FRACTION OF ORDER TWELVEПРО ЛАНЦЮГОВИЙ ДРIБ ДВАНАДЦЯТОГО ПОРЯДКУWe present some new relations between a continued fraction U(q) of order twelve (which is established byM. S. M. Naika et al.) and U(qn) for n = 7, 9, 11 and 13.Наведено деякi новi спiввiдношення мiж ланцюговим дробом U(q) дванадцятого порядку (який описаноМ. С. М. Найка та iншими авторами) i U(qn) для n = 7, 9, 11 та 13.1. Introduction. Throughout the paper, we assume |q| < 1 and for a positive integer n,we use the standard notation(a)0 := (a; q)0 = 1,(a)n := (a; q)n =n−1∏i=0(1− aqi)and(a)∞ := (a; q)∞ =∞∏n=0(1− aqn).In Chapter 16 of his Second Notebook [12, p. 197; 3, p. 34] S. Ramanujan develops thetheory of theta functions and his theta function is defined byf(a, b) :=∞∑n=−∞an(n+1)2 bn(n−1)2 = (−a; ab)∞(−b; ab)∞(ab; ab)∞, |ab| < 1.Following Ramanujan [12, p. 197], we defineφ(q) := f(q, q) = 1 + 2∞∑k=1qk2=(−q;−q)∞(q;−q)∞ ,ψ(q) := f(q, q3) =∞∑k=0qk(k+1)2 =(q2; q2)∞(q; q2)∞andf(−q) := f(−q,−q2) =∞∑k=−∞(−1)kq k(3k−1)2 = (q; q)∞.For convenience, we denotef(−qn) = fn.The celebrated Rogers – Ramanujan continued fraction is defined asc© K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR, 2010ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12 16091610 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMARR(q) :=q1/5f(−q,−q4)f(−q2,−q3) =q1/51 +q1 +q21 +q31 + . . .. (1.1)On p. 365 of his Lost Notebook [13], Ramanujan recorded five identities, which showsthe relations between R(q) and five continued fractions R(−q), R(q2), R(q3), R(q4)and R(q5). He also recorded these identities at the scattered places of his Notebooks[12]. L. J. Rogers [14] found the modular equations relating R(q) and R(qn) for n == 2, 3, 5 and 11, the latter equations is not found in Ramanujan’s works. For a proofof these, one may refer [4]. Recently K. R. Vasuki and S. R. Swamy [19] found themodular equation relating R(q) with R(q7).The Ramanujan’s cubic continued fraction G(q) be defined byG(q) :=q1/3f(−q,−q5)f(−q3,−q3) =q1/31 +q + q21 +q2 + q41 +q3 + q61 + . . .. (1.2)The continued fraction (1.2) was first introduced by Ramanujan in his second letter toG. H. Hardy [10]. He also recorded the continued fraction (1.2) on p. 365 of his LostNotebook [13] and claims that there are many results of G(q) which are analogous tothe famous Rogers – Ramanujan continued fraction (1.1).Motivated by Ramanujan’s claim H. H. Chan [5], N. D. Baruah [1], K. R. Vasukiand B. R. Srivatsakumar [18] have established the modular relations between G(q) andG(qn) for n = 2, 3, 5, 7, 11 and 13.The Ramanujan Go¨llnitz – Gordon continued fraction [13, p. 44; 8; 9] is defined byH(q) :=q1/2f(−q3,−q5)f(−q,−q7) =q1/21 +q21 + q3 +q41 + q5 + . . .. (1.3)Chan and S. S. Hang [6], K. R. Vasuki and B. R. Srivatsakumar [17] have establishedthe identities which gives the relations between H(q) and H(qn) with n = 3, 4, 5, 11by employing the modular equations stated by Ramanujan. Recently B. Cho, J. K. Kooand Y. K. Park [7] have extended the above results on continued fraction (1.3) to all oddprimes p by computing the affine models of modular curves X(Γ) with Γ = Γ1(8) ∩∩ Γ0(16p).Motivated by the above works on the continued fractions (1.1) – (1.3), in this paper,we establish the relation between the following continued fraction U(q) with U(qn) forn = 2, 7, 9, 11 and 13U(q) :=qf(−q,−q11)f(−q5,−q7) =q(1− q)(1− q3) +q3(1− q2)(1− q4)(1− q3)(1 + q6) +q3(1− q8)(1− q10)(1− q3)(1 + q12) + . . . . (1.4)The continued fraction (1.4) was established by M. S. M. Naika et al. [11] as a particu-lar case of fascinating continued fraction identity recorded by Ramanujan in his SecondNotebook [12, p. 74]. Furthermore they have also established the modular relation be-tween the continued fraction U(q) with U(qn) with n = 3 and 5.ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12ON A CONTINUED FRACTION OF ORDER TWELVE 16112. Some preliminary results.Theorem 2.1 [11]. We haveφ(q)φ(q3)=1 + U(q)1− U(q) . (2.1)Theorem 2.2. We haveφ2(−q)φ2(−q3) = 1−4U(q)1 + U2(q). (2.2)Proof. From [2], we have(φ2(q) + φ2(q3))2=4φ(q3)φ3(−q3)φ(q)φ(−q) , (2.3)and also from [16], we have(3φ2(q3)− φ2(q))2 = 4φ(q3)φ(q)φ3(−q)φ(−q3) . (2.4)From (2.3) and (2.4), we deduce thatφ2(−q)φ2(−q3) =3− φ2(q)φ(q3)1 +φ2(q)φ2(q3).Employing (2.1) in the right-hand side of the above identity, we obtain (2.2).Theorem 2.3. We haveψ2(q2)qψ2(q6)= U(q) +1U(q)− 1. (2.5)Proof. From [2], we obtain1− φ2(−q)φ2(−q3) =4qf1f312f4f33, (2.6)and also from [2], we have1 +ψ2(q2)qψ2(q6)=f33 f4qf1f312.Using (2.6) and the above identity, we find that1 +ψ2(q2)qψ2(q6)=41− φ2(−q)φ2(−q3),which upon using (2.2) gives the required result.Theorem 2.4. We haveψ2(q)q1/2ψ2(q3)=1 + U(q)1− U(q)√U(q) +1U(q)− 1. (2.7)ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 121612 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMARProof. Consider,ψ4(q)qψ4(q3)=φ2(q)ψ2(q2)qφ2(q3)ψ2(q6)(2.8)where, we have used the identity φ(q)ψ(q2) = ψ2(q) from Entry 25 [3, p. 40]. Nowusing (2.1) and (2.7) in the right-hand side of the above, we obtain the required result.Theorem 2.5. We haveψ4(−q)qψ4(−q3) =[U2(q) + 1U(q)+4U(q)1 + U2(q)− 5]. (2.9)Proof. Changing q to −q in (2.8) and then using (2.2) and (2.5) on the right-handside of (2.8), we obtain (2.9).To prove our main results, we require following easily deducible identities:φ(q) =f52f21 f24, ψ(q) =f22f1,φ(−q) = f21f2, ψ(−q) = f1f4f2,f(q) =f32f1f4, χ(q) =f22f1f4andχ(−q) = f1f2. (2.10)And also the following identities analogous to Rogers – Ramanujan forty identities:Theorem 2.6 [15]. LetM(q) =f(−q5,−q7)f4andN(q) =f(−q,−q11)f4.ThenM(q2)N(q) + qM(q)N(q2) =f3f24f4f8, (2.11)M(q2)N(q)− qM(q)N(q2) = f21 f6f24f2f3f4f8, (2.12)M(q3)N(q) + q2M(q)N(q3) =f1f56 f9f36f22 f23 f312f18, (2.13)M(q3)N(q)− q2M(q)N(q3) = f1f218f4f9f12, (2.14)ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12ON A CONTINUED FRACTION OF ORDER TWELVE 1613M(q5)N(q)− q4M(q)N(q5) = f1f26 f30f2f3f12f20(2.15)andM(q)M(q5)− q6N(q)N(q5) = f5f6f230f4f10f15f60. (2.16)A proof of (2.11) – (2.16) can be found in [15].3. Main results. In this section, we derive the relation between the continuedfraction U(q) with U(qn) for n = 2, 3, 5, 7, 9, 11 and 13.Theorem 3.1. Let u := U(q) and v := U(q2). Thenu2 − v + 2uv − u2v + v2 = 0.Proof. By (1.4), we haveU(q) =qN(q)M(q). (3.1)Dividing (2.11) by (2.12) and on using (2.10), we obtainM(q2)N(q) + qM(q)N(q2)M(q2)N(q)− qM(q)N(q2) =ϕ(−q3)ϕ(−q) .Squaring on both sides of the above identity and using (3.1) on the left-hand side and(2.2) on the right-hand side of the above identity and then dividing throughout by 4u,we deduce the required result.Theorem 3.2 [11]. Let u = U(q) and v = U(q3). Thenu3 − v3 + v2 − v + 3uv − 3u2v2 + u3v2 − u3v = 0.Proof. Dividing (2.14) by (2.13) and on using (2.10), we find thatM(q3)N(q)− q2M(q)N(q3)M(q3)N(q) + q2M(q)N(q3)=φ(−q2)φ(−q3)φ(−q18)φ(−q6)φ(−q9)φ(−q6) .Squaring on both sides of the above identity, and using (3.1) on the left-hand side andemploying the identityφ(q)φ(−q) = φ2(−q2)on the right-hand side of the above, we obtain(U(q)− U(q3)U(q) + U(q3))2=φ(q)φ(q3)φ(−q)φ(−q3)φ(−q3)φ(−q9)φ(q9)φ(q3).Squaring again on both sides of the above, employing (2.1) and (2.2) on the right-handside of the above identity and then factorizing using maple, we deduce that8(−v3 + v2 − v + 3uv − 3u2v2 + u3v2 − u3v + u3)××(u5v − u4v3 + u4v2 − 3u4v + 5u3v3 − u3v2 − u3v + u3 − u2v5 + u4v2+ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 121614 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR+u2v3 − 5u2v2 + 3v4u− uv3 + uv2 − v4) = 0.Now, from the definition of u and v, we haveu = U(q) = 1− q + q5 − q6 + q7 + . . . (3.2)andv = U(q3) = 1− q3 + q15 − q18 + q21 + . . . .Using (3.2) and the above in the above factors, we see that the first factor becomesq3(3− 6q + 3q2 − 3q3 + 6q4 + 3q5 − 9q6 + . . .),and the second factor becomes−q3(10− 22q + 15q2 − 22q3 + 33q4 − 9q5 + . . .).Thus the second factor does not vanish. Hence by the identity Theorem, we musthavev3 − v2 + v − 3uv + 3u2v2 − u3v2 + u3v − u3 = 0.Theorem 3.2 is proved.Theorem 3.3 [11]. Let u := U(q) and v := U(q5). Thenuv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4 − 10u4v3 − 10u2v3++5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5 = 0.Proof. Dividing (2.15) by (2.16) and using (2.10), we find thatM(q5)N(q)− q4M(q)N(q5)M(q)M(q5)− q6N(q)N(q5) =qψ(−q)ψ(−q3)ψ(−q15)ψ(−q5) .Taking power 4 on both sides of the above and employing (2.9) on the right-handside of the above and then factorizing using Maple, we deduce that(u2 + u2v2 + 1− 4uv + v2)(uv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4−−10u4v3 − 10u2v3 + 5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5) = 0.Changing q to q5 in (3.2), we find thatv = U(q5) = 1− q5 + q25 − q30 + q35 + . . . .Using (3.2) and the above in the above factors, we see that the first factor becomesq2(2− 4q4 + 2q5 − 4q6 + 4q8 − 2q9 + 3q10 + . . .),and the second factor becomes−q7(40− 40q + 20q2 + 10q3 − 55q4 + 110q5 − 180q6 + . . .).Thus the first factor does not vanish. Hence by the identity theorem, we must haveuv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4 − 10u4v3 − 10u2v3++5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5 = 0.Theorem 3.3 is proved.ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12ON A CONTINUED FRACTION OF ORDER TWELVE 1615Theorem 3.4. Let u = U(q) and v = U(q7). Then7u3v2 − 35u3v4 − 7u3v − 7u3v3 + 35u4v3 − u7++7u5v − 7u6v + 7u7v − 28u5v2 + 28u6v2 − 14v2u7 + 7v3u5 − 7v3u6++7v3u7 − 35v4u5 + 7v5u− 7v5u2 + 7v5u3 + 35v5u4 − 7v5u5 + 28v5u6−−7v5u7 − 14v6u+ 28v6u2 − 28v6u3 + 7v6u5 − 28v6u6 + 7v6u7 + 7v7u−−7v7u2 + 7v7u3 − 7v7u5 + 14v7u6 − 7v7u7 + v7u8 − v8u+ v − 7uv++7uv2 + 14u2v − 28u2v2 − 7uv3 + 28u2v3 = 0.Proof. IfP =φ(q)φ(q3)andQ =φ(q7)φ(q21),then (QP)4−(PQ)4+ 14[(QP)2−(PQ)2]=(PQ)3 +27(PQ)3+ 7(PQ+3PQ)[1−(PQ)2−(QP)2]. (3.3)We have deduced the identity (3.3) on the lines remarked by N. D. Baruah [1]. Nowemploying (2.1) in (3.3), we deduce the required result.Theorem 3.5. Let u = U(q) and v = U(q9), then9u4v + 126v4u− 90uv3 + 27v8u3 + v − 9vu+ 45v2u+ 378u7v5++135u8v4 + 54u8v2 − 9u8v + 18u7v + 558v4u6 − 558v5u3 + 561v4u3 − 10u9v2−−333u6v3 + 4u9v − 561u6v5 + 16u9v3 − 19u9v4 + 369v5u2−−99u5v2 − 378v4u2 + 135v7u2 + 99uv6 + 16u9v5 + 333u3v6++243u7v3 − 243u2v6 − 126u8v5 − 27u6v − 54v7u+ 9v8u− 135u7v2++10v3 − 198v6u4 + 369u4v5 − 16v6 + 19v5 − 16v4 − u9 − 4v8++10v7 − 243v4u4 − 153v2u2 + 99v7u4 + 297v3u2 + 9u8v8 + 243u5v5−ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 121616 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR−18v8u2 − 297v6u7 + 30u6v8 + 153v7u7 − 30vu3 − 165v7u3 + 198u5v3−−99u8v3 − 423v3u3 − 9u4v8 − 369u7v4 + 165u6v2++423u6v6 − 153u5v6 − 171u6v7 − 369u5v4 − 9u5v8 + 54u5v7++27vu2 + 90u8v6 + v9 + 9u5v − 10u9v6 − 45u8v7 + 171v2u3 + 153v3u4++4u9v7 − 4v2 − 27u7v8 − 54u4v2 − u9v8 − 135v5u = 0.Proof. Let w = U(q3) then, by Theorem 3.2, we haveu3 − w3 + w2 − w + 3uw − 3u2w2 + u3w2 − u3w = 0. (3.4)Changing q to q3 in (3.4), we obtainw3 − v3 + v2 − v + 3wv − 3w2v2 + w3v2 − w3v = 0. (3.5)Now eliminating w between (3.4) and (3.5), we deduce the required result.Theorem 3.6. Let u = U(q) and v = U(q11), then−363v7u10 − 528v10u8 + 11v10u11 − 1089v9u8 + 1463v4u8 − 1012v7u8++11v11u2 + vu− 968v4u4 + 44v3u− 374v3u2 + 77v4u−−528v4u2 − 11vu2 − 363v5u2 + 55v7u11 + v11u11++759v7u9 + 55v5u+ 55v11u7 − 759v3u7 − 55v7u++363v7u2 − 77v8u+ 528v8u2 + 176v10u2++44v11u9 + 363v2u5 − 44vu9 − 11v2u11 − 11v10u++528v10u4 + 528v4u10 + 374v9u2 − 44v9u−−110v2u2 + 11v2u+ 77vu8 + 374v3u10 − 363v2u7−−77vu4 + 11vu10 + 1089v4u3 − 374v9u10++44vu3 − 759v9u5 + 55vu5 + 759v5u3++363v5u10 + 1012v5u8 − 1012v5u4 − 55vu7 + 1496v5u5−−759v5u9 − 528v2u4 − 704v5u7 − 55v5u11 − 968v8u8++1463v8u4 + 1089v3u8 − 528v8u10 − 1089v3u4−−1089v8u3 − 1012v8u5 + 1496v7u7 + 77v8u11+ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12ON A CONTINUED FRACTION OF ORDER TWELVE 1617+1089v8u9 − 363v10u5 + 1012v8u7 + u12 − 374v2u9++1023v3u3 − 110v10u101012v7u4 − 704v7u5 − 77v11u8++1012v4u5 + 11vu11 − 44v11u3 − 1012v4u7 + 924v6u6++1089v9u4 + 1023v9u9 − 759v7u3 + 759v3u5 + 44v9u11++528v2u8 + 759v9u7 − 803v3u9 − 44v3u11 + 363v10u7++1089v4u9 + 11v11u+ 374v2u3 − 77v4u11 + 77v11u4 + 374v10u9−−55v11u5 + v12 − 374v10u3 − 11v11u10 − 803v9u3 + 176v2u10 = 0.Proof. LetP =ϕ(q)ϕ(q3)andQ =ϕ(q11)ϕ(q33),then from [18], we have(PQ)5 +35(PQ)5− 11[(PQ)3 +33(PQ)3]+ 308[PQ+3PQ]==(PQ)6+(QP)6+ 22[(PQ)4+(QP)4][3−(PQ+3PQ)]++11[(PQ)2+(QP)2][(PQ)3 +33(PQ)3− 15(PQ+3PQ)+ 45]+ 924.Using (2.1) in the above, we deduce the required result.Theorem 3.7. Let u = U(q) and v = U(q13), then−8346u8v5 + 611u11v12 + 6318u4v8 + 8723u8v6 − 8346u5v8 − 78u7v5++910u10v7 + 1508u8v2 − 780u8v7 + 8476u8v9 − 10322u4v9 − 5148u8v8−−481u3v12 − 7228u8v10 + 10894u5v9 + 611u3v2 − 208u6v13 + 910u7v10−−10322u5v10 − 143u12v12 + 6409u3v9 − 1911u5v12 + 1313u10v2−−832u7v11 − 5317u3v10 − 130u7v12 − 2184u11v11 − 182u13v9 + 2652u3v11−−5148u6v6 − 5317u10v3 − 91u13v4 + 1508u6v12 + 78uv7 + 8723u6v8+ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 121618 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR+4706u6v3 + 1846u5v2 + 1508u2v8 + v14 + 9607u10v4 − 5317u4v11−−182uv5 + 78u6v − 1170u6v2 − 2184u3v3 − 7228u4v6 − 6331u11v9++195u9v + 10894u9v5 + 6409u9v3 − 7228u10v8 + 6318u10v6 + 78u7v13++4706u3v6 − 780u6v7 + 13u13v12 − 130u2v7 + 52u2v13 − 1911u9v2++6318u8v4 + 78uv6 + 4407u11v10 + 1846u2v5 − 8346u6v9 − 1768u7v7−−8346u9v6 − 832u7v3 − 10322u10v5 − 7631u4v4 − 3926u8v3 − 780u7v8++910u7v4 − 130u7v2 + 78u7v + 10257u4v5 + u14 − 832u3v7 − 78u7v9−−78u9v7 + 10257u10v9 + 8476u9v8 − 780u7v6 − 12909u9v9 − 7631u10v10++846u12v9 − u13v13 + 156u13v10 − 10322u9v4 + 4407u4v3 + 13uv2++10257u9v10 + 6409u11v5 − 6331u5v3 − 1378u4v2 + 13u2v + 4407u10v11−−3926u11v6 − 3926u3v8 − 143u2v2 − 1911u12v5 + 6409u5v11 + 8476u6v5++156uv4 − 65u3v + 1508u12v6 − 208uv8 + 13u12v13 − 65u11v13 + 156u10v13++195u13v5 − 1378u10v12 + 1313u4v12 − 91u4v13 − 130u12v7 − 3926u6v11−−832u11v7 − 65uv3 − 208u13v6 − 1170u12v8 + 9607u4v10 − 13u3v13++6318u6v10 + 78u13v8 + 611u12v11 + 156u4v + 611u2v3 + 78u13v7++4706u11v8 − 1378u2v4 + 10257u5v4 − 208u8v − 26u13v − 7228u6v4−−91u10v + 52u12v + 4407u3v4 − 13u11v + 8476u5v6 − 481u11v2 − 1378u12v10−−1170u8v12 + 78u8v13 + 4706u8v11 − 182u5v − 13uv11 + 195uv9 − 91uv10−−65u13v11 − uv − 39u2v12 − 481u2v11 − 26uv13 − 1170u2v6 + 52uv12−−1911u2v9 + 1846u9v12 + 910u4v7 − 6331u9v11 − 13u13v3 + 2652u11v3−−481u12v3 − 182u9v13 + 52u13v2 + 195u5v13 + 1313u2v10 − 39u12v2−−6331u3v5 − 5317u11v4 + 1313u12v4 − 12909u5v5 − 78u5v7 = 0.Proof. Let P =ϕ(q)ϕ(q3)and Q =ϕ(q13)ϕ(q39), then from [18], we haveISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12ON A CONTINUED FRACTION OF ORDER TWELVE 1619(QP)7+(PQ)7+ 13[(PQ)6+(QP)6]− 26[(PQ)5+(QP)5]−−13[3(PQ)2 +27(PQ)2+ 10][(QP)4+(PQ)4]+ 13[10(PQ)2 +90(PQ)2+ 68]××[(QP)3+(PQ)3]+ 13[10(PQ)2 +90(PQ)2+ 68][(QP)3+(PQ)3]++13[(PQ)4 +81(PQ)4− 20((PQ)2 +9(PQ)2)− 115][(QP)2+(PQ)2]−−13[(PQ)4 +81(PQ)4− 10((PQ)2 +9(PQ)2)− 131][(QP)+(PQ)]== (PQ)6 +729(PQ)6− 26[(PQ)4 +81(PQ)4]+ 169[(PQ)2 +9(PQ)2]+ 832.Now using (2.1) in the above identity, we deduce the required result.Acknowledgement. The authors are thankful to DST, New Delhi for awardingresearch project [No. SR/S4/MS:517/08] under which this work has been done. Further,authors thank the referee for comments to improve the manuscript.1. Baruah N. D. Modular equations for Ramanujan’s cubic continued fraction // J. Math. Anal. and Appl.– 2002. – 268. – P. 244 – 255.2. Baruah N. D., Barman R. Certain Theta-function identities and Ramanujan’s modular equations of degree3 // Indian J. Math. – 2006. – 48, № 1. – P. 113 – 133.3. Berndt B. C. Ramanujan notebooks. Pt III. – New York: Springer, 1991.4. Berndt B. C. Ramanujan Notebooks. Pt V. – New York: Springer, 1998.5. Chan H. H. On Ramanujan’s cubic continued fraction // Acta arithm. – 1995. – 73, № 4. – P. 343 – 355.6. Chan H. H., Huang S. S. On the Ramanujan – Go¨llnitz – Gordan continued fraction // Ramanujan J. –1997. – 1. – P. 75 – 90.7. Cho B., Koo J. K., Park Y. K. Arithmetic of the Ramanujan Go¨llnitz Gordan continued fraction // J.Number Theory. – 2009. – 4, № 129. – P. 922 – 947.8. Gollnitz H. Partition mit Diffrenzebedingungen // J. reine und angew. Math. – 1967. – 25. – S. 154 – 190.9. Gordon B. Some continued fractions of the Rogers Ramanujan type // Duke Math. J. – 1965. – 32. –P. 741 – 748.10. Hardy G. H. Ramanujan. – 3rd ed. – New York: Chelsea, 1978.11. Naika M. S. M., Dharmendra B. N., Shivashankra K. A continued fraction of order twelve // Cent. Eur.J. Math. – DOI: 10.2478/s 11533-008-0031-y.12. Ramanujan S. Notebooks (2 volumes). – Tata Inst. Fundam. Research. – Bombay, 1957.13. Ramanujan S. The “Lost” Notebook and other unpublished papers. – New Delhi: Narosa, 1988.14. Rogers L. J. On a type of modular relation // Proc. London Math. Soc. – 1920. – 19. – P. 387 – 397.15. Vasuki K. R., Guruprasad P. S. On certian new modular relations for the Rogres – Ramanujan typefunctions of order twelve // Proc. Jangjeon Math. Soc. (to appear).16. Vasuki K. R., Sharath G., Rajanna K. R. Two modular equations for squares of the cubic functions withApplications // Note Math. (to appear).17. Vasuki K. R., Srivastakumar B. R. Certian identities for Ramanujan – Go¨llnitz – Gordan continued fraction// J. Comput. Appl. Math. – 2006. – 187. – P. 87 – 95.18. Vasuki K. R., Srivastakumar B. R. Two identities for Ramanujan’s cubic continued fraction. – Preprint.19. Vasuki K. R., Swamy S. R. A new identity for the Rogers – Ramanujan continued fraction // J. Appl.Math. Anal. and Appl. – 2006. – 2, № 1. – P. 71 – 83.Received 27.10.09,after revision — 08.07.10ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12

On a continued fraction of order twelve

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On a continued fraction of order twelve

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