50. ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION. Dedicated to George E. Andrews on the occasion of his 60th Birthday.
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1 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION Dedicated to George E Andrews on the occasion of his 60th Birthday Introduction The continued fraction R( = has become known as Ramanujan s continued fraction, though he also considered R( 1 = We can exp each of these as a series, find R( = , R( 1 = Thirty years ago, George Szekeres noticed that, apart from some zeros near the beginning, the sign of the coefficients in both expansions is periodic with period 5 Indeed, he Bruce Richmond [3] proved that if we define the c(n d(n by R( = c(n n, R( 1 = d(n n, 2 3 then ( c(n = 21/2 4π5 1/2 5 3/4 n 3/4 exp 25 ( 4π5 1/2 d(n = 21/2 5 3/4 n 3/4 exp 25 ( 2nπ n {cos 1/2 5 4π 25 ( 4nπ n {cos 1/ π 25 } + O(n 1/2, } + O(n 1/2 1 Typeset by AMS-TEX
2 2 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION Notice that the sign of the dominant term exhibits periodicity with period 5 In examining Ramanujan s Lost Notebook, George E Andrews discovered some relevant formulae He managed [1] to find a combinatorial interpretation of these formulae give a proof of Szekeres s observations But both Ramanujan Andrews missed the fact that one can take the formulae a little further obtain beautiful simple formulae from which Szekeres s observations follow easily First of all, let me introduce some fairly stard notation ( a ; b = (1 a+nb, ( = (;, ( a1, a2,, a k ; b = ( a1 ; b ( a2 ; b ( a k ; b, a 1, a2,, a k b1, b2,, b ;, b = (a1 a2,, a k ; b k ( b1, b2,, b k ; b I shall prove the following formulae R( = (125 ( 5 R( 1 = (125 ( 5 {( 40, 85 20, 105 ; ( 10, 115 5, 120 ; 125 { 30, 95 15, 110 ; ( 5, , 65 ; , 65 30, 95 ; ( 15, , 70 ; , , 115 ; ( 45, 80 40, 85 ; ( 35, 90 45, 80 ; 125 } , 70 35, 90 ; 125 } It follows that c(5n n = 1/(, 2, 3, 4, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 21, 22, 23, 24 ; 25,
3 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION 3 c(5n + 1 n = 1/(, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 14, 15, 16, 17, 18, 19, 19, 20, 21, 22, 23, 24 ; 25, c(5n + 2 n = /(, 2, 3, 4, 5, 6, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 16, 17, 19, 20, 21, 22, 23, 24 ; 25, c(5n + 3 n = 1/(,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 24 ; 25, c(5n + 4 n = 2 /(, 2, 4, 5, 6, 7, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24 ; 25, d(5n n = 1/(, 2, 3, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 22, 23, 24 ; 25, d(5n + 1 n = 1/(, 2, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 23, 24 ; 25, d(5n + 2 n = 1/(, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 12, 13, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24 ; 25, d(5n + 3 n = 3 /( 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 ; 25 d(5n + 4 n = 1/(, 2, 3, 4, 5, 6, 7, 8, 8, 10, 11, 12, 13, 14, 15, 17, 17, 18, 19, 20, 21, 22, 23, 24 ; 25 Thus we obtain the following combinatorial interpretations c(5n is the number of partitions of n into parts which are ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±9, ±10, ±11, ±12 (mod 25, where parts ±4 come in two flavours,
4 4 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION c(5n+1 is the number of partitions of n into parts which are ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±11 (mod 25, where parts ±6 come in two flavours, c(5n+2 is the number of partitions of n 1 into parts which are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±9 come in two flavours, c(5n+3 is the number of partitions of n into parts which are ±1, ±3, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±1 come in two flavours, c(5n+4 is the number of partitions of n 2 into parts which are ±1, ±2, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±11 come in two flavours, d(5n is the number of partitions of n into parts which are ±1, ±2, ±3, ±4, ±5, ±7, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±3 come in two flavours, d(5n+1 is the number of partitions of n into parts which are ±1, ±2, ±3, ±5, ±6, ±7, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±2 come in two flavours, d(5n+2 is the number of partitions of n into parts which are ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±12 (mod 25, where parts ±7 come in two flavours, d(5n+3 is the number of partitions of n 3 into parts which are ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±11, ±12 (mod 25, where parts ±12 come in two flavours d(5n+4 is the number of partitions of n into parts which are ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±10, ±11, ±12 (mod 25, where parts ±8 come in two flavours It follows that c(2 = 0, c(4 = 0, c(9 = 0, while otherwise c(5n > 0, c(5n + 1 > 0, c(5n + 2 < 0, c(5n + 3 < 0, c(5n + 4 < 0 d(3 = 0, d(8 = 0, d(13 = 0, d(23 = 0, while otherwise d(5n > 0, d(5n + 1 < 0, d(5n + 2 > 0, d(5n + 3 < 0, d(5n + 4 < 0
5 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION 5 We mention here results euivalent to results stated by Ramanujan proved by Andrews, namely R( = 8, 8, 12, 12 6, 10, 10, 14 ; , 8, 12, 18 4, 10, 10, 16 ; 20 R( 1 = 4, 4, 16, 16 2, 10, 10, 18 ; 20 4, 6, 14, 16 8, 10, 10, 12 ; 20, we also make the following conjectures R( = 8, 8, 12, 28, 32, 32, 32, 32, 48, 48, 48, 48, 52, 68, 72, 72 16, 16, 20, 20, 20, 20, 24, 24, 56, 56, 60, 60, 60, 60, 64, 64 ; , 12, 12, 12, 28, 28, 28, 32, 48, 52, 52, 52, 68, 68, 68, 72 4, 16, 20, 20, 20, 20, 24, 36, 44, 56, 60, 60, 60, 60, 64, 76 ; ( 8, 8, 8, 8, 12, 28, 32, 32, 48, 48, 52, 68, 72, 72, 72, 72 4, 4, 20, 20, 20, 20, 36, 36, 44, 44, 60, 60, 60, 60, 76, 76 ; 80 3 ( 8, 8, 8, 12, 28, 32, 32, 32, 48, 48, 48, 52, 68, 72, 72, 72 4, 16, 20, 20, 20, 20, 24, 36, 44, 56, 60, 60, 60, 60, 64, 76 ; 80 R( 1 = 4, 16, 16, 24, 24, 24, 24, 36, 44, 56, 56, 56, 56, 64, 64, 76 12, 12, 20, 20, 20, 20, 28, 28, 52, 52, 60, 60, 60, 60, 68, 68 ; 80 4, 16, 16, 16, 24, 24, 24, 36, 44, 56, 56, 56, 64, 64, 64, 76 8, 12, 20, 20, 20, 20, 28, 32, 48, 52, 60, 60, 60, 60, 68, 72 ; ( 4, 16, 16, 16, 16, 24, 24, 36, 44, 56, 56, 64, 64, 64, 64, 76 8, 8, 20, 20, 20, 20, 32, 32, 48, 48, 60, 60, 60, 60, 72, 72 ; , 4, 4, 16, 24, 36, 36, 36, 44, 44, 44, 56, 64, 76, 76, 76 8, 12, 20, 20, 20, 20, 28, 32, 48, 52, 60, 60, 60, 60, 68, 72 ; 80
6 6 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION Proofs The main tool in our proofs is the uintuple product identity, a 2, a 2, a, a 1 ; = which we will now prove Proof: ( 1 n a 3n (3n2 +n/2 + ( 1 n a 3n 1 (3n2 n/2, a 2, a 2, a, a 1 ; = ( a, a 1, ; (a 2, a 2, 2 ; 2 ( 2 = 1 ( 2 a r (r2 +r/2 ( 1 s a 2s s2 = a n c n (, where c n ( = 1 ( 2 r+2s=n ( 1 s (r2 +r/2+s 2 If we put r = n 2t, s = n + t, we find Similarly c 3n ( = 1 ( 2 c 3n+1 ( = 1 ( 2 ( 1 n+t ((n 2t2 +(n 2t/2+(n+t 2 t= = ( 1 n (3n2 +n/2 1 ( 2 = ( 1 n (3n2 +n/2 ( 1 t 3t2 t ( 1 n+t ((n+1 2t2 +(n+1 2t/2+(n+t 2 t= = ( 1 n (3n2 +n+2/2 1 ( 2 = 0 ( 1 t 3t2 3t
7 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION 7 c 3n 1 ( = 1 ( 2 ( 1 n+t ((n 1 2t2 +(n 1 2t/2+(n+t 2 t= = ( 1 n (3n2 n/2 1 ( 2 = ( 1 n (3n2 n/2 ( 1 t 3t2 +t Our starting point is the celebrated result, due to LJRogers (see [2], pp , R( = 2, 3, 4 ; 5 Thus we have, by the uintuple product identity with replaced by 5 a = 1, R( = 1 { 2, 3, 5 ( 5, 4 ; 5 = 1 ( 5 ( 1 n (15n2 +n/2 + } ( 1 n (15n2 +11n+2/2 We split each sum into five, according to the residue modulo 5 of n Thus we replace n by 5n, 5n + 1, 5n + 2, 5n 1 5n 2 to obtain { R( = 1 ( 5 ( 1 n (375n2 +5n/2 ( 1 n (375n2 +155n+16/2 + ( 1 n (375n2 +305n+62/2 ( 1 n (375n2 +145n+14/2 + ( 1 n (375n2 +295n+58/2 + ( 1 n (375n2 +55n+2/2 ( 1 n (375n2 +205n+28/2 + ( 1 n (375n2 +355n+84/2 } ( 1 n (375n2 +95n+6/2 + ( 1 n (375n2 +245n+40/2 We group these ten sums into pairs according to the residue modulo 5 of the powers {( R( = 1 ( 5 ( 1 n (375n2 +5n/2 + ( 1 n (375n2 +245n+40/2
8 8 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION ( + ( 1 n (375n2 +55n/2 + ( 1 n (375n2 +305n+60/2 ( 7 ( 1 n (375n2 +145n/2 ( 1 n (375n2 +355n+70/2 ( 3 ( 1 n (375n2 +95n/2 + ( 1 n (375n2 +155n+10/2 ( } 14 ( 1 n (375n2 +205n/2 ( 1 n (375n2 +295n+30/2 Each pair can now be summed by the uintuple product identity, with replaced by 125 a = 20, a = 30, a = 45, a = 5, a = 55 respectively, to give the desired result The proof of the result for R( 1 is similar so is omitted References [1] George E Andrews, Ramanujan s lost notebook III The Rogers-Ramanujan continued fraction, Adv Math, 41 (1981, [2] G H Hardy E M Wright, An Introduction to the Theory of Numbers, Oxford, 1960 [3] B Richmond G Szekeres, The Taylor coefficients of certain infinite products, Acta Sci Math, 40 (1978,
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