# Lectures on Analytic Number Theory

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1 Lectures on Analytic Number Theory By H. Rademacher Tata Institute of Fundamental Research, Bombay

2 Lectures on Analytic Number Theory By H. Rademacher Notes by K. Balagangadharan and V. Venugopal Rao Tata Institute of Fundamental Research, Bombay

3 Contents I Formal Power Series Lecture Lecture 3 Lecture 7 4 Lecture 3 5 Lecture 30 6 Lecture 39 7 Lecture 46 8 Lecture 55 II Analysis 59 9 Lecture 60 0 Lecture 67 Lecture 74 Lecture 8 3 Lecture 89 iii

4 CONTENTS iv 4 Lecture 95 5 Lecture 00 III Analytic theory of partitions 08 6 Lecture 09 7 Lecture 8 8 Lecture 4 9 Lecture 9 0 Lecture 36 Lecture 43 Lecture 50 3 Lecture 55 4 Lecture 60 5 Lecture 65 6 Lecture 69 7 Lecture 74 8 Lecture 79 9 Lecture Lecture 88 3 Lecture 94 3 Lecture 00

5 CONTENTS v IV Representation by squares Lecture Lecture 4 35 Lecture 9 36 Lecture 5 37 Lecture 3 38 Lecture Lecture 4 40 Lecture 46 4 Lecture 5 4 Lecture Lecture 6 44 Lecture Lecture Lecture 7

6 Part I Formal Power Series

8 . Lecture 3 theory. We have to introduce the algebra of formal power series in order to vindicate what Euler did with great tact and insight. A formal power series is an expression a 0 + a x+a x +. Where the symbol x is an indeterminate symbol i.e., it is never assigned a numerical value. Consequently, all questions of convergence are irrelevant. Formal power series are manipulated in the same way as ordinary power series. We build an algebra with these by defining addition and multiplication in the following way. If A a n x n, B b n x n, n0 we define A+ BCwhere C c n x n and ABD where D d n x n, n0 with the stipulation that we perform these operations in such a way that these equations are true modulo x N, whatever be N. (This reauirement stems from the fact that we can assign a valuation in the set of power series by defining the order of A a n x n to be where a is the first non-zero coefficient). n0 Therefore c n and d n may be computed as for finite polynomials; then c n a n + b n, n0 d n a 0 b n + a b n + +a n b + a n b 0. A B means that the two series are equal term by term, A 0 means that all the coefficiants of A are zero. It is easy to verify that the following relations 3 hold: A+ B B+ a AB BA A+(B+C)(A+ B)+C A(B+C)AB+ AC A(BC)(AB)C We summarise these facts by saying that the formal power series form a commutative ring. This will be the case when the coefficients are taen from such a ring, eg. the integres, real numbers, complex numbers. The ring of power series has the additional property that there are no divisors of zero (in case the ring of coefficients is itself an integrity domain), ie. if A, B0, either A0 or B0. We see this as follows: Suppose A0, B0. Let a be the first non-zero coefficient in A, and b j the first non-zero coefficient in B. Let AB d n x n ; then n0 d + j ( a b + j + + a b j+ ) + a b j + ( a + b j + + a + j b 0 ). n0

9 . Lecture 4 In this expression the middle term is not zero while all the other terms are zero. Therefore d + j 0 and so A.B 0, which is a contradiction. From this property follows the cancellation law: If A and A.BA.C, then BC. For, AB ACA(B C). Since A 0, B C or BC. If the ring of coefficients has a unit element so has the ring of power series. As an example of multiplication of formal power series, let, 4 then A x and B+ x+ x + A a n x n, where a 0, a, and a n 0 for n, n0 B b n x n, where b n, n0,,, 3,... n0 C c n x n, where c n a 0 b n + a b n + +a n b 0 ; n0 c 0 a 0 b 0, c n b n b n 0, n,, 3,...; so ( x)(+ x+ x + ). We can very well give a meaning to infinite sums and products in certain cases. Thus A + A + B, C C D, both equations understood in the sense module x N, so that only a finite number of A s or (C ) s can contribute as far as x N. Let us apply our methods to prove the identity: Let + x+ x + x 3 + (+ x)(+ x )(+ x 4 )(+ x 8 ) C (+ x)(+ x )(+ x 4 )... ( x)c ( x)(+ x)(+ x )(+ x 4 )... ( x )(+ x )(+ x 4 )... ( x 4 )(+ x 4 )...

10 . Lecture 5 Continuing in this way, all powers of x on the right eventually disappear, and we have ( x)c. However we have shown that ( x)(+x+x + ), therefore ( x)c ( x)(+ x+ x + ), and by the law of cancellation, C + x+ x + which we were to prove. This identity easily lends itself to an interpretation which gives an example 5 of the application of Euler s idea. Once again we stress the simple fact that x n x m x n+m. We have + x+ x + x 3 + (+ x)(+ x )(+ x 4 )(+ x 8 ) This is an equality between two formal power series (one represented as a product). The coefficients must then be identical. The coefficient of x n on the right hand side is the number of ways in which n can be written as the sum of powers of. But the coefficient of x n on the left side is. We therefore conclude: every natural number can be expressed in one and only one way as the sum of powers of. We have proved that + x+ x + x 3 + (+ x)(+ x )(+ x 4 ) If we replace x by x 3 and repeat the whole story, modulo x 3N, the coefficients of these formal power series will still be equal: Similarly + x 3 + x 6 + x 9 + (+ x 3 )(+ x.3 )(+ x 4.3 ) + x 5 + x.5 + x (+ x 5 )(+ x.5 )(+ x 4.5 ) We continue indefinitely, replacing x by odd powers of x. It is permissible to multiply these infinitely many equations together, because any given power of x comes from only a finite number of factors. On the left appears (+ x + x + x 3 + ). odd On the right side will occur factors of the form (+ x N ). But N can be written uniquely as x λ.m where m is odd. That means for each N, + x N will occur once and only once on the right side. We would lie to rearrange the factors to obtain (+ x)(+ x )(+ x 3 ) This may be done for the following reason. For any N, that part of the 6 formal power series up to x N is a polymial derived from a finite number of

11 . Lecture 6 factors. Rearranging the factors will not change the polynomial. But since this is true for any N, the entire series will be unchanged by the rearrangement of factors. We have thus proved the identity (+ x + x + x 3 + ) odd (+ x n ) () This is an equality of two formal power series and could be written a n x n b n x n. Let us find what a n and b n are. On the left we have n0 n (+ x. + x. + x 3. + )(+ x.3 + x.3 + x ) n0 (+ x.5 + x.5 + x ) x n will be obtainded as many times as n can be expressed as the sum of odd numbers, allowing repetitions. On the right side of (), we have (+ x)(+ x )(+ x 3 ) x n will be obtained as many times as n can be expressed as the sum of integers, no two of which are equal. a n and b n are the number of ways in which n can be expressed respectively in the two manners just stated. But a n b n. Therefore we have proved the following theorem of Euler: Theorem. The number of representations of an integer n as the sum of different parts is the same as the number of representations of n as the sum of odd parts, repetitions permitted. We give now a different proof of the identity (). (+ x n ) ( x n ) n n ( x n )(+ x n ) n ( x n ). Again this interchange of the order of the factors is permissible. For, up to 7 any given power of x, the formal series is a polynomial which does not depend on the order of the factors. (+ x n ) ( x n ) n n n ( x n ), n n n (+ x n ) ( x n ) ( x n ) n ( x n ). n

12 . Lecture 7 Now ( x n ) 0, and by the law of cancellation, we may cancel it from n both sides of the equation obtaining, (+ x n ) ( x n ). n Multiplying both sides by n ( + x n + x (n ) + x + ) 3(n ) n ( (+ x n ) ) ( + x n + x n + x + ) (n ) n n n ( + x n + x + ). (n ) For the same reason as before, we may rearrange the order of the factors on the left. n n ( (+ x n ) )( + x n + x n + x + ) (n ) However, n ( + x n + x + ). (n ) n ( )( + x n + x n + x + ), (n ) n because we have shown that ( x)(+ x+ x + ), and this remains true when x is replaced by x n. Therefore the above equation reduces to (+ x n ) 5 ( ( + x n + x + ) (n ) + x n + x n + x + ) 3n n n which is the identity (). Theorem is easily verified for 0 as follows:0, +9, +8, 3+7, 4+6, 8 ++7, +3+6, +4+5, +3+5, are the unrestricted partitions. Partitions into odd summands with repetitions are n odd

13 . Lecture 8 +9, 3+7, 5+5, +++7, ++3+5, , , , , We have ten partitions in each category. It will be useful to extend the theory of formal power series to allow us to find the reciprocal of the series a 0 + a x+a x + where we assume that a 0 0. (The coefficients are now assumed to form a field). If the series b 0 + b x+b x + a 0 + a x+a x +, we would have (a 0 + a x+a x + )(b 0 + b x+b x + ). This means that a 0 b 0 and since a 0 0, b 0 /a 0. All other coefficients on the left vanish: a 0 b + a b 0 0, a 0 b + a b + a b We may now find b from the first of these equations since all the a s and b 0 are nown. Then b can be found from the next equation, since b will then be nown. Continuing in this, manner all the b s can be computed by successively solving linear equations since the new unnown of any equation is always accompanied by a ν 0. The uniquely determined formal series b 0 + b x+b x + is now called the reciprocal of a 0 + a x+a x + (We 9 can not invert if a 0 0 since in that case we shall have to introduce negative exponents and so shall be going out of our ring of power series). In view of this definition it is meaningful to write x + x+ x + since we have shown that ( x)(+ x+ x + ). Replacing x by x, x + x + x + Using this expression, identity () may be written For any N, ( (+ x n ) + x + x + ) n odd x ( x ) odd N odd, N odd x. Since this is true for any N, we may interchange the order of factors in the entire product and get ( x ) ( x ) n odd odd

14 . Lecture 9 Therefore, in its revised form identity (l) becomes: ( x n ) n ( x n ) n odd In order to determine in how many ways a number n can be split into parts, Wuler introduced a parameter z into his formal power series. (The problem was proposed to Euler in St.Petersburgh: in how many ways can 50 be decomposed into the sum of 7 summands?). He considered such expression as (+ xz)(+ x z) This is a formal power series in x. The coefficients of x are now polynomials in z, and since these polynomials form a ring they porvide an sdmissible set of coefficients. The product is not a formal power series in z 0 however. The coefficient of z for example, is an infinite sum which we do not allow. (+ xz)(+ x z)(+ x 3 z) +zx+zx + (z+z )x 3 + (z+z )x 4 + (z+z )x 5 + +z(x+ x + x 3 + )+z (x 3 + x 4 + x 5 + )+ +za (x)+z A (x)+z 3 A 3 (x)+ () The expressions A (x), A (x), are themselves formal power series in x. They begin with higher and higher powers of x, for the lowest power of x occurring in A m (x) is x m x m(m+)/. This term arises by multiplying (xz)(x z)(x 3 z) (x m z). The advantage in the use of the parameter z is that any power of x multiplyingz m is obtained by multiplying m different powers of x. Thus each term in A m (x) is the product of m powers of x. Thez s therefore record the number of parts we have used in building up a number. Now we consider the finite product P N (z, x) N (+zx n ). P N (z, x) is a polynomial in z: P N (z, x)+za (N) (x)+z A (N) (x)+ + z N A (N) N (x), where A(N) N (x) xn(n+)/. Replacing z by Zx, we have So N (+zx n+ )P N (zx, x) n (+zx)p N (zx, x) ( +zx N+) P N (z, x), n +zxa (N) (x)+z x A (N) (x)+

15 . Lecture 0 (+zx) ( +zxa (N) (x)+ + (zx)n A (N) N (x)) ( +zx N+)( +A (N) (x)+z A (x)+ ) (N) We may now compare powers of z on both sides since these are polynomi- als. Taingz, N, we have x A (N) (x)+ x N+ A (N) (x); (x)+ x A (N) (x) A(N) A (N) (x)( x )a (N) (x)x( x N+ ), A (N) (x) x ( ) x N+ A (N) x (x), A (N) (x) x x A(N) (x) (mod xn ). From this recurrence relation we immediately have Hence A (N) (x) x x (mod x N ), A (N) (x) x x ( x)( x ) x 3 ( x)( x ) A (N) (mod x N ) (mod x N ) x (+)/ ( x)( x ) ( x ) (mod x N ) (+zx n ) + zx x + z x 3 ( x)( x ) + z 3 x 6 ( x)( x )( x 3 ) n + (mod x N )

16 Lecture In the last lecture we proved the identity: where (+zx ) z A (x), () n A (x) 0 x (+)/ ( x)( x ) ( x ) We shall loo upon the right side of () as a power series in x and not as a power-series in z, as otherwise the infinite product on the left side would have no sense in our formalism. Let us inerpret () arithmetically. If we want to decompose m into summands, we have evidently to loo forz and then for x m, and the coefficient ofz x m on the right side of () gives us exactly what we want. We have ( x)( x ) ( x ) say, with n 0 m0 p () 0. x n Therefore m occurs only in the form x n n 0 p () m x m, n 0 mn + n + +n, n j 0, and p m () tells us how often m can be represented by dfferent summands (with possible repetitions). On the other hand the coefficient of x m on the left-side x n ()

17 . Lecture of () gives us the number of partitions of m into summands not exceeding. Hence, Theorem. m can be represented as the sum of different parts as often as 3 m (+) can be expressed as the sum of parts not exceeding (repetition being allowed). (In the first the number of parts is fixed, in the second, the size of parts). In a similar way, we can extablish the identity n ( zx n ) z B (x), (3) with B 0, which again can be interpreted arithmetically as follows. The left side is (zx) n n 0 (zx ) n n 0 0 (zx 3 ) n3 n 3 0 and B (x) x ( x)( x ) ( x ) The left-side of (3) gives m with the representation (4) mn + n + 3n 3 + i.e., as a sum of parts with repetitions allowed. Exactly as above we have: Theorem 3. m can be expressed as the sum of parts (repetitions allowed) as often as m as the sum of parts not exceeding. We shall now consider odd summands which will be of interest in connexion withv-function later. As earlier we can establish the identity (+zx V ) z C (x) (5) V odd with the provide that C (x). The tric is the same. One studies temporatily 4 V a truncated affair (+zx V ), replaces z byzx and evaluates C (x) as in V Lecture. This would be perfectly legitimate. However one could proceed as 0

18 . Lecture 3 Euler did - this is not quite our story. Multiplying both sides by +zx, we have z C (x)(+zx ) z x C (x). 0 Now compare powers of z on both sides - and this was what required some extra thought. C (x) begins with x +3+ +( ) x ; in fact they begin with later and later powers of x and so can be added up. We have C 0, 0 C (x) x C (x)+ x C (x), >0, or C (x) x x C (x) from this recurrence relation we obtain x C (x) x, C (x) x3 x C x 4 (x) 4 ( x )( x 4 ), C 3 (x) x5 x C x 9 (x) 6 ( x )( x 4 )( x 6 ), C (x) x ( x )( x 4 ) ( x ), carrying on the same rigmarole. Now note that all this can be retranslated into something. Let us give the number theoretic interpretation. The coefficient ofz x m 5 gives the number of times m can be expressed as the sum of different odd summands. On the other hand, the coefficients in the expansion of ( x ) ( x ) give the decomposition into even summands, with repetitions. Hence, Theorem 4. m is the sum of different odd parts as often as m is the sum of even parts not exceeding, or what is the same thing, as m is the sum of parts not exceeding. (since m and are obviously of the same parity, it follows that m is an integer). Finally we can prove that V odd( zx V ) z D (x) (6) 0

19 . Lecture 4 Replacingzbyzx, we obtain leading to the D (x) x ( x ) ( x ), Theorem 5. m is the sum of odd parts as often as m is the sum of even parts not exceeding, or m is the sum of even parts not exceeding. ( m again is integral). Some other methods Temporarily we give up power series and mae use of graphs to study partitions. A partition of N may be represented as an array of dots, the number of dots in a row being equal to the magnitude of a summand. Let us arrange the summands according to size. For instance, let us consider a partition of 8 into 4 different parts 6 If we read the diagram by rows we get the partiton On the other hand reading by columns we have the pertition In general it is clear that if we represent a partition of n into parts graphically, then reading the graph vertically yields a partition of n with the largest part, and conversely. This method demonstrates a one-to-one correspondence between partitions of n with parts and partitions sees that the number of partitions of n with largest part is equal to the number of partitions of n into parts not exceeding.

20 . Lecture 5 Draw a diagonal upward starting from the last but one dot in the column on the extreme left. All the dots to the right of this diagonal constitute a partition of into 4 parts. For each partition of 8 into 4 different parts there corresponds thus a partition of into parts. This process wors in general for a 7 partition of n with different parts. If we throw away the dots on and to the left of the diagonal (which is drawn from the last but one point from the bottom in order to exsure that the number of different parts constinues to be exactly ), we are left with a partition of n (++3+ +( ))n ( ). This partition has exactly parts because each row is longer by at least one dot than the row below it, so an entire row is never discarded. Conversely, starting with a partition of n ( ) into parts, we can build up a unique partition of n into different parts. Add to the next to the smallest part, to the next longer, 3 to the next and so on. This one-to-one correspondence proves that the number of partitions of n into different parts equals the number of partitions of n ( ) into parts. We can prove graphically that the number of partitons of n into odd summands is the same as the number of partitions of n into even summands not exceeding. The last row of the diagram contains at least one dot, the next higher at least three, the one above at least five, and so on. Above and on the diagonal there are ( ) dots. When these are removed, an even number of dots is left in each row, althogether adding up to n. This proves the result. Theorem can also be proved graphically, although the proof is not quite 8 as simple. The idea of the proof is examplified by considering the partitons of 35. We have (+)+7 +(8+4+)

21 . Lecture } {{ } 3 times Thus to each unrestricted partition of 35 we can mae correspond a partition into add summands with possible repetitions. Conversely (+)+( ) Now consider the following diagram times 0 Each part is represented by a row of dots with the longest row at ehe top, second longest next to the top, etc. The oddness of the parts allows uo to place the rows symmetrically about a central vertical axis. Now connect the 9 dots in the following way. Connect the dots on this vertical axis with those on the left half of the top row. Then connect the column to the right of this axis to the other half of the top row. We continue in this way as indicated by the diagram drawing right angles first on one side of the centre and then on the other. We now interpret this diagram as a new partition of 35 each part being represented by one of the lines indicated. In this way we obtain the partition of 35 into different parts. It can be proved that this method wors in general. That is, to prove that given a partition of n into odd parts, this method transforms it into a unique partition of n into distinct parts; conversely, given a partation into distinct parts, the process can be reversed to find a unique partition into odd parts. This establishes a one-to-one correspondence between the two sorts of partitions. This proves our theorem.

22 Lecture 3 The series z A (x) that we had last time is itself rather interesting; the A (x) 0 0 have a queer shape: A (x) x ( )/ ( x)( x ) ( x ) Such series are called Euler series. Such expressions in which the factors in the denominator are increasing in this way have been used for wide generalisations of hypergeometric series. Euler indeed solved the problem of computing the coefficients numerically. The coefficient ofz x m is obtained by expanding ( x) ( x ) as a power series. This is rather trivial if we are in the field of complex numbers, since we can then have a decomposition into partial fractiions. Euler did find a nice sort of recursion formula. There is therefore a good deal to be said for a rather elementary treatment. We shall, however, proceed to more important discussions the problem of unrestricted partitions. Consider the infinite product (this is justifiable modulo x N ) m x m m n0 n 0 x n x mn n 0 x n x 3n3 n j 0 + x+x + + p n x n () n 7

23 3. Lecture 8 What does p n signify? p n appeared in collecting the term x n. Following Euler s idea of addition of exponents, we have nn + n + 3n 3 + 4n 4 + n j 0, () so that p n is the number of solutions of a finite Diophantine equation (since the right side of () becomes void after a finite stage) or the number of ways in which n can be expressed in this way, or the number of unrestricted partitions. Euler wrote this as m ( x m ) p(n)x n, (3) with the provide that p(0). We want to find as much as possible about p(n). Let us calculate p(n). Expanding the product, n0 ( x n )( x)( x )( x 3 ) n x x + x 5 + x 7 x x 5 ++ (Note Euler s sill and patience; he calculated up to x n and found to this surprise that the coefficients were always 0, ±, two positive terms followed by two negative terms). We want to find the law of exponents, as every sensible man would. Writing down the first few coeffieicnts and taing differences, we have the sequence of odd numbers interspersed with the sequence of natural numbers. Euler forecast by induction what the general power would be as follows Write down the coefficients by picing up 0, and every other alternate term, and continue the row towards the left by putting in the remaining coefficients. Now we find that the second differences have the constant value 3. But an arithmetical progression of the second order can be expressed as a polynomial of the second degree. The typical coefficient will therefore be given by an expression of the form

24 3. Lecture 9 aλ + bλ+c a(λ+ ) + b(λ+ )+c a(λ+) + b(λ+)+c a(λ+)+b a(λ+ 3)+ b a (the constant second difference) Hence a3 or a3/. Taingλ0 we find that c0 and b, so that the general coefficient has the form λ(3λ ). Observing that whenλis changed to λ, λ(3λ ) becomes λ(3λ+), the coefficient of x λ(3λ )/ is ( ) λ, and hence ( x n ) ( ) λ x λ(3λ )/, (4) n λ which is Euler s theorem. This sequence of numbers λ(3λ ) played a particular role in the middle ages. They are called pentagonal numbers and Euler s theorem is called the pentagonal numbers theorem. We have the so-called triangular numbers: where the second differences are all ; the square-numbers for which the second difference are always ; and so on. The triangular numbers can be represented by dots piled up in the form of equilateral triangles; the square numbers by successively expanding squares.

25 3. Lecture 0 The pentagons however do not fit together lie this. We start with one pentagon; notice that the vertices lie perspectively along rays through the origin. So tae two sides as basic and magnify them and add successive shelves. The second differences now are always 3: In general we can have r-gonal numbers where the last difference are all r. We go bac to equation (4): 4 ( x m ) m ( ) λ x λ(3λ )/ λ It is quite interesting to go into the history of this. It appeared in Euler s Introductio in Analysin Infinitorum, Caput XVI, de Partitio numerorum, 748 (the first boo on the differential and integral calculus). It was actually discovered earlier and was mentioned in a paper communicated to the St. Petersburgh Academy in 74, and in letters to Nicholas Bernoulli (74) and Goldbach (743). The proof that bothered him for nine years was first given in a letter dated 9th june 750 to Goldvach, and was printed in 750. The identity (4) is remarable; it was the first time in history that an identity belonging to the V-functions appeared (later invented and studied systematically by Jocobi). The interesting fact is that we have a power-series in which the exponents are of the second degree in the subscripts. The V-functions have a representation as a series and slso as an infinite porduct. The proof of identity (4) is quite exciting and elementary. By using distributivity we brea up the product in the following way: ( x)( x )( x 3 )( x 4 ) ( x)( x )( x 3 )( x 4 ) x ( x)x ( x)( x )x 3

26 3. Lecture ( x)( x )( x 3 )( x 4 ) ( x) ( x 4 )x 5 which may be re-arranged, opening first parenthesis, as x x ( x )x 3 ( x )( x 3 )x 4 ( x )( x 3 )( x 4 )x 5 +x 3 +( x )x 4 +( x )( x 3 )x 5 So 5 x x + x 5 + x 7 ( x )+ x 9 ( x )( x 3 )+ x x + x ( x3 )x 9 +( x 3 )( x 4 )x x 9 ( x 3 )x x x + x 5 + x 7 x ( x 3 )x 5 ( x 3 )( x 4 )x 8 When this is continued, we get some free terms at the beginning followed by a typical remainder ( x )x m + ( x )( x + )x m+ + ( x )( x + )( x + )x m+, which may be rearranged into x m + ( x + )x m+ + ( x + )( x + )x m+ x m+ ( x + )x m+ (*) x m x m++ ( x + )x m+3+ ( x + )( x + ) x m+4+3 (**) We have two free terms with opposite signs at the beginning. In (*) the difference between exponents in successive terms is, while in (**) this increases to +; this difference is in both cases the exponent of x in the first factor. The remainder after the free terms begine with, so that the sequence of signs is+ ++ This process perpetuates itself and the question 6 remains which powers actually appear. It is sufficient to mar down a scheme for the exponents which completely characterises the expansion. The scheme is illustrated by what follows.

27 3. Lecture We write down the sequence of natural numbers in a row; the sequence less the first two membere is repeated in a parallel row below leaving out the first three placess at the beginning. Adding up we get , below which is placed the original sequence less the first three members, again translating the whole to the right by two places. We again add up and repeat the procedure. A typical stage in the procedure is exhibited below. m m+ m+ m+3 m+4 m m++ m+3+ m+4+3 m+5+4 m+6+5 The free indices then appear successively as 7 and in general: , λ+(λ+)+ +(λ ) λ(3λ ), (λ+)+(λ+)+ + λ λ(3λ+), which are the only exponents appearing. We thus have ( x n ) n ( ) λ x λ(3λ )/ λ

28 Lecture 4 In the last lecture we proved the surprising theorem on pentagonal numbers: 8 ( x m ) m ( ) λ x λ(3λ )/ () λ We do not need these identities for their own sae, but for their applications to number theory. We have the same sort of power-series on both sides; let us compare the coefficients of x n. On the left side n appears as the sum of different exponents. But in contradiction to previous situations, the coefficients appear with both positive and negative signs, so that when we collect the terms there may be cancellations. There are gaps in the powers that appear, but among those which appear with non-zero coefficients, we have a pair of positive terms followed by a pair of negative terms and vice versa. In most cases the coefficients are zero; this is because of cancellations, so that roughly terms with positive and negative signs are in equal number. A positive sign appears if we multiply an even number of times. otherwise a negative sign. So an even number of different summands is as frequent generally as an odd number. Hence the following theorem: The number of decompositions of n into an even number of different parts is the same as the number of decompositions into an odd number, with the exception that there is a surplus of one sort or the other if n is a pentagonal number of the form λ(3λ )/. Before proceeding further let us examine a number of concrete instances. Tae 6 which is not a pentagonal number. The partitions are 6, +5, +4, 9 ++3, so that there are two decompositions into an even number of different parts, and two into an odd number. Next tae 7, which is a pentagonal number, 7 λ(3λ+) withλ. We can actually foresee that the excess will be in the even partitions. The partitions are 7, +6, +5, 3+4, ++4. Tae 8 which 3

29 4. Lecture 4 again is not pentagonal. We have three in each category: 8, + 7, + 6, 3 + 5, ++5, This is a very extraordinary property of pentagonal numbers. One would lie to have a direct proof of this. A proof is due to Fabian Franlin (Comptes Rendus, Paris. 880), a pupil of the famous Sylvester. The proof is combinatorial. We want to establish a one-one correspondence between partitions containing an even number of summands and those containing an odd number - except for pentagonal numbers. Consider a partition with the summands arranged in increasing order, each summand being denoted by a horizontal row of dots. Mar specifically the first row, with r dots, and the last slope, with s dots i.e., points on or below a segnment starting from the dot on the extreme right of the last row and inclined at 45 (as in the diagram). We mae out two cases.. s<r. Transfer the last slope to a position immediately above the first row. The diagram is now as shown below: The uppermost row is still shorter than the others. (because in our case 30 s<r). By this procedure the number of rows is changed by. This establishes the one-one correspondence between partition of the odd type and even type.

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