# A Rademacher-Type Formula forpod(n)

Authors: Vyacheslav Kiria-Kaiserberg
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

A Rademacher-type formula for the Fourier coefficients of the generating function for the partitions ofn where no odd part is repeated is presented.


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## Body

## 1. Partitions

Apartition of a positive integer n is a representation of n as a sum of positive integers where order of summands (parts) does not matter. Let p(n) represent the number of partitions of n. In 1937, Rademacher [1, 2] was able to express p(n) as a convergent series:(1.1)p(n)=1π2∑k=1∞Ak(n)kddn(sinh⁡((π/k)(2/3)(n-1/24))n-1/24),
where(1.2)Ak(n)=∑0≤h<kgcd(h,k)=1eπi(s(h,k)-2nh/k)
is a Kloosterman sum and(1.3)s(h,k)=∑r=1k-1rk(hrk-⌊hrk⌋-12)
is a Dedekind sum.In 2011, Bruinier and Ono [3] announced a new formula that expresses p(n) as a finite sum.


### 1.1. Formula forp̃(n)

Let(1.4)f(q):=∑n=0∞p(n)qn=∏m=1∞11-qm
be Euler's generating function for p(n). H. Rademacher used the classical circle method to find the coefficients of qn. There are many other infinite products to which this method could be applied. We introduce one of these infinite products here and derive the formula for the coefficients of qn. Define(1.5)G(q):=∏m=1∞(1-qm)(1-q2m)2=(f(q2))2f(q).
Let p̃(j) denote the coefficient of qj in the expansion of G(q), that is,(1.6)G(q)=∑j=0∞p̃(j)qj.We will find a closed expression forp̃(j). Note that(1.7)G(-q)=∏m=1∞1+q2m-11-q2m=∑n=0∞pod(n)qn,
where pod(n) equals the number of partitions of n where no odd part is repeated. Thus (1.8)pod(n)=(-1)np̃(n)=(-1)n2π∑k=1(k,2)=2∞k∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kddn×(sinh⁡(π8n-1/2k)8n-1)
which is simpler than the one given by Sills [4, page 4, Equation (1.13)]  in 2010:(1.9)pod(n)=2π∑k=1∞k(1-(-1)k+⌊(4,k)4⌋)∑0≤h<k(h,k)=1ω(h,k)ω(4h/(k,4),k/(k,4))ω(2h/(k,2),k/(k,2))×e-2πinh/kddn(sinh⁡(π(k,4)(8n-1)/4k)8n-1),
where ω(h,k) is defined as(1.10)ω(h,k)=exp⁡(πi∑r=1k-1rk(hrk-⌊hrk⌋-12)).


## 1.1. Formula forp̃(n)

Let(1.4)f(q):=∑n=0∞p(n)qn=∏m=1∞11-qm
be Euler's generating function for p(n). H. Rademacher used the classical circle method to find the coefficients of qn. There are many other infinite products to which this method could be applied. We introduce one of these infinite products here and derive the formula for the coefficients of qn. Define(1.5)G(q):=∏m=1∞(1-qm)(1-q2m)2=(f(q2))2f(q).
Let p̃(j) denote the coefficient of qj in the expansion of G(q), that is,(1.6)G(q)=∑j=0∞p̃(j)qj.We will find a closed expression forp̃(j). Note that(1.7)G(-q)=∏m=1∞1+q2m-11-q2m=∑n=0∞pod(n)qn,
where pod(n) equals the number of partitions of n where no odd part is repeated. Thus (1.8)pod(n)=(-1)np̃(n)=(-1)n2π∑k=1(k,2)=2∞k∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kddn×(sinh⁡(π8n-1/2k)8n-1)
which is simpler than the one given by Sills [4, page 4, Equation (1.13)]  in 2010:(1.9)pod(n)=2π∑k=1∞k(1-(-1)k+⌊(4,k)4⌋)∑0≤h<k(h,k)=1ω(h,k)ω(4h/(k,4),k/(k,4))ω(2h/(k,2),k/(k,2))×e-2πinh/kddn(sinh⁡(π(k,4)(8n-1)/4k)8n-1),
where ω(h,k) is defined as(1.10)ω(h,k)=exp⁡(πi∑r=1k-1rk(hrk-⌊hrk⌋-12)).


## 2. Evaluation of the Path Integral

### 2.1. Convergence and Cauchy Residue Theorem

Consideringq as a complex variable in(2.1)∏m=1∞(1-qm)(1-q2m)2=∏m=1∞1(1+qm)(1-q2m)=∏m=1∞1(1-(-qm))(1-q2m),
we see from the right-hand side that infinite product and thus also infinite series are convergent for |q|<1 since(2.2)∑n=0∞(qk)n=1(1-qk),
is a geometric series which converges for |q|<1 for any fixed k≥1.Next, we note that from(2.3)G(q)=∑j=0∞p̃(j)qj,
we get that(2.4)G(q)qn+1=∑j=0∞p̃(j)qjqn+1if0<|q|<1.The series on the right side of (2.4) is a Laurent series of G(q)/q(n+1). It has a pole of order n+1 at q=0 with residue p̃(n). Applying Cauchy's Residue Theorem we get that (2.5)p̃(n)=12πi∫CG(q)qn+1dq=12πi∫C(f(q2))2f(q)qn+1dq,
where C is any positively oriented simple closed countour lying inside the unit circle.


### 2.2. Change of the Variable

The change of the variableq=e2πiτ maps the unit disk |q|<1 into an infinite vertical strip of width 1 in the τ-plane. To see this we note that from q=e2πiτ we get logq=2πiτ, so τ=logq/2πi. Choosing the branch cut to be [0,1], we get(2.6)τ=log⁡|q|2πi+Arg(q)2π.Asq traverses a circle centered at q=0 of radius e-2π in the positive direction, the point τ varies from i to i+1 along a horizontal segment as could be easly deduced from (2.6).Replacing the segment by the Rademacher path composed of upper arcs of the Ford circles formed by the Farey seriesℱN, (2.5) becomes(2.7)p̃(n)=12πi∫ii+1(f(e4πiτ))22πie2πiτf(e2πiτ)e2πiτ(n+1)dτ,
which simplifies to(2.8)p̃(n)=∫ii+1(f(e4πiτ))2f(e2πiτ)e-2πiτndτ,=∫P(N)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ.
The above can be written as(2.9)∫P(N)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ=∑k=1N∑0≤h<k(h,k)=1∫γ(h,k)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ,
where γ(h,k) is the upper arc of the Ford circle C(h,k).


### 2.3. Another Change of the Variable

Consider another change of variable(2.10)τ=hk+izk,so that(2.11)z=-ik(τ-hk)(2.12)dz=-ikdτ.Under this transformation the Ford circleC(h,k) in the τ-plane with center at h/k+i1/2k2 and radius 1/2k2 is mapped to a negatively oriented circle Ck in the z-plane with center at 1/2k and radius 1/2k. This follows from the fact that any point on the Ford circle C(h,k) is given by(2.13)τ=(hk+i12k2)+12k2eiθ,0≤θ<2π.Substitution of (2.13) into (2.11) gives(2.14)z=12k+12k(-ieiθ),
which is a circle centered at 1/2k with radius 1/2k. Now we make change of variable in (2.9). This gives(2.15)p̃(n)=i∑k=1Nk-1∑0≤h<k(h,k)=1e-2πinh/k∫sh,kth,k(f(e4πih/k-4πiz/k))2f(e2πih/k-2πz/k)e2πnz/kdz,
where(2.16)sh,k=kk2+kp2+kpk2+kp2i,th,k=kk2+ks2-ksk2+ks2iare initial and terminal points, respectively.


### 2.4. Modular Transformation

Next, we note that(2.17)f(q)=f(e2πiτ)=eπiτ/12η(τ),
where η(τ) is the Dedekind eta function. Rewriting modular functional equation [5, page 96]  for η(τ) in terms of f(q)=f(e2πiτ)=f(e2πih/k-2πz/k) we get(2.18)f(e2πih/k-2πz/k)=ω(h,k)exp⁡(π(z-1-z)12k)zf(exp⁡(2πiiz-1+Hk)),
with hH≡-1(modk), (h,k)=1.To evaluate (2.15) we would like to express(2.19)G(q)=G(e2πiτ)=G(e2πih/k-2πz/k)=(f(e4πih/k-4πiz/k))2f(e2πih/k-2πz/k),
in the same way we did for f(q) above. Two cases have to be considered: (k,2)=1 and (k,2)=2. When (k,2)=1 we will replace h by 2h and z by 2z, and when (k,2)=2, k will be replaced by k/2 in order to obtain f(q2) from f(q). Hence, we have(2.20)G(e2πih/k-2πz/k)={ω2(2h,k)eπ((2z)-1-2z)/6k2zf2(e2πi(i(2z)-1+H2)/k)ω(h,k)eπ(z-1-z)/12kzf(e2πi(iz-1+H2)/k),if(k,2)=1,ω2(h,k/2)eπ(z-1-z)/3kzf2(e4πi(iz-1+H1)/k)ω(h,k)eπ(z-1-z)/12kzf(e2πi(iz-1+H1)/k),if(k,2)=2,
which simplifies to(2.21)G(e2πih/k-2πz/k)={2ω2(2h,k)ω(h,k)e-πz/4kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k),if(k,2)=1,ω2(h,k/2)ω(h,k)eπ(z-1-z)/4kzG(e2πi(iz-1+H1)/k),if(k,2)=2,
where hHj≡-1(modk) and j∣Hj for j=1,2.We return to evaluation of (2.15). To proceed we note that(2.22)G(e2πi(iz-1+H1)/k)=1+{G(e2πi(iz-1+H1)/k)-1}.
Rewriting (2.15) in terms of (2.21) and (2.22) we obtain(2.23)p̃(n)=i∑k=1(k,2)=1Nk-1∑0≤h<k(h,k)=12ω2(2h,k)ω(h,k)e-2πinh/k∫sh,kth,kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k)eπz/k(2n-1/4)dz+i∑k=1(k,2)=2Nk-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k∫sh,kth,kz{1+(G(e2πi(iz-1+H1)/k)-1)}×e((πz/k)(2n-1/4)+π/4zk)dz=2i∑k=1(k,2)=1Nk-1∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫sh,kth,kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k)eπz/k(2n-1/4)dz+i∑k=1(k,2)=2Nk-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k(J1(h,k)+J2(h,k)),
where(2.24)J1(h,k)=∫sh,kth,kze(πz/k(2n-1/4)+π/4zk)dz,J2(h,k)=∫sh,kth,kz{G(e2πi(iz-1+H1)/k)-1}e(πz/k(2n-1/4)+π/4zk)dz.


### 2.5. Estimation of the First Term

We will estimate the first term in (2.23) and will show that it is small for large N. To do this we change variable again by letting ξ=zk. Then the first term in (2.23) becomes(2.25)2i∑k=1(k,2)=1Nk-5/2∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫sh,k*th,k*ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)eπξ/k2(2n-1/4)dξ,
where(2.26)sh,k*=k2k2+kp2+kkpk2+kp2i,(2.27)th,k*=k2k2+ks2-kksk2+ks2i,
are initial and terminal points obtained from (2.16), respectively. Under this change of variable circle Ck in z-plane with center at 1/2k and radius 1/2k is mapped to a circle Ck* in ξ-plane centered at 1/2 with radius 1/2. Note also that the mapping w=1/ξ maps the circle Ck* and its interior onto a half-plane ℜ(w)≥1 (where ℜ(w) denotes the real part of complex variable w and ℑ(w) is the imaginary part). From elementary complex analysis we have that ℜ(w)=x/(x2+y2) and ℑ(w)=-y/(x2+y2), where x+iy=ξ. It is readily seen that the segment 0<x≤1 in the ξ-plane is mapped to an infinite strip [1,∞) in the w-plane. So, it follows that inside and on the circle Ck* we have that 0<ℜ(ξ)≤1 and ℜ(1/ξ)≥1. We now show that ℜ(1/ξ)=1 on the circle Ck*. To see this note that in the polar form ξ=1/2+(1/2)eiθ on Ck*, 0≤θ≤2π. From this we get that(2.28)1ξ=21+eiθ=2(1+cos⁡θ)+isin⁡θ=2[(1+cos⁡θ)-isin⁡θ](1+cos⁡θ)2+sin⁡2θ=2(1+cos⁡θ)2+2cos⁡θ-i2sin⁡θ2+2cos⁡θ=1-isin⁡θ1+cos⁡θ.
So, ℜ(1/ξ)=1.Furthermore, we may move path of integration from the arc joiningsh,k* and th,k* to a segment connecting these two points on the circle Ck*. By [5, page 104], Theorem  5.9  the length of the path of integration is bounded by 22k/N, and on the segment connecting sh,k* and th,k*, |ξ|<2k/N.Next, let us definep̃*(m) by(2.29)∑m=0∞p̃*(m)qm=f2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k),
which is a part of the integrand in (2.25). Then, estimating the integrand in (2.25) we get
(2.30)|ξe(πξ/k2)(2n-1/4)|×|-1+f2(e2πi(i(2ξ/k-1)+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)|=|ξ|1/2|eπξ/k2(2n-1/4)|×|∑m=1∞p̃*(m)exp⁡(2πim(i(ξ/k)-1+H2)k)|=|ξ|1/2|eπξ/k2(2n-1/4)|×|∑m=1∞p̃*(m)exp⁡(-2πmξ)exp⁡(2πimH2k)|≤|ξ|1/2eπ/k2(2n-1/4)R(ξ)|∑m=1∞p̃*(m)exp⁡(-2πmξ)exp⁡(2πimH2k)|≤|ξ|1/2e2πn∑m=1∞|p̃*(m)|exp⁡(-2πmR(1ξ))≤|ξ|1/2e2πn∑m=1∞|p̃*(m)|e-2πm=|ξ|1/2e2πn∑m=1∞|p̃*(m)|ym,(wherey=e-2π)=c|ξ|1/2,
where(2.31)c=e2πn∑m=1∞|p̃*(m)|ym.
Note that c does not depend on ξ or N. It depends on n, but n remains fixed in the above analysis. So,(2.32)|∫s*(h,k)t*(h,k)ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)e(πξ/k2)(2n-1/4)dξ|≤c|ξ|1/2≤c(2kN)1/222NN<αk3/2N-3/2,
for some constant α, and we have that(2.33)|2i∑k=1(k,2)=1Nk-5/2∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫s*(h,k)t*(h,k)ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)eπξ/k2(2n-1/4)(dξ)|≤2∑k=1(k,2)=1N∑0≤h<k(h,k)=1αk-1N-3/2≤2αN-3/2∑k=1(k,2)=1N1=2αN-1/2.This completes the estimation of the first term in (2.23). We proceed to the second term.


### 2.6. Estimation of the Second Term

First, we will show that(2.34)J2(h,k)=∫sh,kth,kz{G(e2πi(iz-1+H1)/k)-1}e(πz/k(2n-1/4)+π/4zk)dz
is small for large N. Making change of variable ξ=zk as before, we get that(2.35)J2(h,k)=k-3/2∫sh,k*th,k*ξ{G(e2πi(i(ξ/k)-1+H1)/k)-1}e(πξ/k2(2n-1/4)+π/4ξ)dξ,
where sh,k* and th,k* are as in (2.26), respectively. As before, we define p̃**(m) by(2.36)∑m=0∞p̃**(m)qm=G(e2πi(i(ξ/k-1)+H1)/k)-1.Then, estimating the integrand, we see that(2.37)|ξe((πξ/k2)(2n-1/4)+π/4ξ)|×|G(e2πi(i(ξ/k)-1+H1)/k)-1|=|ξ|1/2|eπξ/k2(2n-1/4)eπ/4ξ|×|∑m=0∞p̃**(m)exp⁡(2πim(ik/ξ+H1)k)-1|≤|ξ|1/2e(π/k2)(2n-1/4)R(ξ)eπ/4R(1/ξ)|∑m=1∞p̃**(m)exp⁡(-2πmξ)exp⁡(2πimH1k)|≤|ξ|1/2e2πneπ/4R(1/ξ)∑m=1∞|p̃**(m)|exp⁡(-2πmR(1ξ))=|ξ|1/2e2πn∑m=1∞|p̃**(m)|exp⁡((-2πm+π4)R(1ξ))≤|ξ|1/2e2πn∑m=1∞|p̃**(m)|exp⁡⁡(-2πm+π4)=|ξ|1/2e2πn∑m=1∞|p̃**(m)|e-π/4(8m-1)≤|ξ|1/2e2πn∑m=1∞|p̃**(8m-1)|e-π/4(8m-1)=|ξ|1/2e2πn∑m=1∞|p̃**(8m-1)|x8m-1,wherex=e-π/4=b|ξ|1/2,
where(2.38)b=e2πn∑m=1∞|p̃**(8m-1)|x8m-1.
Note that b does not depend on ξ or N. It depends on n, but n is fixed. It follows, therefore, that(2.39)|J2(h,k)|≤b(2kN)1/222NN<βk3/2N-3/2,
for some constant β. Then we have that(2.40)|i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinhkJ2(h,k)|<∑k=1(k,2)=2N∑0≤h<k(h,k)=1βk-1N-3/2≤βN-3/2∑k=1(k,2)=2N1=βN-1/2.Combining the results from (2.33) and (2.40) we have that(2.41)p̃(n)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kJ1(h,k)+O(βN-1/2+2αN-1/2)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kJ1(h,k)+O(N-1/2).Finally, we turn our attention to(2.42)J1(h,k)=k-3/2∫sh,k*th,k*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ.We note that(2.43)J1(h,k)=∫Ck*-∫sh,k*0-∫0th,k*=∫Ck*-S1-S2,
where Ck* is a circle in the ξ-plane centered at 1/2 with radius 1/2, as before. It is easily seen that the length of the arc connecting 0 and sh,k* is less then(2.44)2π|sh,k*|2≤π|sh,k*|≤π2kN.
From the discussion above we know that ℜ(1/ξ)=1 and 0<ℜ(ξ)≤1 on Ck*. So, the integrand in S1 could be estimated as(2.45)|ξe(πξ/k2(2n-1/4)+π/4ξ)|=|ξ|1/2|eπξ/k2(2n-1/4)||eπ/4ξ|=|ξ|1/2eπ/k2(2n-1/4)R(ξ)eπ/4R(1/ξ)≤21/4k1/2N1/2e2πneπ/4.


### 2.7. Combining the Results

We combine the results in (2.44) and (2.45) to get(2.46)|S1|<γk3/2N-3/2,
where γ is a constant. We can obtain similar estimate for S2 and, as before, we get an error term O(N-1/2) in the formula for p̃(n). Therefore, we can write(2.47)p̃(n)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫Ck*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ+O(N-1/2).
Letting N→∞ we have that(2.48)p̃(n)=i∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫Ck*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ.We introduce another change of variable(2.49)ξ=1w,dξ=-1w2.Then (2.48) becomes(2.50)p̃(n)=1i∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫1-∞i1+∞iw-5/2e((π/k2)(2n-1/4)1/w+πw/4)dw.
Let t=πw/4 in (2.50), then the above becomes(2.51)p̃(n)=2π(π3/28)∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k12πi×∫π/4-∞iπ/4+∞it-5/2e(t+(π2/4k2)(2n-1/4)(1/t))dt.


### 2.8. Bessel Function

In Watson's Treatise on Bessel functions [6, page 181], we find a formula equivalent to the following:(2.52)Iν(z)=(1/2z)ν2πi∫c-∞ic+∞it-ν-1et+(z2/4t)dt,(ifc>0,R(ν)>0).
Let(2.53)z2={π24k2(2n-14)}1/2andν=3/2. Then we have(2.54)p̃(n)=2π(π3/28)∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×π-3/2(2n-1/4)-3/44-3/4k-3/2I3/2(πk2n-14)=2π(2n-1/4)-3/48∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kI3/2(πk2n-14).Note that Bessel functions of this order can be expressed as(2.55)I3/2(z)=2zπddz(sinh⁡zz).
Expanding (2.55) we have that(2.56)I3/2(z)=2zπ(cosh⁡zz-sinh⁡zz2).Substituting (2.53) into (2.56), we get(2.57)I3/2(z)=I3/2(πk(2n-14)1/2)=I3/2(π8n-12k)=2(π8n-1/2k)π(cosh⁡(π8n-1/2k)(π8n-1/2k)-sinh⁡(π8n-1/2k)(π8n-1/2k)2)=(8n-1)1/4k(2cosh⁡(π8n-1/2k)πk8n-1-4k/πsinh⁡(π8n-1/2k)πk(8n-1))=1π8n-1/k(2cosh⁡(π8n-12k)-4ksinh⁡(π8n-1/2k)π8n-1).
Multiplying (2.57) by(2.58)2π(2n-1/4)-3/48=2π(8n-1)3/4,
we get(2.59)2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k.


### 2.9. Final Form

Finally, we rewrite (2.54) in terms of (2.59) to get(2.60)p̃(n)=∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k.
Thus,(2.61)pod(n)=(-1)n∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k,
or equivalently(2.62)pod(n)=(-1)n2π∑k=1(k,2)=2∞k∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kddn(sinh⁡(π8n-1/2k)8n-1).


## 2.1. Convergence and Cauchy Residue Theorem

Consideringq as a complex variable in(2.1)∏m=1∞(1-qm)(1-q2m)2=∏m=1∞1(1+qm)(1-q2m)=∏m=1∞1(1-(-qm))(1-q2m),
we see from the right-hand side that infinite product and thus also infinite series are convergent for |q|<1 since(2.2)∑n=0∞(qk)n=1(1-qk),
is a geometric series which converges for |q|<1 for any fixed k≥1.Next, we note that from(2.3)G(q)=∑j=0∞p̃(j)qj,
we get that(2.4)G(q)qn+1=∑j=0∞p̃(j)qjqn+1if0<|q|<1.The series on the right side of (2.4) is a Laurent series of G(q)/q(n+1). It has a pole of order n+1 at q=0 with residue p̃(n). Applying Cauchy's Residue Theorem we get that (2.5)p̃(n)=12πi∫CG(q)qn+1dq=12πi∫C(f(q2))2f(q)qn+1dq,
where C is any positively oriented simple closed countour lying inside the unit circle.


## 2.2. Change of the Variable

The change of the variableq=e2πiτ maps the unit disk |q|<1 into an infinite vertical strip of width 1 in the τ-plane. To see this we note that from q=e2πiτ we get logq=2πiτ, so τ=logq/2πi. Choosing the branch cut to be [0,1], we get(2.6)τ=log⁡|q|2πi+Arg(q)2π.Asq traverses a circle centered at q=0 of radius e-2π in the positive direction, the point τ varies from i to i+1 along a horizontal segment as could be easly deduced from (2.6).Replacing the segment by the Rademacher path composed of upper arcs of the Ford circles formed by the Farey seriesℱN, (2.5) becomes(2.7)p̃(n)=12πi∫ii+1(f(e4πiτ))22πie2πiτf(e2πiτ)e2πiτ(n+1)dτ,
which simplifies to(2.8)p̃(n)=∫ii+1(f(e4πiτ))2f(e2πiτ)e-2πiτndτ,=∫P(N)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ.
The above can be written as(2.9)∫P(N)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ=∑k=1N∑0≤h<k(h,k)=1∫γ(h,k)(f(e4πiτ))2f(e2πiτ)e-2πiτndτ,
where γ(h,k) is the upper arc of the Ford circle C(h,k).


## 2.3. Another Change of the Variable

Consider another change of variable(2.10)τ=hk+izk,so that(2.11)z=-ik(τ-hk)(2.12)dz=-ikdτ.Under this transformation the Ford circleC(h,k) in the τ-plane with center at h/k+i1/2k2 and radius 1/2k2 is mapped to a negatively oriented circle Ck in the z-plane with center at 1/2k and radius 1/2k. This follows from the fact that any point on the Ford circle C(h,k) is given by(2.13)τ=(hk+i12k2)+12k2eiθ,0≤θ<2π.Substitution of (2.13) into (2.11) gives(2.14)z=12k+12k(-ieiθ),
which is a circle centered at 1/2k with radius 1/2k. Now we make change of variable in (2.9). This gives(2.15)p̃(n)=i∑k=1Nk-1∑0≤h<k(h,k)=1e-2πinh/k∫sh,kth,k(f(e4πih/k-4πiz/k))2f(e2πih/k-2πz/k)e2πnz/kdz,
where(2.16)sh,k=kk2+kp2+kpk2+kp2i,th,k=kk2+ks2-ksk2+ks2iare initial and terminal points, respectively.


## 2.4. Modular Transformation

Next, we note that(2.17)f(q)=f(e2πiτ)=eπiτ/12η(τ),
where η(τ) is the Dedekind eta function. Rewriting modular functional equation [5, page 96]  for η(τ) in terms of f(q)=f(e2πiτ)=f(e2πih/k-2πz/k) we get(2.18)f(e2πih/k-2πz/k)=ω(h,k)exp⁡(π(z-1-z)12k)zf(exp⁡(2πiiz-1+Hk)),
with hH≡-1(modk), (h,k)=1.To evaluate (2.15) we would like to express(2.19)G(q)=G(e2πiτ)=G(e2πih/k-2πz/k)=(f(e4πih/k-4πiz/k))2f(e2πih/k-2πz/k),
in the same way we did for f(q) above. Two cases have to be considered: (k,2)=1 and (k,2)=2. When (k,2)=1 we will replace h by 2h and z by 2z, and when (k,2)=2, k will be replaced by k/2 in order to obtain f(q2) from f(q). Hence, we have(2.20)G(e2πih/k-2πz/k)={ω2(2h,k)eπ((2z)-1-2z)/6k2zf2(e2πi(i(2z)-1+H2)/k)ω(h,k)eπ(z-1-z)/12kzf(e2πi(iz-1+H2)/k),if(k,2)=1,ω2(h,k/2)eπ(z-1-z)/3kzf2(e4πi(iz-1+H1)/k)ω(h,k)eπ(z-1-z)/12kzf(e2πi(iz-1+H1)/k),if(k,2)=2,
which simplifies to(2.21)G(e2πih/k-2πz/k)={2ω2(2h,k)ω(h,k)e-πz/4kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k),if(k,2)=1,ω2(h,k/2)ω(h,k)eπ(z-1-z)/4kzG(e2πi(iz-1+H1)/k),if(k,2)=2,
where hHj≡-1(modk) and j∣Hj for j=1,2.We return to evaluation of (2.15). To proceed we note that(2.22)G(e2πi(iz-1+H1)/k)=1+{G(e2πi(iz-1+H1)/k)-1}.
Rewriting (2.15) in terms of (2.21) and (2.22) we obtain(2.23)p̃(n)=i∑k=1(k,2)=1Nk-1∑0≤h<k(h,k)=12ω2(2h,k)ω(h,k)e-2πinh/k∫sh,kth,kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k)eπz/k(2n-1/4)dz+i∑k=1(k,2)=2Nk-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k∫sh,kth,kz{1+(G(e2πi(iz-1+H1)/k)-1)}×e((πz/k)(2n-1/4)+π/4zk)dz=2i∑k=1(k,2)=1Nk-1∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫sh,kth,kzf2(e2πi(i(2z)-1+H2)/k)f(e2πi(iz-1+H2)/k)eπz/k(2n-1/4)dz+i∑k=1(k,2)=2Nk-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k(J1(h,k)+J2(h,k)),
where(2.24)J1(h,k)=∫sh,kth,kze(πz/k(2n-1/4)+π/4zk)dz,J2(h,k)=∫sh,kth,kz{G(e2πi(iz-1+H1)/k)-1}e(πz/k(2n-1/4)+π/4zk)dz.


## 2.5. Estimation of the First Term

We will estimate the first term in (2.23) and will show that it is small for large N. To do this we change variable again by letting ξ=zk. Then the first term in (2.23) becomes(2.25)2i∑k=1(k,2)=1Nk-5/2∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫sh,k*th,k*ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)eπξ/k2(2n-1/4)dξ,
where(2.26)sh,k*=k2k2+kp2+kkpk2+kp2i,(2.27)th,k*=k2k2+ks2-kksk2+ks2i,
are initial and terminal points obtained from (2.16), respectively. Under this change of variable circle Ck in z-plane with center at 1/2k and radius 1/2k is mapped to a circle Ck* in ξ-plane centered at 1/2 with radius 1/2. Note also that the mapping w=1/ξ maps the circle Ck* and its interior onto a half-plane ℜ(w)≥1 (where ℜ(w) denotes the real part of complex variable w and ℑ(w) is the imaginary part). From elementary complex analysis we have that ℜ(w)=x/(x2+y2) and ℑ(w)=-y/(x2+y2), where x+iy=ξ. It is readily seen that the segment 0<x≤1 in the ξ-plane is mapped to an infinite strip [1,∞) in the w-plane. So, it follows that inside and on the circle Ck* we have that 0<ℜ(ξ)≤1 and ℜ(1/ξ)≥1. We now show that ℜ(1/ξ)=1 on the circle Ck*. To see this note that in the polar form ξ=1/2+(1/2)eiθ on Ck*, 0≤θ≤2π. From this we get that(2.28)1ξ=21+eiθ=2(1+cos⁡θ)+isin⁡θ=2[(1+cos⁡θ)-isin⁡θ](1+cos⁡θ)2+sin⁡2θ=2(1+cos⁡θ)2+2cos⁡θ-i2sin⁡θ2+2cos⁡θ=1-isin⁡θ1+cos⁡θ.
So, ℜ(1/ξ)=1.Furthermore, we may move path of integration from the arc joiningsh,k* and th,k* to a segment connecting these two points on the circle Ck*. By [5, page 104], Theorem  5.9  the length of the path of integration is bounded by 22k/N, and on the segment connecting sh,k* and th,k*, |ξ|<2k/N.Next, let us definep̃*(m) by(2.29)∑m=0∞p̃*(m)qm=f2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k),
which is a part of the integrand in (2.25). Then, estimating the integrand in (2.25) we get
(2.30)|ξe(πξ/k2)(2n-1/4)|×|-1+f2(e2πi(i(2ξ/k-1)+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)|=|ξ|1/2|eπξ/k2(2n-1/4)|×|∑m=1∞p̃*(m)exp⁡(2πim(i(ξ/k)-1+H2)k)|=|ξ|1/2|eπξ/k2(2n-1/4)|×|∑m=1∞p̃*(m)exp⁡(-2πmξ)exp⁡(2πimH2k)|≤|ξ|1/2eπ/k2(2n-1/4)R(ξ)|∑m=1∞p̃*(m)exp⁡(-2πmξ)exp⁡(2πimH2k)|≤|ξ|1/2e2πn∑m=1∞|p̃*(m)|exp⁡(-2πmR(1ξ))≤|ξ|1/2e2πn∑m=1∞|p̃*(m)|e-2πm=|ξ|1/2e2πn∑m=1∞|p̃*(m)|ym,(wherey=e-2π)=c|ξ|1/2,
where(2.31)c=e2πn∑m=1∞|p̃*(m)|ym.
Note that c does not depend on ξ or N. It depends on n, but n remains fixed in the above analysis. So,(2.32)|∫s*(h,k)t*(h,k)ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)e(πξ/k2)(2n-1/4)dξ|≤c|ξ|1/2≤c(2kN)1/222NN<αk3/2N-3/2,
for some constant α, and we have that(2.33)|2i∑k=1(k,2)=1Nk-5/2∑0≤h<k(h,k)=1ω2(2h,k)ω(h,k)e-2πinh/k∫s*(h,k)t*(h,k)ξf2(e2πi(i(2ξ/k)-1+H2)/k)f(e2πi(i(ξ/k)-1+H2)/k)eπξ/k2(2n-1/4)(dξ)|≤2∑k=1(k,2)=1N∑0≤h<k(h,k)=1αk-1N-3/2≤2αN-3/2∑k=1(k,2)=1N1=2αN-1/2.This completes the estimation of the first term in (2.23). We proceed to the second term.


## 2.6. Estimation of the Second Term

First, we will show that(2.34)J2(h,k)=∫sh,kth,kz{G(e2πi(iz-1+H1)/k)-1}e(πz/k(2n-1/4)+π/4zk)dz
is small for large N. Making change of variable ξ=zk as before, we get that(2.35)J2(h,k)=k-3/2∫sh,k*th,k*ξ{G(e2πi(i(ξ/k)-1+H1)/k)-1}e(πξ/k2(2n-1/4)+π/4ξ)dξ,
where sh,k* and th,k* are as in (2.26), respectively. As before, we define p̃**(m) by(2.36)∑m=0∞p̃**(m)qm=G(e2πi(i(ξ/k-1)+H1)/k)-1.Then, estimating the integrand, we see that(2.37)|ξe((πξ/k2)(2n-1/4)+π/4ξ)|×|G(e2πi(i(ξ/k)-1+H1)/k)-1|=|ξ|1/2|eπξ/k2(2n-1/4)eπ/4ξ|×|∑m=0∞p̃**(m)exp⁡(2πim(ik/ξ+H1)k)-1|≤|ξ|1/2e(π/k2)(2n-1/4)R(ξ)eπ/4R(1/ξ)|∑m=1∞p̃**(m)exp⁡(-2πmξ)exp⁡(2πimH1k)|≤|ξ|1/2e2πneπ/4R(1/ξ)∑m=1∞|p̃**(m)|exp⁡(-2πmR(1ξ))=|ξ|1/2e2πn∑m=1∞|p̃**(m)|exp⁡((-2πm+π4)R(1ξ))≤|ξ|1/2e2πn∑m=1∞|p̃**(m)|exp⁡⁡(-2πm+π4)=|ξ|1/2e2πn∑m=1∞|p̃**(m)|e-π/4(8m-1)≤|ξ|1/2e2πn∑m=1∞|p̃**(8m-1)|e-π/4(8m-1)=|ξ|1/2e2πn∑m=1∞|p̃**(8m-1)|x8m-1,wherex=e-π/4=b|ξ|1/2,
where(2.38)b=e2πn∑m=1∞|p̃**(8m-1)|x8m-1.
Note that b does not depend on ξ or N. It depends on n, but n is fixed. It follows, therefore, that(2.39)|J2(h,k)|≤b(2kN)1/222NN<βk3/2N-3/2,
for some constant β. Then we have that(2.40)|i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinhkJ2(h,k)|<∑k=1(k,2)=2N∑0≤h<k(h,k)=1βk-1N-3/2≤βN-3/2∑k=1(k,2)=2N1=βN-1/2.Combining the results from (2.33) and (2.40) we have that(2.41)p̃(n)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kJ1(h,k)+O(βN-1/2+2αN-1/2)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kJ1(h,k)+O(N-1/2).Finally, we turn our attention to(2.42)J1(h,k)=k-3/2∫sh,k*th,k*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ.We note that(2.43)J1(h,k)=∫Ck*-∫sh,k*0-∫0th,k*=∫Ck*-S1-S2,
where Ck* is a circle in the ξ-plane centered at 1/2 with radius 1/2, as before. It is easily seen that the length of the arc connecting 0 and sh,k* is less then(2.44)2π|sh,k*|2≤π|sh,k*|≤π2kN.
From the discussion above we know that ℜ(1/ξ)=1 and 0<ℜ(ξ)≤1 on Ck*. So, the integrand in S1 could be estimated as(2.45)|ξe(πξ/k2(2n-1/4)+π/4ξ)|=|ξ|1/2|eπξ/k2(2n-1/4)||eπ/4ξ|=|ξ|1/2eπ/k2(2n-1/4)R(ξ)eπ/4R(1/ξ)≤21/4k1/2N1/2e2πneπ/4.


## 2.7. Combining the Results

We combine the results in (2.44) and (2.45) to get(2.46)|S1|<γk3/2N-3/2,
where γ is a constant. We can obtain similar estimate for S2 and, as before, we get an error term O(N-1/2) in the formula for p̃(n). Therefore, we can write(2.47)p̃(n)=i∑k=1(k,2)=2Nk-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫Ck*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ+O(N-1/2).
Letting N→∞ we have that(2.48)p̃(n)=i∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫Ck*ξe((πξ/k2)(2n-1/4)+π/4ξ)dξ.We introduce another change of variable(2.49)ξ=1w,dξ=-1w2.Then (2.48) becomes(2.50)p̃(n)=1i∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×∫1-∞i1+∞iw-5/2e((π/k2)(2n-1/4)1/w+πw/4)dw.
Let t=πw/4 in (2.50), then the above becomes(2.51)p̃(n)=2π(π3/28)∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k12πi×∫π/4-∞iπ/4+∞it-5/2e(t+(π2/4k2)(2n-1/4)(1/t))dt.


## 2.8. Bessel Function

In Watson's Treatise on Bessel functions [6, page 181], we find a formula equivalent to the following:(2.52)Iν(z)=(1/2z)ν2πi∫c-∞ic+∞it-ν-1et+(z2/4t)dt,(ifc>0,R(ν)>0).
Let(2.53)z2={π24k2(2n-14)}1/2andν=3/2. Then we have(2.54)p̃(n)=2π(π3/28)∑k=1(k,2)=2∞k-5/2∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×π-3/2(2n-1/4)-3/44-3/4k-3/2I3/2(πk2n-14)=2π(2n-1/4)-3/48∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kI3/2(πk2n-14).Note that Bessel functions of this order can be expressed as(2.55)I3/2(z)=2zπddz(sinh⁡zz).
Expanding (2.55) we have that(2.56)I3/2(z)=2zπ(cosh⁡zz-sinh⁡zz2).Substituting (2.53) into (2.56), we get(2.57)I3/2(z)=I3/2(πk(2n-14)1/2)=I3/2(π8n-12k)=2(π8n-1/2k)π(cosh⁡(π8n-1/2k)(π8n-1/2k)-sinh⁡(π8n-1/2k)(π8n-1/2k)2)=(8n-1)1/4k(2cosh⁡(π8n-1/2k)πk8n-1-4k/πsinh⁡(π8n-1/2k)πk(8n-1))=1π8n-1/k(2cosh⁡(π8n-12k)-4ksinh⁡(π8n-1/2k)π8n-1).
Multiplying (2.57) by(2.58)2π(2n-1/4)-3/48=2π(8n-1)3/4,
we get(2.59)2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k.


## 2.9. Final Form

Finally, we rewrite (2.54) in terms of (2.59) to get(2.60)p̃(n)=∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k.
Thus,(2.61)pod(n)=(-1)n∑k=1(k,2)=2∞k-1∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/k×2(2cosh⁡(π8n-1/2k)-4ksinh⁡(π8n-1/2k)/π8n-1)(8n-1)3/48n-1/k,
or equivalently(2.62)pod(n)=(-1)n2π∑k=1(k,2)=2∞k∑0≤h<k(h,k)=1ω2(h,k/2)ω(h,k)e-2πinh/kddn(sinh⁡(π8n-1/2k)8n-1).



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*Source: 976723-2011-10-29.xml*