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LL-Functions of Quadratic Characters at Negative Integers

Henri Cohen

Alex Best

Casper Putz

Jana Sotáková

Joey van Langen

drawing

The beginning of the project is the following theorem due to Siegel, and Cohen and Zagier:

Theorem 2.1: Let Fk,DF_{k,D} be defined as follows: Fk,D(τ)=Bk4kL(χD,1k)+n1ck,D(n)qnF_{k,D}(\tau) = -\frac{B_k}{4k} L(\chi_D, 1-k) + \sum_{n \geq 1} c_{k,D}(n)q^n

for ck,D(n)=dNdk1(Dd)s<(n/d)(D),s(n/d)Dmod2σk1((n/d)2Ds24) c_{k,D}(n) = \sum_{d|N} d^{k-1} \left( \frac{D}{d} \right) \sum_{|s|< (n/d)\sqrt(D), s \equiv (n/d)D \mod 2} \sigma_{k-1}\left( \frac{(n/d)^2 D - s^2}{4} \right) then Fk,DM2k(Γ)F_{k,D} \in M_{2k}(\Gamma).

With the notation that σk1(α)=0\sigma_{k-1}(\alpha) =0 if α\alpha is not an integer, we can simplify the formula using Sk(m,N)=sZσk1(ms2N)S_k(m,N) = \sum_{s\in \mathbf Z} \sigma_{k-1}\left(\frac{m - s^2}{N}\right)

to the following: Fk,D(τ)=Bk4kL(χD,1k)+Sk(D,N)q+O(q2)F_{k,D}(\tau) = -\frac{B_k}{4k} L(\chi_D, 1-k) + S_{k}(D,N) \cdot q + O(q^2)

For small values of kk, we know that M2k(Γ)=CE2kM_{2k}(\Gamma) = \mathbb{C} \cdot E_{2k} for the Eisenstein series E2k=1+22kB2kn1σk1(n)qn E_{2k} = 1 + \frac{- 2 \cdot 2k }{B_{2k}} \sum_{n \geq 1} \sigma_{k-1}(n)q^n



For example, for k=2k=2, M4(Γ)M_4(\Gamma) is generated by E4=1+240q+O(q2) E_4 = 1 + 240 q + O(q^2) and so we have F2,NF_{2,N} is a multiple of E4E_4, say F2,N=αE4F_{2,N} = \alpha E_4.

Comparing coefficients, a0(F2,N)=168L(χD,1)=αa_0(F_{2,N}) = \frac{-1}{6 \cdot 8} L(\chi_D, -1) = \alpha a1(F2,N)=S2(D,4)=α240 a_1(F_{2,N}) = S_2(D,4) = \alpha \cdot 240

Comparing coefficients, we deduce that L(χD,1)=68240S2(D,4)=15S2(D,4) L(\chi_D, -1) = \frac{- \cdot 6 \cdot 8}{240} \cdot S_2(D,4) = \frac{-1}{5} S_2(D,4)

In [1]:
def Sk(k,m,N):
    o = sigma(ZZ(m).divide_knowing_divisible_by(N), k = k-1) if m%N == 0 else 0
    cur = m-1
    diff = 3
    while cur > 0:
        if (cur)%N ==0:
            o += 2*sigma((cur).divide_knowing_divisible_by(N), k=k-1)
        cur = cur - diff
        diff = diff + 2
    return o

@cached_function
def fund_discs(B,sqmod=1,cong=1,cl=0):
    return [d for d in range(1, B, 1) if is_fundamental_discriminant(d) and Integers(sqmod)(d).is_square() and d % cong == cl]

In case the space M2k(Γ)M_{2k}(\Gamma) has dimension larger than 11, writing Fk,NF_{k,N} in an appropriate basis, we can get other formulas:

L(χD,3)=S4(D,4)L(\chi_D, -3) = S_4(D,4)

L(χD,5)=8S6(D,4)65S6(D,1)+25(D2)S6(D,4)195L(\chi_D,-5) = -\frac{8S_6(D,4)}{65} -\frac{S_6(D,1) + 2^5 \left(\frac{D}{2}\right)S_6(D,4)}{195} =23(3+22(D2))S6(D,4)+S6(D,1)195 = -\frac{2^3(3+2^2 \left(\frac{D}{2}\right))S_6(D,4) + S_6(D,1)}{195}

L(χD,7)=24S8(D,4)17+S8(D,1)+281(D2)S8(D,4)153L(\chi_D,-7) = -\frac{24S_8(D,4)}{17} +\frac{S_8(D,1) + 2^{8-1} \left(\frac{D}{2}\right)S_8(D,4)}{153} =23(33+25(D2))S8(D,4)+S8(D,1)153 = -\frac{2^3(3^3 +2^5\left(\frac{D}{2}\right))S_8(D,4) + S_8(D,1)}{153}


...

L(χD,11)=9783S12(D,4)13820+3S12(D,1)+2121(D2)S12(D,4)17275+S12(32D,4)+3121(D3)S12(D,4)276400L(\chi_D,-11) = \frac{-9783S_{12}(D,4)}{13820} +3\frac{S_{12}(D,1) + 2^{12-1} \left(\frac{D}{2}\right)S_{12}(D,4)}{17275} + \frac{S_{12}(3^2D,4) + 3^{12-1} \left(\frac{D}{3}\right)S_{12}(D,4)}{276400} =(209783+3162121(D2)+3121(D3))S12(D,4)+S12(32D,4)+316S12(D,1)276400 = \frac{(-20\cdot 9783 +3\cdot16\cdot2^{12-1} \left(\frac{D}{2}\right) + 3^{12-1} \left(\frac{D}{3}\right))S_{12}(D,4)+ S_{12}(3^2D,4)+ 3\cdot16S_{12}(D,1)}{276400}

There exists a more general version of Theorem 2.1 which works for general kk.

Theorem 2.3: Let D>1D>1 be a fundamental discriminant and let k2k\geq 2 be any integer. Let ψ\psi be a primitive character modulo FF with ψ(1)=(1)k\psi(-1) = (-1)^k. Let NN be a squarefree integer such that gcd(F,ND)=1\gcd(F,ND) = 1. Set ck,D(0)=L(ψ,1k)L(ϕχD,1k)4pN(1ϕχD(p)pk1)c_{k,D}(0) = \frac{L(\psi,1-k)L(\phi\chi_D,1-k)}{4}\prod_{p\mid N} (1-\phi\chi_D(p)p^{k-1}) and for n1n\geq 1: ck,D(n)=dn,gcd(d,N)=1ϕχD(d)dk1sZσk1,ψ((n/d)2Ds24N)c_{k,D}(n) = \sum_{d\mid n,\gcd(d,N)=1}\phi\chi_D(d)d^{k-1}\sum_{s\in\mathbf Z}\sigma_{k-1,\psi}\left(\frac{(n/d)^2D-s^2}{4N}\right) where σk1,ψ(m)=0\sigma_{k-1,\psi}(m) = 0 if mm is not a positive integer and otherwise σk1,ψ(m)=dψ(d)dk1\sigma_{k-1,\psi}(m) = \sum_{d\mid }\psi(d)d^{k-1}. Then n0ck,D(n)qnM2k(Γ0(FN),ψ2).\sum_{n\geq 0}c_{k,D}(n)q^n\in M_{2k}(\Gamma_0(FN),\psi^2).

A consequence of this theorem is that there always exist relations as we found before between L(χD,1k)L(\chi_D,1-k) and Sk(mD)S_k(mD) for appropriate mm.

Corollary: Let k2k\geq 2 be even and N1N\geq 1 with 4N4\mid N. Then there exists tt and constants cnc_n for 1nt1\leq n\leq t such that whenever DD is a square modulo NN we have

L(χD,1k)=1ntcnSk(n2D,N)L(\chi_D,1-k) = \sum_{1\leq n\leq t}c_n S_k(n^2D,N)

So instead of finding relations theoretically we can search for them, knowing that they must exist.

In [3]:
N = 4*4
pts=[]
for D in fund_discs(3000, sqmod=N, cong=8, cl=1):
    pts +=[(quadratic_L_function__exact(-1,D), Sk(2,D,N))]
points(pts)
Out[3]:
In [4]:
Matrix(pts).right_kernel()
Out[4]:
Vector space of degree 2 and dimension 1 over Rational Field
Basis matrix:
[1 2]
In [5]:
N = 4*12
pts=[]
for D in fund_discs(3000, sqmod=N, cong=3, cl=1):
    pts +=[(quadratic_L_function__exact(-1,D), Sk(2,D,N))]
points(pts)
Out[5]:
In [6]:
N = 4*12
pts=[]
for D in fund_discs(1000, sqmod=N, cong=3, cl=1):
    pts +=[(quadratic_L_function__exact(-1,D), Sk(2,D,N),Sk(2,9*D,N))]

points(pts)
Out[6]:

Our next goal is to find such relations for different NN and determine which relations are ''best'' to compute the special value L(χD,1k)L(\chi_D,1-k) (i.e. find some cost function for these relations).

In [7]:
class QuadraticLValueRelation():
    def __init__(self,N,Dcong,Dmod,k,dic):
        self._k=  k
        self._N=  N
        self._Dcong = Dcong
        self._Dmod= Dmod
        self._dic= dic
        pass
    
    def __repr__(self):
        return "Quadratic L-value relation given by sigma_%d(%d)L(chi_D,%d) = "%(self._k-1,self._N/4,1-self._k) + "+".join(r"%d/%d*S_%d(%d^2 D,%d)" %(v.numerator(),v.denominator(),self._k,ke,self._N) for ke,v in self._dic.iteritems()) + " for D = %d (mod %d)" % (self._Dmod,self._Dcong)
        
    def _latex_(self):
        return "\sigma_{%d}(%d)L(\chi_D,%d) = "%(self._k-1, self._N/4,1-self._k) + "+".join(r"%sS_%d\left(%d^2 D,%d\right)" %(latex(v),self._k,ke,self._N) for ke,v in self._dic.iteritems())+ " \\text{ when }D \equiv %d \pmod{%d}" % (self._Dmod,self._Dcong)
        
    def cost(self):
        N = ZZ(self._N)
        Dcong= ZZ(self._Dcong)
        Dmod = ZZ(self._Dmod)
        mx = 0
        gc = gcd(Dcong,N)
        for D in Integers(N):
            if Integers(Dcong)(D.lift()) != Dmod:
                continue
            new = sum([1 for n in Integers(N) if n^2 == D])
            if new > mx:
                mx = new

        return max(self._dic.keys())*mx/N

    def probapplies(self):
        Dcong= ZZ(self._Dcong)
        Dmod = ZZ(self._Dmod)
        produ = 1
        if Dcong % 2 == 0 and Dmod % 4 in [2,3]:
            return 0
        for (p,e) in factor(Dcong):
            if p == 2:
                mod16 =[1,5,9,13,8,12]
                if e >=4:
                    if Dmod % 16 in mod16:
                        produ *= 1/(6*2^(e-4))
                    else: return 0
                if e == 3:
                    if Dmod % 8 == 0 or Dmod % 8 ==4:
                        produ *= 1/6
                    else: produ *=1/3
                if e == 2 or e ==1:
                    if Dmod % 4 ==0:
                        produ *= 1/3
                    else: produ *= 2/3
                continue
            if e >= 2 and Dmod % p^2 == 0:
                return 0
            if e == 1:
                if Dmod % p == 0:
                    produ *= (p-1)/((p^2-1))
                    continue
                produ *= p/(p^2-1)
            else:
                produ *= 1/(p^e - p^(e-2))

        return produ

    def test(self):
        for D in fund_discs(1000, self._N, self._Dcong, self._Dmod):
            if (sigma(self._N/4, k = self._k - 1)*quadratic_L_function__exact(1-self._k,D) != sum(v*Sk(self._k,ke^2*D,self._N) for ke,v in self._dic.iteritems())):
                print quadratic_L_function__exact(1-self._k,D) , sum(v*Sk(self._k,ke^2*D,self._N) for ke,v in self._dic.iteritems()),self._dic
                raise Exception
In [8]:
def find_relations(N, k=2, sample=10000, tmax=5, confidence=40):
    fails = []
    rell = []
    divs = N.divisors()

    for c in range(N):
        found = False
        discs = fund_discs(sample, sqmod=N, cong=N, cl=c)
        if len(discs) >= confidence:
            for t in range(len(divs)):
                pts = [tuple([(sigma(N/4,k-1) if True else 1) * quadratic_L_function__exact(1-k,D)] + [Sk(k,d*d*D,N) for d in divs[:t]]) for D in discs[:confidence]]

            #for t in range(tmax):
            #    pts = [tuple([sigma(N/4,k-1) * quadratic_L_function__exact(1-k,D)] + [Sk(k,n*n*D,N) for n in range(1,t+1)]) for D in discs[:confidence]]
                if len(pts) >= confidence:
                    M = Matrix(pts)
                    R = M.right_kernel()
                    if R.dimension() == 1:
                        rel = R.basis()[0]
                        if (rel[0] != 0):
                            found = True
                            di = {n:-v for n,v in zip(divs[:t],rel[1:]) if n != 0 and v != 0}
                            ne = QuadraticLValueRelation(N,N,c,k,di)
                            #print rel,ne
                            rell += [ne]
                            break
            if not found:
                fails.append(c)
        elif discs:
            print "not enough discriminants: ",c
    if fails:
        print "failed: ",fails
    return rell

We can now find relations for different NN. For example for N=12N=12:

In [9]:
relations = find_relations(12, 2, sample=6000, tmax=50, confidence=30)
for rel in sorted(relations, key=lambda rel: rel.cost()):
    show(rel)
    rel.test()
    print "                        Cost of this relation:  ", rel.cost()
    print "Probability this relation applies to random D:  ", rel.probapplies()
Out[9]:
σ1(3)L(χD,1)=8S2(12D,12) when D0(mod12)\sigma_{1}(3)L(\chi_D,-1) = -8S_2\left(1^2 D,12\right) \text{ when }D \equiv 0 \pmod{12}
                        Cost of this relation:   1/6
Probability this relation applies to random D:   1/12
Out[9]:
σ1(3)L(χD,1)=8S2(12D,12) when D9(mod12)\sigma_{1}(3)L(\chi_D,-1) = -8S_2\left(1^2 D,12\right) \text{ when }D \equiv 9 \pmod{12}
                        Cost of this relation:   1/6
Probability this relation applies to random D:   1/6
Out[9]:
σ1(3)L(χD,1)=4S2(12D,12) when D1(mod12)\sigma_{1}(3)L(\chi_D,-1) = -4S_2\left(1^2 D,12\right) \text{ when }D \equiv 1 \pmod{12}
                        Cost of this relation:   1/3
Probability this relation applies to random D:   1/4
Out[9]:
σ1(3)L(χD,1)=4S2(12D,12) when D4(mod12)\sigma_{1}(3)L(\chi_D,-1) = -4S_2\left(1^2 D,12\right) \text{ when }D \equiv 4 \pmod{12}
                        Cost of this relation:   1/3
Probability this relation applies to random D:   1/8

N=28N=28:

The cost of computing Sk(m,N)=sZσk1(ms2N)S_k(m,N) = \sum_{s\in \mathbf Z} \sigma_{k-1}\left(\frac{m - s^2}{N}\right)

is

{sm:s2D (mod N)}={sm:s2m (mod N)}{sm}m \{ s \leq \sqrt{m} : s^2 \equiv D \text{ (mod } N )\} = \frac{\{ s \leq \sqrt{m} : s^2 \equiv m \text{ (mod } N )\}}{\{ s \leq \sqrt{m}\}} \sqrt{m}

approximated by

{sN:s2m (mod N)}{sN}m \frac{\{ s \leq N : s^2 \equiv m \text{ (mod } N)\}}{\{ s \leq N \}} \sqrt{m}

In [10]:
relations = find_relations(28, 2, sample=6000, tmax=50, confidence=30)
for rel in sorted(relations, key=lambda rel: rel.cost()):
    show(rel)
    rel.test()
    print "                        Cost of this relation:  ", rel.cost()
    print "Probability this relation applies to random D:  ", rel.probapplies()
Out[10]:
σ1(7)L(χD,1)=8S2(12D,28)+8S2(22D,28) when D0(mod28)\sigma_{1}(7)L(\chi_D,-1) = -8S_2\left(1^2 D,28\right)+-8S_2\left(2^2 D,28\right) \text{ when }D \equiv 0 \pmod{28}
                        Cost of this relation:   1/7
Probability this relation applies to random D:   1/24
Out[10]:
σ1(7)L(χD,1)=4S2(12D,28)+4S2(22D,28) when D4(mod28)\sigma_{1}(7)L(\chi_D,-1) = -4S_2\left(1^2 D,28\right)+-4S_2\left(2^2 D,28\right) \text{ when }D \equiv 4 \pmod{28}
                        Cost of this relation:   2/7
Probability this relation applies to random D:   7/144
Out[10]:
σ1(7)L(χD,1)=4S2(12D,28)+4S2(22D,28) when D8(mod28)\sigma_{1}(7)L(\chi_D,-1) = -4S_2\left(1^2 D,28\right)+-4S_2\left(2^2 D,28\right) \text{ when }D \equiv 8 \pmod{28}
                        Cost of this relation:   2/7
Probability this relation applies to random D:   7/144
Out[10]:
σ1(7)L(χD,1)=4S2(12D,28)+4S2(22D,28) when D16(mod28)\sigma_{1}(7)L(\chi_D,-1) = -4S_2\left(1^2 D,28\right)+-4S_2\left(2^2 D,28\right) \text{ when }D \equiv 16 \pmod{28}
                        Cost of this relation:   2/7
Probability this relation applies to random D:   7/144
Out[10]:
σ1(7)L(χD,1)=75S2(12D,28)+15S2(72D,28) when D21(mod28)\sigma_{1}(7)L(\chi_D,-1) = -\frac{7}{5}S_2\left(1^2 D,28\right)+-\frac{1}{5}S_2\left(7^2 D,28\right) \text{ when }D \equiv 21 \pmod{28}
                        Cost of this relation:   1/2
Probability this relation applies to random D:   1/12
Out[10]:
σ1(7)L(χD,1)=75S2(12D,28)+15S2(72D,28) when D1(mod28)\sigma_{1}(7)L(\chi_D,-1) = -\frac{7}{5}S_2\left(1^2 D,28\right)+-\frac{1}{5}S_2\left(7^2 D,28\right) \text{ when }D \equiv 1 \pmod{28}
                        Cost of this relation:   1
Probability this relation applies to random D:   7/72
Out[10]:
σ1(7)L(χD,1)=75S2(12D,28)+15S2(72D,28) when D9(mod28)\sigma_{1}(7)L(\chi_D,-1) = -\frac{7}{5}S_2\left(1^2 D,28\right)+-\frac{1}{5}S_2\left(7^2 D,28\right) \text{ when }D \equiv 9 \pmod{28}
                        Cost of this relation:   1
Probability this relation applies to random D:   7/72
Out[10]:
σ1(7)L(χD,1)=75S2(12D,28)+15S2(72D,28) when D25(mod28)\sigma_{1}(7)L(\chi_D,-1) = -\frac{7}{5}S_2\left(1^2 D,28\right)+-\frac{1}{5}S_2\left(7^2 D,28\right) \text{ when }D \equiv 25 \pmod{28}
                        Cost of this relation:   1
Probability this relation applies to random D:   7/72

N=124N=124:

In [11]:
relations = find_relations(124, 2, sample=10000, tmax=50, confidence=30)
for rel in sorted(relations, key=lambda rel: rel.cost()):
    show(rel)
    rel.test()
    print "                        Cost of this relation:  ", rel.cost()
    print "Probability this relation applies to random D:  ", rel.probapplies()
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D0(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 0 \pmod{124}
                        Cost of this relation:   1/2
Probability this relation applies to random D:   1/96
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D93(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 93 \pmod{124}
                        Cost of this relation:   1/2
Probability this relation applies to random D:   1/48
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D1(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 1 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D4(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 4 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D5(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 5 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D8(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 8 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D9(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 9 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D16(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 16 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D20(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 20 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D25(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 25 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D28(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 28 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D32(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 32 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D33(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 33 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D36(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 36 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D40(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 40 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D41(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 41 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D45(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 45 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D49(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 49 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D56(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 56 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D64(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 64 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D69(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 69 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D72(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 72 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D76(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 76 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D80(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 80 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D81(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 81 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D97(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 97 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D100(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 100 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D101(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 101 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D109(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 109 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D112(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 112 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/2880
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D113(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 113 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440
Out[11]:
σ1(31)L(χD,1)=315S2(12D,124)+15S2(312D,124) when D121(mod124)\sigma_{1}(31)L(\chi_D,-1) = -\frac{31}{5}S_2\left(1^2 D,124\right)+-\frac{1}{5}S_2\left(31^2 D,124\right) \text{ when }D \equiv 121 \pmod{124}
                        Cost of this relation:   1
Probability this relation applies to random D:   31/1440