A229294 Number of solutions to x^2 + y^2 + z^2 + t^2 == n (mod 2*n) for x,y,z,t in [0, 2*n).

8, 96, 264, 384, 1160, 3168, 3080, 1536, 7560, 13920, 11528, 12672, 18824, 36960, 38280, 6144, 41480, 90720, 57608, 55680, 101640, 138336, 101384, 50688, 149000, 225888, 208008, 147840, 201608, 459360, 245768, 24576, 380424, 497760, 446600, 362880, 415880

Offset: 1

Author

José María Grau Ribas, Sep 22 2013

Keywords

• nonn

Comments

All values are divisible by a(1)=8 and the sequence a(n)/8 is multiplicative. - Andrew Howroyd, Aug 07 2018

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A229295, A229296, A229297, A240547.

Programs

Mathematica

A[n_] := Sum[If[Mod[a^2 + b^2 + c^2 + d^2, 2n] == n, 1, 0], {d, 0, 2n - 1}, {a, 0, 2n - 1}, {b, 0, 2n - 1}, {c, 0, 2n - 1}];Table[Print[aaa = A[n]]; aaa, {n, 1, 40}]

PARI

a(n)={my(m=2*n); my(p=Mod(sum(i=0, m-1, x^(i^2%m)), x^m-1)^4); polcoeff( lift(p), n)} \\ Andrew Howroyd, Aug 07 2018

PARI

a(n)={my(f=factor(n)); 8*prod(i=1, #f~, my([p,e]=f[i,]); if(p==2, 3*2^(2*e), p^(2*e-1)*(p^(e+1)+p^e-1)))} \\ Andrew Howroyd, Aug 07 2018

Python

def A229294(n):
    ndict = {}
    n2 = 2*n
    for i in range(n2):
        i3 = pow(i,2,n2)
        for j in range(i+1):
            j3 = pow(j,2,n2)
            m = (i3+j3) % n2
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (n-i) % n2
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 07 2017

Formula

a(n) = 8*A240547(n) for odd n, a(2^k) = 24*2^(2*k). - Andrew Howroyd, Aug 07 2018