A288104 Number of solutions to x^9 + y^9 = z^9 mod n.

1, 4, 9, 20, 25, 36, 55, 112, 189, 100, 121, 180, 109, 220, 225, 704, 289, 756, 487, 500, 495, 484, 529, 1008, 725, 436, 5103, 1100, 841, 900, 1081, 4864, 1089, 1156, 1375, 3780, 973, 1948, 981, 2800, 1681, 1980, 1513, 2420, 4725, 2116, 2209, 6336, 2989, 2900, 2601

Offset: 1

Author

Seiichi Manyama, Jun 05 2017

Keywords

• nonn
• mult

Comments

Equivalently, the number of solutions to x^9 + y^9 + z^9 == 0 (mod n). - Andrew Howroyd, Jul 17 2018

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Seiichi Manyama)

Crossrefs

Number of solutions to x^k + y^k = z^k mod n: A062775 (k=2), A063454 (k=3), A288099 (k=4), A288100 (k=5), A288101 (k=6), A288102 (k=7), A288103 (k=8), this sequence (k=9), A288105 (k=10).

Programs

Mathematica

Table[cnt=0; Do[If[Mod[x^9 + y^9 - z^9, n]==0, cnt++], {x, 0, n-1}, {y, 0, n-1}, {z, 0, n-1}]; cnt, {n, 50}] (* Vincenzo Librandi, Jul 18 2018 *)

PARI

a(n)={my(p=Mod(sum(i=0, n-1, x^lift(Mod(i,n)^9)), 1-x^n)); polcoeff(lift(p^3), 0)} \\ Andrew Howroyd, Jul 17 2018

Python

def A288104(n):
    ndict = {}
    for i in range(n):
        m = pow(i,9,n)
        if m in ndict:
            ndict[m] += 1
        else:
            ndict[m] = 1
    count = 0
    for i in ndict:
        ni = ndict[i]
        for j in ndict:
            k = (i+j) % n
            if k in ndict:
                count += ni*ndict[j]*ndict[k]
    return count # Chai Wah Wu, Jun 05 2017

Extensions

Keyword:mult added by Andrew Howroyd, Jul 17 2018