A137401 a(n) is the number of ordered solutions (x,y,z) to x^3 + y^3 == z^3 mod n with 1 <= x,y,z <= n-1.
0, 0, 2, 7, 12, 20, 0, 63, 116, 72, 90, 131, 0, 108, 182, 339, 240, 602, 324, 415, 326, 420, 462, 839, 604, 216, 1808, 763, 756, 812, 810, 1735, 992, 1056, 1092, 3311, 648, 1620, 650, 2511, 1560, 1640, 1134, 2227, 4328, 1980, 2070, 3683, 2484, 2644, 2450, 1519
Offset: 1
Keywords
Examples
a(4)=7 because (1, 2, 1), (1, 3, 2), (2, 1, 1), (2, 2, 2), (2, 3, 3), (3, 1, 2), (3, 2, 3) are solutions for n=4.
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..425 from Robert G. Wilson v)
Crossrefs
Cf. A063454.
Programs
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Mathematica
f[n_] := Block[ {c = 0}, Do[ If[ Mod[x^3 + y^3, n] == Mod[z^3, n], c++ ], {x, n - 1}, {y, n - 1}, {z, n - 1}]; c]; Table[Length[Select[Tuples[Range[n - 1], 3], Mod[ #[[1]]^3 + #[[2]]^3 - #[[3]]^3, n] == 0 &]], {n, 2, 50}] (* Stefan Steinerberger, Apr 12 2008 *)
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Python
def A137401(n): ndict = {} for i in range(1,n): m = pow(i,3,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 06 2017
Formula
Extensions
More terms from Stefan Steinerberger and Robert G. Wilson v, Apr 12 2008
Comments