A254073
Number of solutions to x^3 + y^3 + z^3 == 1 (mod n) for 1 <= x, y, z <= n.
Original entry on oeis.org
1, 4, 9, 16, 25, 36, 90, 64, 162, 100, 121, 144, 252, 360, 225, 256, 289, 648, 468, 400, 810, 484, 529, 576, 625, 1008, 1458, 1440, 841, 900, 1143, 1024, 1089, 1156, 2250, 2592, 1602, 1872, 2268, 1600, 1681, 3240, 2115, 1936, 4050, 2116, 2209, 2304, 4410
Offset: 1
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a[n_] := Sum[ If[ Mod[x^3 + y^3 + z^3, n] == 1, 1, 0], {x, n}, {y, n}, {z, n}]; a[1]=1; Table[a[n], {n, 2,22}]
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a(n) = {nb = 0; for (x=1, n, for (y=1, n, for (z=1, n, if ((Mod(x^3,n) + Mod(y^3,n) + Mod(z^3,n)) % n == Mod(1, n), nb ++);););); nb;} \\ Michel Marcus, Mar 11 2015
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a(n)={my(p=Mod(sum(i=0, n-1, x^(i^3%n)), 1-x^n)^3); polcoeff(lift(p), 1%n)} \\ Andrew Howroyd, Jul 18 2018
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def A254073(n):
ndict = {}
for i in range(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 = (1-i-j) % n
if k in ndict:
count += ni*ndict[j]*ndict[k]
return count # Chai Wah Wu, Jun 06 2017
A276919
Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.
Original entry on oeis.org
1, 8, 27, 64, 125, 216, 336, 512, 1296, 1000, 1331, 1728, 1794, 2688, 3375, 4096, 4913, 10368, 7410, 8000, 9072, 10648, 12167, 13824, 15625, 14352, 34992, 21504, 24389, 27000, 30225, 32768, 35937, 39304, 42000, 82944, 48396, 59280, 48438, 64000, 68921, 72576, 77529, 85184, 162000, 97336
Offset: 1
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JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 1], {n, 1, 50}]
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a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 1)))); \\ Michel Marcus, Oct 11 2016
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qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1, 4, Mod(v[i], n)^3)==1, print1(v", "); t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016
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def A276919(n):
ndict = {}
for i in range(n):
i3 = pow(i,3,n)
for j in range(i+1):
j3 = pow(j,3,n)
m = (i3+j3) % n
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 = (1-i) % n
if j in ndict:
count += ndict[i]*ndict[j]
return count # Chai Wah Wu, Jun 06 2017
A276920
Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.
Original entry on oeis.org
1, 8, 27, 72, 125, 216, 595, 704, 1539, 1000, 1331, 1944, 3133, 4760, 3375, 5632, 4913, 12312, 8911, 9000, 16065, 10648, 12167, 19008, 16125, 25064, 45927, 42840, 24389, 27000, 35371, 47104, 35937, 39304, 74375, 110808, 58645, 71288, 84591, 88000
Offset: 1
For n = 3, we see that all nondecreasing solutions {x, y, z, t} are in {{1, 1, 1, 3}, {1, 1, 2, 2}, {1, 2, 3, 3}, {2, 2, 2, 3}, {3, 3, 3, 3}}. The numbers in the sets can be ordered in 4, 6, 12, 4 and 1 ways respectively. Therefore, a(3) = 4 + 6 + 12 + 4 + 1 = 27. - _David A. Corneth_, Oct 11 2016
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CF:= table([[false, false, true] = 12, [true, false, false] = 12, [true, false, true] = 6, [false, false, false] = 24, [true, true, true] = 1, [false, true, true] = 4, [false, true, false] = 12, [true, true, false] = 4]):
f1:= proc(n)
option remember;
local count, t, x,y,z,signature;
if isprime(n) and n mod 3 = 2 then return n^3 fi;
count:= 0;
for t from 1 to n do
for x from 1 to t do
for y from 1 to x do
for z from 1 to y do
if t^3 + x^3 + y^3 + z^3 mod n = 0 then
signature:= map(evalb,[z=y,y=x,x=t]);
count:= count + CF[signature];
fi
od od od od;
count
end proc:
f:= proc(n) local t;
mul(f1(t[1]^t[2]),t=ifactors(n)[2])
end proc:
map(f, [$1..40]); # Robert Israel, Oct 13 2016
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JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 0], {n, 1, 50}]
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a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 0)))); \\ Michel Marcus, Oct 11 2016
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qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i, t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1,4,Mod(v[i],n)^3)==0, t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016
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def A276920(n):
ndict = {}
for i in range(n):
i3 = pow(i,3,n)
for j in range(i+1):
j3 = pow(j,3,n)
m = (i3+j3) % n
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 = (-i) % n
if j in ndict:
count += ndict[i]*ndict[j]
return count # Chai Wah Wu, Jun 06 2017
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