A000189 -id:A000189 - cp's OEIS frontend

A337111 Number of length four 1..n vectors that contain their geometric mean.

1, 2, 3, 16, 17, 18, 19, 56, 105, 106, 107, 144, 145, 146, 147, 208, 209, 306, 307, 320, 321, 322, 323, 432, 529, 530, 723, 736, 737, 738, 739, 968, 969, 970, 971, 1176, 1177, 1178, 1179, 1288, 1289, 1290, 1291, 1304, 1401, 1402, 1403, 1608, 1777, 2018, 2019, 2032

Offset: 1

Hywel Normington, Aug 16 2020

From David A. Corneth, Aug 27 2020: (Start)
Let (a, b, g, n) be a tuple where g is the geometric mean of a*b*g*n and a, b, g <= n. Then there are two cases: g < n and g = n.
If g = n then (a, b, g, n) = (n, n, n, n) adding 1 to the number of permutations.
If g < n then a*b = g^4 / (g*n) = g^3 / n. Furthermore let s be the squarefree part of n (Cf. A007947). Then s | g and so candidates for g are (s*i) where 1 <= i < floor(n/s), depending on whether g^3 / n = (s*i)^3 / n is an integer.
It follows that for suitable values of i, (a, b) are a pair of divisors of (s*i)^3 / n where a*b = (s*i)^3 / n and max(a, b) <= n. (End)
Bounds: n + 12*(floor(n/4) + 2*floor(n/8) + 4*floor(n/9) + 2*floor(n/12) + 2*floor(n/16) + 4*floor(n/18)) <= a(n) <= n^4 - 14*(n-1). - Hywel Normington, Jan 25 2021

			For n = 2 the a(2) = 2 solutions are: (1,1,1,1) and (2,2,2,2).
For n = 4 the a(4) = 16 solutions are:
  (1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4),
  and the 12 permutations of (1, 2, 2, 4).
For n = 40, the a(40)-a(39) = 109 new solutions are:
  (40,40,40,40),
  the 24 permutations of (1, 10, 25, 40),
  the 12 permutations of (5, 5, 10, 40),
  the 12 permutations of (5, 20, 40, 40),
  the 24 permutations of (8, 20, 25, 40),
  the 12 permutations of (10, 20, 20, 40),
  and the 24 permutations of (25, 27, 30, 40).

David A. Corneth, Table of n, a(n) for n = 1..10000
Hywel Normington, Python code, 2020.
Hywel Normington, Julia code, 2023.

Cf. A007947, A000189, A248435, A337110.

first(n) = { my(res = vector(n)); s = 0; for(i = 1, n, s += b(i);  res[i] = s; ); res }
b(n) = {my(resa = 1); my(s = factorback(factor(n)[, 1])); for(i = 1, n \ s - 1, s4 = (s*i)^3; if(s4 % n == 0, c = tuples((s*i)^3/n, s*i, n); for(i = 1, #c, resa+=qperms(c[i]) ) ) ); resa }
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)}
tuples(n, s, u) = {my(res = List(), u4n, d, i); d = divisors(n); i = (#d + 1) \ 2; while(i > 0 && d[#d - i + 1] <= u, c = vecsort([d[i], d[#d - i + 1], s, u]); listput(res, c); i--); res} \\ David A. Corneth, Aug 28 2020

Empirical: if A000189(n) = 1 then a(n) = a(n-1) + 1.
From David A. Corneth, Aug 25 2020: (Start)
The above holds. That is: if x^3 == 0 (mod n) has only one solution then a(n) = a(n-1) + 1. Proof:
Let (a, b, c, n) be such a tuple. Let without loss of generality c be the geometric mean of the tuple. Then a*b*c*n = c^4 and as c is not 0 we have c^3 = a*b*n. So then c^3 == 0 (mod n). If c^3 == 0 (mod n) has only 1 solution then c = n. This gives the tuple (n, n, n, n) which has 1 permutation. So giving a(n) = a(n-1) + 1. (End)
a(n) - a(n-1) == 1 (mod 12). - Hywel Normington, Sep 28 2020

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.

Original entry on oeis.org

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

Neven Juric (neven.juric(AT)apis-it.hr), Apr 11 2008

nonn

Record values of A137401: 0, 2, 7, 12, 20, 63, 116, 131, 182, 339, 602, 839, 1808, 3311, 4328, 7964, 8864, 9231, 19583, 21986, 41363, 52676, 81467, 87596, 92087, 112616, 236951, 247940, 378071, 386423, 521135, ... - Robert G. Wilson v

			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.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..425 from Robert G. Wilson v)

Cf. A063454.

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 *)

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

a(n) = A063454(n)-3*A087786(n)+3*A000189(n)-1. - Vladeta Jovovic, Apr 11 2008

More terms from Stefan Steinerberger and Robert G. Wilson v, Apr 12 2008

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

José María Grau Ribas, Sep 22 2016

It appears that a(n) = n^3 for n in A088232. See also A066498. - Michel Marcus, Oct 11 2016

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

Cf. A000189, A047726, A060839, A063454, A087412, A087786, A254073, A276920.

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}]

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

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

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

José María Grau Ribas, Sep 22 2016

a(n) = n^3 if n is in A074243. - Robert Israel, Oct 13 2016

			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

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..242 from Robert Israel)

Cf. A000189, A047726, A060839, A063454, A074243, A087412, A087786, A254073, A276919.

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

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}]

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

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

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

A372882 a(n) = Sum_{k=1..n} gcd(k^3,n).

Original entry on oeis.org

1, 3, 5, 10, 9, 15, 13, 36, 33, 27, 21, 50, 25, 39, 45, 104, 33, 99, 37, 90, 65, 63, 45, 180, 145, 75, 261, 130, 57, 135, 61, 336, 105, 99, 117, 330, 73, 111, 125, 324, 81, 195, 85, 210, 297, 135, 93, 520, 385, 435, 165, 250, 105, 783, 189, 468, 185, 171, 117, 450

Offset: 1

Seiichi Manyama, May 15 2024

Seiichi Manyama, Table of n, a(n) for n = 1..10000
László Tóth, Menon's identity and arithmetical sums representing functions of several variables, Rend. Sem. Mat. Univ. Politec. Torino, Vol. 69, No. 1 (2011), pp. 97-110.

Cf. A018804, A078430.

Cf. A000010, A000189.

a[n_] := Sum[GCD[k^3, n], {k, 1, n}]; Array[a, 100] (* Amiram Eldar, May 24 2024 *)

PARI
```
a(n) = sum(k=1, n, gcd(k^3, n));
```

From Amiram Eldar, May 24 2024: (Start)
a(n) = n * Sum_{d|n} A000010(d)*A000189(d)/d (Tóth, 2011).
Multiplicative with a(p^e) = p^e * (1 + ((p-1)/p) * Sum_{i=1..e} p^(floor(2*i/3))). (End)

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A337111 Number of length four 1..n vectors that contain their geometric mean.

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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.

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A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

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A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.

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A372882 a(n) = Sum_{k=1..n} gcd(k^3,n).

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