A260965 -id:A260965 - cp's OEIS frontend

A261029 Number of ways to write n in the form F(x,y,z) = x^3 + y^3 + z^3 - 3xyz, where 0 <= x <= y <= z and z >= x+1.

0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 1, 0, 2, 1, 1, 1, 2, 0, 1, 1, 0, 1, 1, 2, 3, 1, 0, 1, 2, 0, 1, 2, 1, 1, 1, 0, 2, 1, 0, 1, 2, 1, 1, 1, 0, 2, 1, 0, 2, 1, 3, 1, 3, 0, 1, 1, 0, 1, 1, 1, 3, 2, 0, 1, 2, 0, 2, 1, 2, 1, 1, 0, 2, 2, 0, 1, 2, 3, 1, 1, 0, 1, 1

Offset: 0

Vladimir Shevelev, Aug 22 2015

nonn

The following is a short proof of the corresponding 1915 result of R. D. Carmichael for a weaker restriction.
If n is in A074232, then a(n) >= 1, in view of the following identities: if n == 1 (mod 3), then n = F((n-1)/3, (n-1)/3, (n+2)/3); if n == 2 (mod 3), then n = F((n-2)/3, (n+1)/3, (n+1)/3); if n == 0 (mod 9), then n = F(n/9-1, n/9, n/9+1). QED
Further, if n > 1 is the cube of a positive number or the sum of two positive cubes, except for 2 and 9, then a(n) >= 2.
The sequence is unbounded.
Proof. We use the homogeneity of F(x,y,z) of degree 3. By induction, show that a(8^k) >= k+1. It is evident for k=0. Suppose that it is true for some value of k. Take k+1 triples (x_i,y_i,z_i) such that 8^k = F(x_i, y_i, z_i), i=1,...,k+1. Then for k+1 triples of even numbers (2*x_i, 2*y_i, 2*z_i) we have 8^(k+1) = F(2*x_i, 2*y_i, 2*z_i). But there is always a triple of not all even numbers (x=(n-1)/3, y=(n-1)/3, z=(n+2)/3) or (x=(n-2)/3, y=(n+1)/3, z=(n+1)/3), where n = 8^(k+1), for which 8^(k+1) = F(x,y,z). So a(8^(k+1)) >= k+2. QED
Theorem. For every n there exists k such that a(k)=n. For a proof, see [Shevelev] link.
Smallest such k are presented in sequence A260935.

Peter J. C. Moses and Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms for n = 0..999 from Peter J. C. Moses)
R. D. Carmichael, On the representation of numbers in the form x^3+y^3+z^3-3xyz, Bull. Amer. Math. Soc. 22 (1915), 111-117.
Peter J. C. Moses, List of triples up to a(2000)
Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A072670, A074232, A260803, A260804, A260965.

r[n_] := Reduce[0 <= x <= y <= z && z >= x+1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Array[a, 100, 0] (* Jean-François Alcover, Nov 06 2018 *)

For positive n, a(n)=0, if and only if n == 3 or 6 (mod 9); if p is prime, other than 3, then a(p) = a(2*p) = 1.

For n >= 1, a(8^(n-1)) = n.

More terms from Peter J. C. Moses, Aug 22 2015

A260804 Number of ways to write n as n = x * y * z * t + x + y + z + t where 1 <= x <= y <= z <= t <= n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 2, 1, 2, 0, 2, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 3, 0, 3, 2, 2, 1, 5, 0, 1, 2, 4, 1, 4, 0, 3, 3, 2, 1, 4, 2, 3, 2, 2, 0, 5, 1, 4, 1, 2, 3, 6, 1, 2, 2, 5, 1, 4, 0, 4, 3, 3, 1, 6, 1, 2, 4, 4, 2, 4, 1, 4, 2, 2, 1, 8, 2, 4, 2, 4, 2, 5, 1, 4, 2, 2, 3, 8, 1, 3, 4, 4, 0, 4, 1, 6, 4, 3, 0

Offset: 0

David A. Corneth, Jul 31 2015

nonn

a(n) = A071689(n) - A001399(n) = A071689(n) - round((n+3)^2/12).
From Vladimir Shevelev, Aug 03 2015: (Start)
Is the set of n for which a(n)=0 finite?
Note that this set contains only numbers n of the form prime + 1. Indeed, if n-1>=4 is a composite number, then n = p*q + 1, p>=2, q>=2. If p <= q, then, for x=1, y=1, z = p-1, t = q-1, we have
x*y*z*t + x + y + z + t = 1*1*(p-1)*(q-1) + 1 + 1 + (p-1) + (q-1) = p*q + 1 = n; so a(n) >= 1. If p > q, then we set x=1, y=1, z = q-1, t = p-1, and again a(n) >= 1.
Note also that limsup_{n->infinity} (a(n)) = infinity. Indeed, this limit is realized, say, on n = primorials +1 (A002110), since, when m goes to infinity, the number of representations of n - 1 = A002110(m) of the form p*q tends to infinity. On primorials +1 > 2 we have a subsequence: 0,1,3,8,27,... .
A generalization. For k>=2, let b_k(n) be the number of ways to write n as n = x_1 * x_2 *...* x_k + x_1 + x_2 + ... + x_k, where 1 <= x_1 <= x_2 <= ... <= x_k <= n.
Then, for n >= k-1, b_k(n) = 0 yields that n - k + 3 is prime with similar other comments. In particular, only b_2(n) = 0 if and only if n+1 is 1 or prime (cf. A072670). (End)

David A. Corneth, Table of n, a(n) for n = 0..9999
Vladimir Shevelev, Representation of positive integers by the form x1...xk+x1+...+xk, arXiv:1508.03970 [math.NT], 2015.

Cf. A001399, A071689, A260803, A260965.

xmax = 9; ymax = 21; zmax = 98; (* When extending data, terms where maxima for x, y or z are reached have to be checked one by one. *)
r[n_] := r[n] = Module[{r1, r2, r3, rn}, r1 = Reap[Do[rn = Reduce[n == x y z t + x + y + z + t && 1 <= x <= y <= z <= t <= n, t, Integers]; If[rn =!= False, Sow[{x, y, z, t} /. {ToRules[rn]}]], {x, 1, xmax}, {y, 1, ymax}, {z, 1, zmax}]]; If[r1 == {Null, {}} , {}, r2 = r1[[2, 1]]; r3 = Flatten[r2, 1]; If[Max[r3[[All, 1]]] == xmax, Print[ "xmax reached at n = ", n]]; If[Max[r3[[All, 2]]] == ymax, Print["ymax reached at n = ", n]]; If[Max[r3[[All, 3]]] == zmax, Print["zmax reached at n = ", n]]; r3]];
a[n_] := Length[r[n]];
Table[Print["a(", n, ") = ", a[n], " ", r[n]]; a[n], {n, 0, 109}] (* Jean-François Alcover, Nov 19 2018 *)

If A260803(n) > 0, then a(n+1) > 0. So if a(n+1) = 0, then A260803(n) = 0. Converse statement is not true. For example, a(24) > 0, while A260803(23) = 0. - Vladimir Shevelev, Aug 14 2015

cp's OEIS Frontend

A261029 Number of ways to write n in the form F(x,y,z) = x^3 + y^3 + z^3 - 3xyz, where 0 <= x <= y <= z and z >= x+1.

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