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

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

Author

David A. Corneth, Jul 31 2015

Keywords

• nonn

Comments

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)

Links

• 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.

Crossrefs

Cf. A001399, A071689, A260803, A260965.

Programs

Mathematica

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 *)

Formula

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