A389142 Lower (1/3)-midsequence of square numbers and triangular numbers; see Comments.

0, 0, 2, 5, 8, 13, 19, 25, 33, 42, 51, 62, 74, 86, 100, 115, 130, 147, 165, 183, 203, 224, 245, 268, 292, 316, 342, 369, 396, 425, 455, 485, 517, 550, 583, 618, 654, 690, 728, 767, 806, 847, 889, 931, 975, 1020, 1065, 1112, 1160, 1208, 1258, 1309, 1360, 1413

Offset

0,3

Comments

Suppose that s = (s(n)) and t = (t(n)) are sequences of numbers and r > 0. The lower (r)-midsequence of s and t is given by u = floor(r*(s + t)); the upper r-midsequence of s and t is given by v = ceiling(r*(s + t)). If s and t are linearly recurrent and r is rational, then u and v are linearly recurrent. Conjecture: a(n+2) = A111097(n+2) for n>=0.

Links

Table of n, a(n) for n=0..53.

Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Formula

a(n) = floor((3*n^2 + n + 1)/6).

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).

G.f.: -x^2*(2 + x)/((-1 + x)^3*(1 + x + x^2)).

3*a(n) = A049347(n)+(n+1)*(3*n-2)/2 . - R. J. Mathar, Oct 05 2025

Example

s = (n^2) = A000290 = (0, 1, 4, 9, 16, 25, ...).
t = (n(n+1)/2) = A000217 = (0, 1, 3, 6, 10, 15, ...).
u(n) = floor((1/3)(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 0, 2, 5, 8, ...).
v(n) = ceiling((1/3)(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 1, 3, 5, 9, ...).

Mathematica

s[n_] := n^2; t[n_] := n (n + 1)/2; r = 1/3;
u[n_] := Floor[r*(s[n] + t[n])];
v[n_] := Ceiling[r*(s[n] + t[n])];
Table[u[n], {n, 0, 60}]  (* A389142 *)
Table[v[n], {n, 0, 60}]  (* A389143 *)

Crossrefs

Cf. A000217, A000290, A111097, A389355, A389143.

Sequence in context: A122221 A083704 A111097 * A027916 A054254 A025216

Adjacent sequences: A389139 A389140 A389141 * A389143 A389144 A389145

Keyword

nonn,easy

Author

Clark Kimberling, Sep 24 2025

Status

approved