A389143 Upper (1/3)-midsequence of square numbers and triangular numbers; see Comments.

0, 1, 3, 5, 9, 14, 19, 26, 34, 42, 52, 63, 74, 87, 101, 115, 131, 148, 165, 184, 204, 224, 246, 269, 292, 317, 343, 369, 397, 426, 455, 486, 518, 550, 584, 619, 654, 691, 729, 767, 807, 848, 889, 932, 976, 1020, 1066, 1113, 1160, 1209, 1259, 1309, 1361, 1414

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.

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) = ceiling((3*n^2 + n)/6).

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

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

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, A387355, A389142.

Sequence in context: A118002 A069533 A054066 * A081946 A166709 A310040

Adjacent sequences: A389140 A389141 A389142 * A389144 A389145 A389146

Keyword

nonn,easy

Author

Clark Kimberling, Sep 29 2025

Status

approved