A389317 Upper (2/3)-midsequence of square numbers and triangular numbers; see Comments.

0, 2, 5, 10, 18, 27, 38, 52, 67, 84, 104, 125, 148, 174, 201, 230, 262, 295, 330, 368, 407, 448, 492, 537, 584, 634, 685, 738, 794, 851, 910, 972, 1035, 1100, 1168, 1237, 1308, 1382, 1457, 1534, 1614, 1695, 1778, 1864, 1951, 2040, 2132, 2225, 2320, 2418

Offset

0,2

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..49.

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

Formula

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

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

G.f.: x*(-2 - x - 2*x^2 - 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((2/3)(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 1, 4, 10, 17, ...).
v(n) = ceiling((2/3)(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 2, 5, 10, 18, ...).

Mathematica

s[n_] := n^2; t[n_] := n (n + 1)/2; r = 2/3;
u[n_] := Floor[r*(s[n] + t[n])];
v[n_] := Ceiling[r*(s[n] + t[n])];
Table[u[n], {n, 0, 60}]  (* A389316 *)
Table[v[n], {n, 0, 60}]  (* A389317 *)

Prog

(Python)
def A389317(n): return n**2+1+(n-1)//3 # Chai Wah Wu, Oct 08 2025

Crossrefs

Cf. A000217, A000290, A387355, A389142, A389316.

Sequence in context: A077631 A381604 A350878 * A390342 A354246 A381100

Adjacent sequences: A389314 A389315 A389316 * A389318 A389319 A389320

Keyword

nonn,easy

Author

Clark Kimberling, Oct 03 2025

Status

approved