A389320 Upper (3/4)-midsequence of square numbers and negative triangular numbers; see Comments.

0, 0, 1, 3, 5, 8, 12, 16, 21, 27, 34, 42, 50, 59, 69, 79, 90, 102, 115, 129, 143, 158, 174, 190, 207, 225, 244, 264, 284, 305, 327, 349, 372, 396, 421, 447, 473, 500, 528, 556, 585, 615, 646, 678, 710, 743, 777, 811, 846, 882, 919, 957, 995, 1034, 1074, 1114

Offset

0,4

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..55.

Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-4,4,-4,4,-3,1).

Formula

a(n) = ceiling((3/8)*(n^2 - n)).

a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 4*a(n-4) + 4*a(n-5) - 4*a(n-6) + 4*a(n-7) - 3*a(n-8) + a(n-9), with (a(0),...,a(8)) = (0, 0, 1, 3, 5, 8, 12, 16, 21).

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((3/4)*(0-0, 1-1, 4-3, 9-6, 16-10, ...)) = (0, 0, 0, 2, 4, 7, 11, 15, ...).
v(n) = ceiling((3/4)*(0-0, 1-1, 4-3, 9-6, 16-10, ...)) = (0, 0, 1, 3, 5, 8, 12, 16, ...).

Mathematica

s[n_] := n^2; t[n_] := - n (n + 1)/2; r = 3/4;
u[n_] := Floor[r*(s[n] + t[n])];
v[n_] := Ceiling[r*(s[n] + t[n])];
Table[u[n], {n, 0, 60}]  (* A389319 *)
Table[v[n], {n, 0, 60}]  (* A389320 *)

Prog

(Python)
def A389320(n): return (3*n*(n-1)-1>>3)+1 # Chai Wah Wu, Nov 08 2025

Crossrefs

Cf. A000217, A000290, A387355, A389318, A389319.

Sequence in context: A194180 A000212 A183139 * A094913 A265060 A320849

Adjacent sequences: A389317 A389318 A389319 * A389321 A389322 A389323

Keyword

nonn,easy

Author

Clark Kimberling, Nov 02 2025

Status

approved