A396835 Upper (2/3,1/3) midsequence of (floor[n/2]^2) and (ceiling[n/2]^2); see Comments.

0, 1, 1, 2, 4, 6, 9, 12, 16, 19, 25, 29, 36, 41, 49, 54, 64, 70, 81, 88, 100, 107, 121, 129, 144, 153, 169, 178, 196, 206, 225, 236, 256, 267, 289, 301, 324, 337, 361, 374, 400, 414, 441, 456, 484, 499, 529, 545, 576, 593, 625, 642, 676, 694, 729, 748, 784, 803, 841, 861, 900

Offset

0,4

Comments

Suppose that s = (s(n)) and t = (t(n)) are sequences of numbers and h > 0 and k > 0. The lower (h, k)-midsequence of s and t is floor(h*s + k*t); the upper (h, k)-midsequence of s and t is ceiling(h*s + k*t).

Links

Table of n, a(n) for n=0..60.

Index entries for linear recurrences with constant coefficients, signature (1,1,-1,0,0,1,-1,-1,1).

Formula

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

G.f.: x*(-1 - 2*x^3 - x^4 - x^5 - x^7)/((-1 + x)^3*(1 + x)^2*(1 + x^2 + x^4)).

a(n) = floor((6*n^2 - 2*n + 15 + (-1)^n*(2*n - 7))/24). - Hoang Xuan Thanh, Jun 20 2026

Example

s(n) = A008794(n+1):  (0, 1, 1, 4, 4, 9, 9, 16, 16, ...).
t(n) = A008794(n+2):  (1, 1, 4, 4, 9, 9, 16, 16, 25, ...).
(u(n)) = (0, 0, 1, 2, 4, 5, 9, 11, 16, 19, 25, 28, 36, 40, 49, ...).
(v(n)) = (0, 1, 1, 2, 4, 6, 9, 12, 16, 19, 25, 29, 36, 41, 49, ...).

Mathematica

z = 60; f[n_] := Floor[n/2]^2; g[n_] := Ceiling[n/2]^2;
r = 2/3; s = 1/3;
v[n_] := Ceiling[r*f[n] + s*g[n]]
Table[v[n], {n, 0, z}]
(* Alternative: *)
LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 1, 1, 2, 4, 6, 9, 12, 16}, 60]

Crossrefs

Cf. A389123, A396811, A396812, A396813, A396834.

Sequence in context: A053096 A155752 A145801 * A033291 A105434 A145196

Adjacent sequences: A396831 A396833 A396834 * A396836 A396837 A396838

Keyword

nonn,easy

Author

Clark Kimberling, Jun 19 2026

Status

approved