A387355 Upper (1/2)-midsequence of square numbers and triangular numbers; see Comments.

0, 1, 4, 8, 13, 20, 29, 39, 50, 63, 78, 94, 111, 130, 151, 173, 196, 221, 248, 276, 305, 336, 369, 403, 438, 475, 514, 554, 595, 638, 683, 729, 776, 825, 876, 928, 981, 1036, 1093, 1151, 1210, 1271, 1334, 1398, 1463, 1530, 1599, 1669, 1740, 1813, 1888, 1964

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..51.

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

Formula

a(n) = ceiling((3*n^2 + n)/4).

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

G.f.: -x*(1 + x + x^3)/((-1 + x)^3*(1 + 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((1/2)*(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 1, 3, 7, 13, ...).
v(n) = ceiling((1/2)*(0+0, 1+1, 4+3, 9+6, 16+10, ...)) = (0, 1, 4, 8, 13, ...).

Mathematica

s[n_] := n^2; t[n_] := n (n + 1)/2; r = 1/2;
u[n_] := Floor[r*(s[n] + t[n])];
v[n_] := Ceiling[r*(s[n] + t[n])];
Table[u[n], {n, 0, 60}]  (* A330707 *)
Table[v[n], {n, 0, 60}]  (* A387355 *)

Prog

(Python)
def A387355(n): return (n*(3*n+1)-1>>2)+1 # Chai Wah Wu, Nov 08 2025

Crossrefs

Cf. A000217, A000290, A330707.

Sequence in context: A170907 A143978 A362016 * A071994 A023661 A157130

Adjacent sequences: A387352 A387353 A387354 * A387356 A387357 A387358

Keyword

nonn,easy

Author

Clark Kimberling, Sep 24 2025

Status

approved