A390343 Upper (3/2)-midsequence of binomial(n,3) and binomial(n,2); see Comments.

0, 0, 2, 6, 15, 30, 53, 84, 126, 180, 248, 330, 429, 546, 683, 840, 1020, 1224, 1454, 1710, 1995, 2310, 2657, 3036, 3450, 3900, 4388, 4914, 5481, 6090, 6743, 7440, 8184, 8976, 9818, 10710, 11655, 12654, 13709, 14820, 15990, 17220, 18512, 19866, 21285, 22770

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..45.

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

Formula

a(n) = ceiling((n^3 - n)/4).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-4) - 3*a(n-5) + 3*a(n-6) - a(n-7), with (a(0),...,a(6)) = (0,0,2,6,15,30,53).

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

Example

s = binomial(n,3) = (0, 0, 0, 1, 4, 10, 20, 35, 56, ...).
t = binomial(n,2) = (0, 0, 1, 3, 6, 10, 15, 21, 28, ...).
u(n) = floor((3/2)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 1, 6, 15, 30, 52, ...).
v(n) = ceiling((3/2)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 2, 6, 15, 30, 53, ...).

Mathematica

s[n_] := Binomial[n, 3] ; t[n_] := Binomial[n, 2]; r = 3/2;
u[n_] := Floor[r*(s[n] + t[n])];
v[n_] := Ceiling[r*(s[n] + t[n])];
Table[u[n], {n, 0, 60}]  (* A011896 *)
Table[v[n], {n, 0, 60}]  (* A390343 *)

Prog

(Python)
def A390343(n): return ((m:=n*(n**2-1))>>2)+bool(m&3) # Chai Wah Wu, Nov 09 2025

Crossrefs

Cf. A000217, A000292, A011896, A390342.

Sequence in context: A342163 A098651 A087427 * A370440 A031126 A289373

Adjacent sequences: A390340 A390341 A390342 * A390344 A390345 A390346

Keyword

nonn,easy

Author

Clark Kimberling, Nov 06 2025

Status

approved