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A390345
Upper (2/3)-midsequence of binomial(n,3) and binomial(n,2); see Comments.
1
0, 0, 1, 3, 7, 14, 24, 38, 56, 80, 110, 147, 191, 243, 304, 374, 454, 544, 646, 760, 887, 1027, 1181, 1350, 1534, 1734, 1950, 2184, 2436, 2707, 2997, 3307, 3638, 3990, 4364, 4760, 5180, 5624, 6093, 6587, 7107, 7654, 8228, 8830, 9460, 10120, 10810, 11531
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.
FORMULA
a(n) = ceiling((n^3 - n)/9).
a(n) = 4*a(n-1) - 6*a(n-2) + 3*a(n-3) + 3*a(n-4) - 6*a(n-5) + 3*a(n-6) + 3*a(n-7) - 6*a(n-8) + 4*a(n-9) - a(n-10), with (a(0),...,a(9)) = (0, 0, 1, 3, 7, 14, 24, 38, 56, 80).
G.f.: x^2*(1 - x + x^2 + x^3 - 2*x^4 + 2*x^5)/((1 - x)^4*(1 + x^3 + x^6)). - Stefano Spezia, Dec 27 2025
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((2/3)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 0, 2, 6, 13, 23, 37, ...).
v(n) = ceiling((2/3)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 1, 3, 7, 14, 24, 38, ...).
MATHEMATICA
s[n_] := Binomial[n, 3] ; t[n_] := Binomial[n, 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}] (* A011849 *)
Table[v[n], {n, 0, 60}] (* A390345 *)
PROG
(Python)
def A390345(n): return (lambda d:d[0]+bool(d[1]))(divmod(n*(n**2-1), 9)) # Chai Wah Wu, Nov 11 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 09 2025
STATUS
approved