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A390344
Upper (1/3)-midsequence of binomial(n,3) and binomial(n,2); see Comments.
2
0, 0, 1, 2, 4, 7, 12, 19, 28, 40, 55, 74, 96, 122, 152, 187, 227, 272, 323, 380, 444, 514, 591, 675, 767, 867, 975, 1092, 1218, 1354, 1499, 1654, 1819, 1995, 2182, 2380, 2590, 2812, 3047, 3294, 3554, 3827, 4114, 4415, 4730, 5060, 5405, 5766, 6142, 6534, 6942
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)/18).
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, 2, 4, 7, 12, 19, 28, 40).
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((1/3)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 0, 1, 3, 6, 11, 18, ...).
v(n) = ceiling((1/3)*(0+0, 0+0, 0+1, 1+3, 4+6, 10+10, ...)) = (0, 0, 1, 2, 4, 7, 12, 19, ...).
MATHEMATICA
s[n_] := Binomial[n, 3] ; t[n_] := Binomial[n, 2]; r = 1/3;
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}] (* A390344 *)
PROG
(Python)
def A390344(n): return (lambda d:d[0]+bool(d[1]))(divmod(n*(n**2-1), 18)) # Chai Wah Wu, Nov 09 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 07 2025
STATUS
approved