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A390561
Lower (1/5,1/2) midsequence of (2^n) and (5*n); see Comments.
2
0, 2, 5, 9, 13, 18, 27, 43, 71, 124, 229, 437, 849, 1670, 3311, 6591, 13147, 26256, 52473, 104905, 209765, 419482, 838915, 1677779, 3355503, 6710948, 13421837, 26843613, 53687161, 107374254, 214748439, 429496807, 858993539, 1717987000, 3435973921, 6871947761
OFFSET
0,2
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).
FORMULA
a(n) = 4*a(n-1) - 6*a(n-2) + 6*a(n-3) - 5*a(n-4) + 2*a(n-5), with (a(0),...,a(4)) = (0, 2, 5, 9, 13).
G.f.: x*(-2 + 3*x - x^2 + 5*x^3)/((-1 + x)^2*(-1 + 2*x - x^2 + 2*x^3)).
EXAMPLE
s = A000079 = (1, 2, 4, 8, 16, 32, 64, ...).
t = A000351 = (0, 5, 10, 15, 20, 25, 30, ...).
u(n) = (0, 2, 5, 9, 13, 18, 27, 43, ...).
v(n) = (1, 3, 6, 10, 14, 19, 28, 44, ...).
MATHEMATICA
s[n_] := 2^n ; t[n_] := 5*n; h = 1/5; k = 1/2;
u[n_] := Floor[h*s[n] + k*t[n]];
v[n_] := Ceiling[h*s[n] + k*t[n]];
Table[u[n], {n, 0, 60}] (* A390561 *)
Table[v[n], {n, 0, 60}] (* A390562 *)
(* Also *)
LinearRecurrence[{4, -6, 6, -5, 2}, {0, 2, 5, 9, 13}, 30]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 11 2025
STATUS
approved