A211786 n^3 + floor(n^3/2).
1, 12, 40, 96, 187, 324, 514, 768, 1093, 1500, 1996, 2592, 3295, 4116, 5062, 6144, 7369, 8748, 10288, 12000, 13891, 15972, 18250, 20736, 23437, 26364, 29524, 32928, 36583, 40500, 44686, 49152, 53905, 58956, 64312, 69984, 75979, 82308
Offset: 1
Links
- Bruno Berselli, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3, -2, -2, 3, -1).
Programs
-
Magma
[n^3+Floor(n^3/2): n in [1..38]]; // Bruno Berselli, May 06 2012
-
Mathematica
f[n_, m_] := Sum[Floor[n^3/k], {k, 1, m}] t = Table[f[n, 2], {n, 1, 90}] FindLinearRecurrence[t] LinearRecurrence[{3, -2, -2, 3, -1},{1, 12, 40, 96, 187},38] (* Ray Chandler, Aug 02 2015 *)
Formula
a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5).
G.f.: x*(1+9*x+6*x^2+2*x^3)/((1+x)*(1-x)^4). [Bruno Berselli, May 06 2012]
a(n) = floor(3*n^3/2) = (6*n^3+(-1)^n-1)/4. [Bruno Berselli, May 06 2012]
Comments