A127873 a(n) = (n^3)/2 + (3*n^2)/2 + 3*n + 3.
8, 19, 39, 71, 118, 183, 269, 379, 516, 683, 883, 1119, 1394, 1711, 2073, 2483, 2944, 3459, 4031, 4663, 5358, 6119, 6949, 7851, 8828, 9883, 11019, 12239, 13546, 14943, 16433, 18019, 19704, 21491, 23383, 25383, 27494, 29719, 32061, 34523, 37108
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..5000
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A127874.
Programs
-
GAP
A127873 := List([1..10^3],n->(n^3)/2+(3*n^2)/2+3*n+3); # Muniru A Asiru, Jan 13 2018
-
Magma
[(n^3)/2 + (3*n^2)/2 + 3*n + 3: n in [0..50]]; // Vincenzo Librandi, Jan 14 2018
-
Maple
A127873 := [seq((n^3)/2+(3*n^2)/2+3*n+3,n=1..10^3)]; # Muniru A Asiru, Jan 13 2018
-
Mathematica
Table[3 + 3 x + (3 x^2)/2 + x^3/2, {x, 41}] Rest@ CoefficientList[ Series[-x (3 x^3 -11 x^2 +13 x - 8)/(x -1)^4, {x, 0, 41}], x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {8, 19, 39, 71}, 41] (* Robert G. Wilson v, Jan 06 2018 *)
-
PARI
a(n)=n^3/2+3*n*(n+2)/2+3 \\ Charles R Greathouse IV, May 15 2013
-
Python
def A127873(n): return (n*(n*(n+3)+6)>>1)+3 # Chai Wah Wu, Jul 12 2025
Formula
G.f.: x*(8-13*x+11*x^2-3*x^3)/(1-x)^4. - Colin Barker, Apr 17 2012
E.g.f.: (1+x+x^2/2)*(1+x)*exp(x) - 3x - 1. - Enrique Navarrete, Apr 16 2025
Comments