OFFSET
1,4
COMMENTS
The number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron is given by A000332(n+3).
Create a sequence wherein the sum of three consecutive numbers is a triangular number: 0,0,0,1,2,3,5,7...; then find the partial sums of this sequence: 0,0,0,1,3,6,11,18...; then take the partial sums of this sequence: 0,0,0,1,4,10,21,39,66... and after dropping the first two zeros, you get this sequence. - J. M. Bergot, Apr 14 2016
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-5,6,-4,1).
FORMULA
a(n) = (1/72)*(-6*n -5*n^2 +2*n^3 +n^4 +4 -4*(-1)^(n mod 3)).
G.f.: x^3/((1-x)^5*(1+x+x^2)). - Bruno Berselli, Sep 11 2012
a(3*n-1) = A000217(A115067(n)); a(3*n) = A000217(A095794(n)); a(3*n+1) = A000217(A143208(n+2)) + A000217(n). - J. M. Bergot, Apr 14 2016
E.g.f.: (1/216)*(8 - 24*x + 24*x^2 + 24*x^3 + 3*x^4)*exp(x) - (1/27)*(cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2))*exp(-x/2). - Ilya Gutkovskiy, Apr 14 2016
EXAMPLE
MATHEMATICA
nnn = 100; Tev[n_] := (n - 2) (n - 1) n/6; Table[Sum[Tev[n - nn], {nn, 0, n - 1, 3}], {n, nnn}]
Table[(1/72) (-6 n - 5 n^2 + 2 n^3 + n^4 + 4 - 4 (-1)^Mod[n, 3]), {n, 50}]
CoefficientList[Series[x^2 / ((1 - x)^5*(1 + x + x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 12 2012 *)
LinearRecurrence[{4, -6, 5, -5, 6, -4, 1}, {0, 0, 1, 4, 10, 21, 39}, 50] (* Harvey P. Dale, Feb 18 2018 *)
PROG
(Magma) I:=[0, 0, 1, 4, 10, 21, 39]; [n le 7 select I[n] else 4*Self(n-1)-6*Self(n-2)+5*Self(n-3)-5*Self(n-4)+6*Self(n-5)-4*Self(n-6)+Self(n-7): n in [1..50]]; // Vincenzo Librandi, Sep 12 2012
(PARI) a(n)=(n^4+2*n^3-5*n^2-6*n+4-4*(-1)^(n%3))/72 \\ Charles R Greathouse IV, Sep 12 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
V.J. Pohjola, Sep 03 2012
STATUS
approved