A216172 Number of all possible tetrahedra of any size, having reverse orientation to the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts.

0, 0, 1, 4, 10, 21, 39, 66, 105, 159, 231, 325, 445, 595, 780, 1005, 1275, 1596, 1974, 2415, 2926, 3514, 4186, 4950, 5814, 6786, 7875, 9090, 10440, 11935, 13585, 15400, 17391, 19569, 21945, 24531, 27339, 30381, 33670, 37219, 41041, 45150, 49560, 54285, 59340

Offset: 1

Author

V.J. Pohjola, Sep 03 2012

Keywords

• nonn
• easy

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

Examples

			For n=9 the numbers of the reversely oriented tetrahedra, starting from the smallest size, are A000292(7)=84, A000292(4)=20, and A000292(1)=1, the sum being a(9)=105.
		

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).

Crossrefs

Cf. A000292, A000332, A216173, A216175.

Programs

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

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 *)

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

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