A350236 a(n) is the sum of the entries in an n X n X n 3D matrix whose elements start at 1 in the corner cells and increase by 1 with each step towards the center.

1, 8, 54, 160, 425, 864, 1666, 2816, 4617, 7000, 10406, 14688, 20449, 27440, 36450, 47104, 60401, 75816, 94582, 116000, 141561, 170368, 204194, 241920, 285625, 333944, 389286, 450016, 518897, 594000, 678466, 770048, 872289, 982600, 1104950, 1236384, 1381321

Offset

1,2

Comments

The 2D version of this problem is discussed in A317614.

Links

Table of n, a(n) for n=1..37.

Saeed Barari, In depth (pdf)

Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Formula

a(n) = (3/4)*n^2 * (n^2 - 2/3*n + (n mod 2)).

From Stefano Spezia, May 19 2022: (Start)

O.g.f.: x*(1 + 6*x + 36*x^2 + 42*x^3 + 45*x^4 + 12*x^5 + 2*x^6)/((1 - x)^5*(1 + x)^3).

E.g.f.: x*((4 + 15*x + 16*x^2 + 3*x^3)*cosh(x) + (1 + 18*x + 16*x^2 + 3*x^3)*sinh(x))/4.

a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5)- 2*a(n-6) - 2*a(n-7) + a(n-8) for n > 8. (End)

Example

For n=3: we have the following 3D matrix: (sliced for each Z surface)
(z=1): 1 2 1
       2 3 2
       1 2 1
(z=2): 2 3 2
       3 4 3
       2 3 2
(z=3): 1 2 1
       2 3 2
       1 2 1
The sum of all elements is: (3/4)*n^2 * (n^2 - 2/3*n + (n mod 2)) = 54.

Maple

a:=n->(3/4)*n^2 * (n^2 - (2/3)*n + modp(n, 2)): seq(a(n), n=1..50);

Mathematica

LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {1, 8, 54, 160, 425, 864, 1666, 2816}, 35] (* Stefano Spezia, May 19 2022 *)

Prog

(Python)
for n in range(1, nmax):
  sum = round(3/4*n**2 * (n**2 - 2/3*n + n % 2))
  print(sum, end=', ')

Crossrefs

Cf. A317614.

Sequence in context: A180095 A234955 A189393 * A254951 A085537 A085540

Adjacent sequences: A350233 A350234 A350235 * A350237 A350238 A350239

Keyword

nonn,easy

Author

Saeed Barari, Dec 21 2021

Extensions

Python program and a(23), a(34) corrected by Georg Fischer, Sep 30 2022

Status

approved