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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
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.
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
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