OFFSET
0,2
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).
FORMULA
a(n) + A210379(n) = (n+1)^4.
From Chai Wah Wu, Nov 27 2016: (Start)
a(n) = (n + 1)^2*((2*n + 1 -(-1)^n)^2 + (2*n + 3 + (-1)^n)^2)/16.
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 > 7.
G.f.: (-x^6 - 6*x^5 - 27*x^4 - 28*x^3 - 27*x^2 - 6*x - 1)/((x - 1)^5*(x + 1)^3). (End)
From Amiram Eldar, Mar 15 2024: (Start)
a(n) = (n+1)^2*floor(((n+1)^2+1)/2).
Sum_{n>=0} 1/a(n) = Pi^4/720 + (Pi-2*tanh(Pi/2))*Pi/4. (End)
E.g.f.: ((2 + 15*x + 26*x^2 + 10*x^3 + x^4)*cosh(x) + (1 + 18*x + 25*x^2 + 10*x^3 + x^4)*sinh(x))/2. - Stefano Spezia, Jul 15 2024
EXAMPLE
Writing the matrices as 4-letter words, the 8 for n=1 are as follows:
0000, 0100, 0010, 0110, 1001, 1101, 1011, 1111
MATHEMATICA
a = 0; b = n; z1 = 35;
t[n_] := t[n] = Flatten[Table[w + z, {w, a, b}, {x, a, b}, {y, a, b}, {z, a, b}]]
c[n_, k_] := c[n, k] = Count[t[n], k]
u[n_] := Sum[c[n, 2 k], {k, 0, 2*n}]
v[n_] := Sum[c[n, 2 k - 1], {k, 1, 2*n - 1}]
Table[u[n], {n, 0, z1}] (* A210378 *)
Table[v[n], {n, 0, z1}] (* A210379 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 20 2012
STATUS
approved