OFFSET
0,4
COMMENTS
a(n) mod 8 = 0.
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).
FORMULA
a(n) = (1/4)*(n-3)*(n-2)*(2*n^2 - 4*n + 1 - (-1)^n) for n>=0.
From Colin Barker, Dec 25 2016: (Start)
a(n) = (n^4 - 3*n^3 + 2*n^2)/2 for n even.
a(n) = (n^4 - 3*n^3 + n^2 + 3*n - 2)/2 for n odd.
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.: 8*x^3*(1 + 3*x)*(1 + x + x^2) / ((1 - x )^5*(1 + x)^3).
(End)
These formulas are true. a(n) = ((-1)^n + 2*n^2 - 1)*(n-1)*(n-2)/4 = (n^2 - p(n))*C(n-1,2), where p(n) is the parity of n, i.e., p(n) = 0 if n is even and p(n) = 1 if n is odd. - Chai Wah Wu, Dec 25 2016
E.g.f.: (1/4)*((-6 -4*x - x^2)*exp(-x) + (6- 8*x + 5*x^2 + 2*x^3 + 2*x^4 )*exp(x)). - G. C. Greubel, Dec 26 2016
MATHEMATICA
Table[(1/4)*(n - 3)*(n - 2)*(2*n^2 - 4*n + 1 - (-1)^n), {n, 0, 50}] (* G. C. Greubel, Dec 26 2016 *)
PROG
(Python)
def a(n):
s=0
for a in range(0, n+1):
for b in range(0, n+1):
if a!=b:
for c in range(0, n+1):
if a!=c and b!=c:
for d in range(0, n+1):
if d!=a and d!=b and d!=c:
if (a+d)%2==0:
s+=1
return s
for i in range(0, 201):
print str(i)+" "+str(a(i))
(Python)
def A280056(x):
return -((x**2-5*x+6)*(-2*x**2+4*x+(-1)**x-1))/4
(Python)
from __future__ import division
def A280056(n):
return (n**2 - (n % 2))*(n-1)*(n-2)//2 # Chai Wah Wu, Dec 25 2016
(PARI) concat(vector(3), Vec(8*x^3*(1 + 3*x)*(1 + x + x^2) / ((1 - x )^5*(1 + x)^3) + O(x^30))) \\ Colin Barker, Dec 25 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Indranil Ghosh, Dec 24 2016
STATUS
approved