OFFSET
0,4
COMMENTS
a(n) mod 8 = 0.
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).
FORMULA
From Colin Barker and Charles R Greathouse IV, Dec 12 2016: (Start)
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n>8.
a(n) = (5*n^4 - 8*n^3 + 4*n^2 - 16*n)/8 for n even.
a(n) = (5*n^4 - 12*n^3 + 2*n^2 + 12*n - 7)/8 for n odd.
G.f.: 8*x^3*(2 + 10*x + 7*x^2 + 8*x^3 + 3*x^4) / ((1 - x)^5*(1 + x)^4).
(End)
PROG
(Python)
def t(n):
s=0
for a in range(0, n+1):
for b in range(0, n+1):
for c in range(0, n+1):
for d in range(0, n+1):
if (a!=b and a!=d and b!=d and c!=a and c!=b and c!=d):
if (a*d-b*c)%2==0:
s+=1
return s
for i in range(0, 201):
print str(i)+" "+str(t(i))
(PARI) F(n, {r=0})={my(s=vector(2), v); forvec(y=vector(4, j, [0, n]), for(k=23*!!r, 23, v=numtoperm(4, k); s[1+(y[v[1]]*y[v[4]]-y[v[3]]*y[v[2]])%2]++), 2*!r); return(s)} \\ Use r=1 for A210369;
a(n)=F(n, 0)[1]; \\ Also works for A210370 if F(n, 1)[2] is used instead. - R. J. Cano, Dec 12 2016
(PARI) a(n)=my(e=n\2+1, o=(n+1)\2); 24*binomial(o, 4) + 16*binomial(e, 2)*binomial(o, 2) + 24*o*binomial(e, 3) + 24*binomial(e, 4) \\ Charles R Greathouse IV, Dec 12 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Indranil Ghosh, Dec 12 2016
STATUS
approved