OFFSET
0,4
COMMENTS
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = 2*((n+1)^3 - 6*(n+1)^2 + 11*(n+1) - 6), for n>0.
a(n) == 12 (mod 12).
From G. C. Greubel, Dec 25 2016: (Start)
G.f.: (12*x^3)/(1 - x)^4.
E.g.f.: 2*x^3*exp(x).
a(n) = 2*n*(n-1)*(n-2).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
a(n) = 12 * A000292(n-2) for n>1. - Alois P. Heinz, Jan 30 2017
MATHEMATICA
Table[2*n*(n-1)*(n-2), {n, 0, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 0, 0, 12}, 50] (* G. C. Greubel, Dec 25 2016 *)
PROG
(Python)
def t(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-b*c)==(a*d+b*c):
s+=1
return s
for i in range(0, 201):
print str(i)+" "+str(t(i))
(PARI) for(n=0, 50, print1(2*n*(n-1)*(n-2), ", ")) \\ G. C. Greubel, Dec 25 2016
(PARI) a(n)=12*binomial(n, 3) \\ Charles R Greathouse IV, Dec 25 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Indranil Ghosh, Dec 25 2016
STATUS
approved