OFFSET
0,2
COMMENTS
For n>0, a(n) is the perimeter of a primitive Pythagorean triangle. - Torlach Rush, Jul 11 2019
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..995
FORMULA
a(n) = (((n-2)*a(n-1))/(n-4)) - (6*(3*(n-1)+1)/(n-4)) for n>=4.
Conjectures from Colin Barker, Jan 01 2017: (Start)
a(n) = 2 + 6*n + 4*n^2 for n>0.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
G.f.: 2*x*(6 - 3*x + x^2) / (1 - x)^3.
(End)
From Torlach Rush, Jul 11 2019: (Start)
a(n) = (2*n+1)*(2*n+2), n>0.
a(n) = 2*((n+1)^2 + ((n+1)*n)), n>0.
(End)
MATHEMATICA
Table[Boole[n != 0] 2 # (2 # - 1) &[n + 1], {n, 0, 50}] (* or *)
CoefficientList[Series[2 x (6 - 3 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* Michael De Vlieger, Jan 01 2017 *)
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*d-b*c)==(a*d+b*c)**n:
s+=1
return s
for i in range(0, 41):
print str(i)+" "+str(t(i))
CROSSREFS
KEYWORD
nonn
AUTHOR
Indranil Ghosh, Jan 01 2017
STATUS
approved