OFFSET
0,3
COMMENTS
a(n) mod 4 = 0 for n > 1.
a(n) is also the number of 2 X 2 matrices with all elements in {-1,..,n-1} and permanent = 2. - Chai Wah Wu, Jan 11 2017
LINKS
Indranil Ghosh and Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms for n = 0..200 from Indranil Ghosh)
FORMULA
a(n) = a(n-1) + 4*A000005(n+1) + 8 for n > 3. - Chai Wah Wu, Jan 11 2017
EXAMPLE
For n = 3, the possible matrices are [0,0,0,0], [0,2,2,0], [0,2,3,1],[0,3,2,1], [0,3,3,3], [1,2,3,0], [1,2,3,1], [1,2,3,2], [1,2,3,3], [1,3,2,0], [1,3,2,1], [1,3,2,2], [1,3,2,3], [2,0,0,2], [2,0,1,3], [2,1,0,3], [2,1,1,3], [2,1,2,3], [2,1,3,3], [2,2,1,3], [2,2,2,2], [2,2,3,1], [2,3,1,3], [2,3,2,1], [3,0,1,2], [3,0,3,3], [3,1,0,2], [3,1,1,2], [3,1,2,2], [3,1,3,2], [3,2,1,2], [3,2,3,1], [3,3,0,3], [3,3,1,2], [3,3,2,1] and [3,3,3,0]. There are 36 possibilities.
Here each of the matrices is defined as M = [a,b,c,d] where a = M[1][1], b = M[1][2], c = M[2][1], d = M[2][2]. So, for n = 3, a(n) = 36.
PROG
(Python)
def t(n):
s=0
for a in range(n+1):
for b in range(n+1):
for c in range(n+1):
for d in range(n+1):
if (a+b+c+d)==(a*d+b*c):
s+=1
return s
for i in range(201):
print(str(i)+" "+str(t(i)))
(Python)
from sympy import divisor_count
A280934_list = [1, 1, 4, 36]
for i in range(4, 100):
CROSSREFS
KEYWORD
nonn
AUTHOR
Indranil Ghosh, Jan 11 2017
STATUS
approved