A280934 Number of 2 X 2 matrices with all elements in {0,..,n} and (sum of terms) = permanent.
1, 1, 4, 36, 52, 76, 92, 116, 136, 160, 176, 208, 224, 248, 272, 300, 316, 348, 364, 396, 420, 444, 460, 500, 520, 544, 568, 600, 616, 656, 672, 704, 728, 752, 776, 820, 836, 860, 884, 924, 940, 980, 996, 1028, 1060, 1084, 1100, 1148, 1168, 1200, 1224, 1256, 1272, 1312, 1336, 1376, 1400, 1424, 1440, 1496, 1512, 1536
Offset: 0
Keywords
Examples
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.
Links
- Indranil Ghosh and Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms for n = 0..200 from Indranil Ghosh)
- Chai Wah Wu, Proof of formula a(n) = a(n-1) + 4*tau(n+1) + 8 for n > 3.
Programs
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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)))
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Python
from sympy import divisor_count A280934_list = [1,1,4,36] for i in range(4,100): A280934_list.append(A280934_list[-1]+4*divisor_count(i+1)+8) # Chai Wah Wu, Jan 11 2017
Formula
a(n) = a(n-1) + 4*A000005(n+1) + 8 for n > 3. - Chai Wah Wu, Jan 11 2017
Comments