cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A280934 Number of 2 X 2 matrices with all elements in {0,..,n} and (sum of terms) = permanent.

Original entry on oeis.org

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

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Author

Indranil Ghosh, Jan 11 2017

Keywords

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

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.
		

Crossrefs

Programs

  • 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):
        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