A281550 Number of 2 X 2 matrices with all elements in 0..n such that the sum of the elements is prime.

0, 10, 46, 114, 234, 458, 826, 1370, 2090, 3010, 4174, 5658, 7534, 9930, 12954, 16662, 21074, 26242, 32246, 39182, 47186, 56386, 66874, 78798, 92290, 107434, 124282, 142942, 163550, 186266, 211250, 238626, 268526, 301134, 336610, 375086, 416678, 461454, 509434, 560662, 615182, 673106

Offset: 0

Author

Indranil Ghosh, Jan 23 2017

Keywords

• nonn

Examples

			For n = 4, a few of the possible matrices are [0,4;2,1], [0,4;3,0], [0,4;3,4], [0,4;4,3], [1,0;0,1], [1,0;0,2], [1,0;0,4], [1,0;1,0], [1,0;1,1], [1,0;1,3], [2,2;3,0], [2,2;3,4], [2,2;4,3], [2,3;0,0], [2,3;0,2], [3,4;3,3], [3,4;4,0], [3,4;4,2], [4,0;0,1], [4,0;0,3], [4,0;1,0], ... There are 234 possibilities.
Here each of the matrices M 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, a(4) = 234.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000 (terms 0..200 from Indranil Ghosh, terms 201..3000 from Chai Wah Wu)

Crossrefs

Cf. A210000, A281090, A281315.

Programs

PARI

a(n)=my(X=Pol(vector(n+1,i,1))+O('x^(4*n)),Y=X^4,s); forprime(p=2,4*n, s+=polcoeff(Y,p)); s \\ Charles R Greathouse IV, Feb 15 2017

Python

from sympy import isprime
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 isprime(a+b+c+d)==True:
                        s+=1
    return s
for i in range(0, 201):
    print(str(i)+" "+str(t(i)))

Formula

a(n) = Sum_{p prime} Sum_{k=0..4} (-1)^k * binomial(4, k) * binomial(p+3-k*(n+1), 3). - David Radcliffe, Jun 13 2025