A280058 Number of 2 X 2 matrices with entries in {0,1,...,n} with determinant = permanent with no entries repeated.

0, 0, 0, 12, 48, 120, 240, 420, 672, 1008, 1440, 1980, 2640, 3432, 4368, 5460, 6720, 8160, 9792, 11628, 13680, 15960, 18480, 21252, 24288, 27600, 31200, 35100, 39312, 43848, 48720, 53940, 59520, 65472, 71808, 78540, 85680, 93240, 101232, 109668, 118560

Offset: 0

Author

Indranil Ghosh, Dec 25 2016

Keywords

• nonn
• easy

Comments

Consider all Pythagorean triples (X,Y,Z=Y+2) ordered by increasing Z; A005843, A005563, A002522 and A007531 give the X, Y, Z and area A values of related triangles; for n >= 2 altitude h(n) = a(n+1)/A002522(n) or h(n)/2 is irreducible fraction in Q\Z. - Ralf Steiner, Mar 29 2020

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000292, A015237 (where the entries can be repeated), A005843, A005563, A002522, A016742, A099761, A007531.

Programs

Mathematica

Table[2*n*(n-1)*(n-2), {n, 0, 50}] (* G. C. Greubel, Dec 25 2016 *)

PARI

for(n=0, 50, print1(2*n*(n-1)*(n-2), ", ")) \\ G. C. Greubel, Dec 25 2016

PARI

a(n)=12*binomial(n,3) \\ Charles R Greathouse IV, Dec 25 2016

Python

def t(n):
    s=0
    for a in range(0,n+1):
        for b in range(0,n+1):
            if a!=b:
                for c in range(0,n+1):
                    if a!=c and b!=c:
                        for d in range(0,n+1):
                            if d!=a and d!=b and d!=c:
                                if (a*d-b*c)==(a*d+b*c):
                                    s+=1
    return s
for i in range(0,201):
    print(str(i)+" "+str(t(i)))

Python

a = lambda n: 2*n*(n-1)*(n-2) # David Radcliffe, Jun 14 2025

Formula

a(n) = 2*((n+1)^3 - 6*(n+1)^2 + 11*(n+1) - 6), for n>0.

a(n) = 2*n*(n-1)*(n-2). - David Radcliffe, Jun 14 2025

a(n) == 0 (mod 12).

From G. C. Greubel, Dec 25 2016: (Start)

G.f.: (12*x^3)/(1 - x)^4.

E.g.f.: 2*x^3*exp(x).

a(n) = 2*n*(n-1)*(n-2).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

a(n) = 12 * A000292(n-2) for n>1. - Alois P. Heinz, Jan 30 2017

a(n+1) = sqrt(A016742(n)*A099761(n-1)) for n>=2. - Ralf Steiner, Mar 29 2020

From Amiram Eldar, Jun 30 2025: (Start)

Sum_{n>=3} 1/a(n) = 1/8.

Sum_{n>=3} (-1)^(n+1)/a(n) = log(2) - 5/8. (End)