A280321 Number of 2 X 2 matrices with all elements in {0,..,n} having determinant = n*permanent.

1, 12, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025, 9409, 9801, 10201

Offset

0,2

Comments

Same as A016754, except for n=1. Here a(1)=12 but A016754(1)=9.

Links

Indranil Ghosh, Table of n, a(n) for n = 0..990

Formula

a(n+1) = (((n-2)*a(n))/(n-1)) + ((12*(n)^2-12*(n)+1)/(n-1)) for n>=1.

Conjectures from Colin Barker, Jan 01 2017: (Start)

a(n) = (1 + 2*n)^2 = A273789(n) = A273743(n) for n>1.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>4.

G.f.: (1 + 9*x - 8*x^2 + 9*x^3 - 3*x^4) / (1 - x)^3.

(End)

From Sela Fried, Feb 01 2026: (Start)

Barker's conjectures are true. The recursion and g.f. follow from the formula for a(n). To see why the formula is true, write the condition as (a*d-b*c)=n*(a*d+b*c) and assume that n>=2. Rearranging, (1-n)*a*d=(1+n)*b*c. Since 1-n<0 and 1+n>0, necessarily a*d=b*c=0. Thus, at least one of a,d and, independently, at least one of b,c must be 0. This gives 2*n+1 possibilities for each of the two pairs, proving that a(n)=(2*n+1)^2. The cases n=0,1 are verified directly. (End)

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*d-b*c)==n*(a*d+b*c):
                        s+=1
    return s
for i in range(41):
    print(str(i)+" "+str(t(i)))

Crossrefs

Cf. A016754.

Sequence in context: A224669 A164577 A195143 * A198274 A229447 A175523

Adjacent sequences: A280318 A280319 A280320 * A280322 A280323 A280324

Keyword

nonn

Author

Indranil Ghosh, Jan 01 2017

Status

approved