A375512 - OEIS

%I #11 Aug 28 2024 11:17:26

%S 0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,2,1,0,2,0,2,3,0,0,4,3,0,4,3,0,6,0,3,5,

%T 0,6,9,0,0,6,8,0,9,0,5,13,0,0,13,6,7,8,6,0,11,10,12,9,0,0,23,0,0,19,

%U 10,12,15,0,8,11,18,0,27,0,0,23,9,15,18,0,25,19

%N a(n) is the number of distinct compositions of four positive integers x, u, v, y (x < u <= v < y) for which x + u + v + y = n and u*v = x*y.

%C (bin(4,0) + bin(4,2) + bin(4,4))*a(n) = 8*a(n) is the number of distinct compositions of four integers x, u, v, y (abs(x) < abs(u) <= abs(v) < abs(y)) for which abs(x) + abs(u) + abs(v) + abs(y) = n and u*v = x*y.

%C a(n) is also the number of 2X2 matrices having the determinant 0 whose elements [x,u;v,y] are positive integers with x < u <= v < y and x + u + v + y = n.

%C a(n) is also the number of distinct linear 2X2 equation systems that do not have exactly one solution and whose coefficients [x,u;v,y] are positive integers with x < u <= v < y and x + u + v + y = n.

%H Felix Huber, <a href="/A375512/b375512.txt">Table of n, a(n) for n = 0..1000</a>

%H Felix Huber, <a href="/A375512/a375512_1.txt">Maple codes</a>

%F Conjecture: a(p) = 0 for primes p.

%F From _Robert Israel_, Aug 23 2024: (Start)

%F The conjecture is true, in fact for any x,y,u,v as in the definition, n has proper divisor gcd(x,u) + gcd(v,y).

%F Proof: Suppose x,y,u,v are positive integers with x + y + u + v = n and x*y = u*v = m.  Let g = gcd(x,u).  Then x = g*X and u = g*U where X and U are coprime.  Since X*y = U*v = m/g, we must have y = h*U and v = h*X where h = gcd(v,y).  Then n = g*X + h*U + g*U + h*X = (g+h)*(U+X).

%F (End)

%e a(9) = 1 because only (1, 2, 2, 4) satisfies the conditions: 1 + 2 + 2 + 4 = 9 and 2*2 = 1*4.

%e a(24) = 4 because (1, 2, 7, 14), (1, 3, 5, 15), (2, 4, 6, 12), (3, 5, 6, 10) satisfy the conditions: 1 + 2 + 7 + 14 = 24 and 2*7 = 1*14, 1 + 3 + 5 + 15 = 24 and 3*5 = 1*15, 2 + 4 + 6 + 12 = 24 and 4*6 = 2*12, 3 + 5 + 6 + 10 = 24 and 5*6 = 3*10.

%e See also linked Maple code.

%p See Huber link.

%o (Python)

%o def A375512(n): return sum(1 for x in range(1,(n>>2)+1) for y in range(x+1,(n-x)//3+1) for z in range(y,(n-y>>1)+1) if x<y<=z<(n-x-y-z) and y*z==x*(n-x-y-z)) # _Chai Wah Wu_, Aug 23 2024

%Y Cf. A038548, A134506, A357259.

%K nonn

%O 0,16

%A _Felix Huber_, Aug 19 2024