%I #31 Jun 22 2024 22:24:31

%S -1,0,2,7,15,74,767

%N x such that y^2 = C(x,0) + C(x,1) + C(x,2) + C(x,3) is solvable in integers.

%C n such that A000125(n) is a perfect square. - _Frank M Jackson_, Mar 13 2013

%C From Pascal's triangle, C(x,0) + C(x,1) = C(x+1,1) = x + 1 and C(x,2) + C(x,3) = C(x+1,3), so the formula in the definition is equivalent to y^2 = C(x+1,3) + x + 1 = A000125(n). - See A047695 for the corresponding y-values. - _M. F. Hasler_, Jun 22 2024

%D R. K. Guy, Unsolved Problems in Number Theory, Section D3.

%H Andrew Bremner, <a href="https://doi.org/10.1090/S0025-5718-1975-0374019-7">An equation of Mordell,</a> Math. Comp., 29 (1975), pp. 925-928.

%H W. Ljunggren, <a href="https://doi.org/10.1112/jlms/s2-3.3.385">A diophantine problem</a>, J. London Math. Soc. (2), 3 (1971), pp. 385-391.

%F x such that 6y^2 = (x + 1)(x^2 - x + 6) has solutions in integers.

%e For x = -1, C(x, k) := x*(x-1)*...*(x-k+1)/k! = (-1)^k for all integers k >= 0, so the right hand side is zero and y = 0 is a solution. For smaller x values, the r.h.s. is negative and there can't be a solution, therefore the first term is a(1) = -1.

%e For x = 0, C(x, k) = { 1 if k=0, else 0 }, so the r.h.s. is 1 and y = +-1 is a solution, whence a(2) = 0.

%e For x = 1, the r.h.s. equals 1 + 1 + 0 + 0 = 2 and there is no integer solution y.

%e For x = 2, the r.h.s. equals 1 + 2 + 1 + 0 = 4 and we have the solutions y = +-2, whence a(3) = 2.

%t Select[Range[-10, 10^3], IntegerQ[Sqrt[((# + 1)(#^2 - # + 6))/6]] &] (* _Alonso del Arte_, Sep 13 2011 *)

%o (PARI) A047694=[n|n<-[-1..777],issquare(binomial(n+1,3)+n+1)] \\ _M. F. Hasler_, Jun 22 2024

%Y Cf. A047695 (the corresponding y values).

%K sign,fini,full,nice

%O 0,3

%A _N. J. A. Sloane_