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 A047694 x such that y^2 = C(x,0) + C(x,1) + C(x,2) + C(x,3) is solvable in integers. 4
 -1, 0, 2, 7, 15, 74, 767 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS n such that A000125(n) is a perfect square. - Frank M Jackson, Mar 13 2013 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 REFERENCES R. K. Guy, Unsolved Problems in Number Theory, Section D3. LINKS Table of n, a(n) for n=0..6. Andrew Bremner, An equation of Mordell, Math. Comp., 29 (1975), pp. 925-928. W. Ljunggren, A diophantine problem, J. London Math. Soc. (2), 3 (1971), pp. 385-391. FORMULA x such that 6y^2 = (x + 1)(x^2 - x + 6) has solutions in integers. EXAMPLE 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. 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. For x = 1, the r.h.s. equals 1 + 1 + 0 + 0 = 2 and there is no integer solution y. For x = 2, the r.h.s. equals 1 + 2 + 1 + 0 = 4 and we have the solutions y = +-2, whence a(3) = 2. MATHEMATICA Select[Range[-10, 10^3], IntegerQ[Sqrt[((# + 1)(#^2 - # + 6))/6]] &] (* Alonso del Arte, Sep 13 2011 *) PROG (PARI) A047694=[n|n<-[-1..777], issquare(binomial(n+1, 3)+n+1)] \\ M. F. Hasler, Jun 22 2024 CROSSREFS Cf. A047695 (the corresponding y values). Sequence in context: A282197 A050612 A120110 * A338399 A262016 A129666 Adjacent sequences: A047691 A047692 A047693 * A047695 A047696 A047697 KEYWORD sign,fini,full,nice AUTHOR N. J. A. Sloane STATUS approved

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Last modified August 7 22:54 EDT 2024. Contains 375018 sequences. (Running on oeis4.)