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A047694
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x such that y^2 = C(x,0) + C(x,1) + C(x,2) + C(x,3) is solvable in integers.
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4
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, Section D3.
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LINKS
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FORMULA
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x such that 6y^2 = (x + 1)(x^2 - x + 6) has solutions in integers.
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[-10, 10^3], IntegerQ[Sqrt[((# + 1)(#^2 - # + 6))/6]] &] (* Alonso del Arte, Sep 13 2011 *)
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PROG
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CROSSREFS
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Cf. A047695 (the corresponding y values).
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KEYWORD
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sign,fini,full,nice
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AUTHOR
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STATUS
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approved
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