

A047694


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


4




OFFSET

0,3


COMMENTS

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 yvalues.  M. F. Hasler, Jun 22 2024


REFERENCES

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


LINKS



FORMULA

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


EXAMPLE

For x = 1, C(x, k) := x*(x1)*...*(xk+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



CROSSREFS

Cf. A047695 (the corresponding y values).


KEYWORD

sign,fini,full,nice


AUTHOR



STATUS

approved



