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Rounded value of 1/(x[n] + n), where x[n] is the n-th negative solution to Gamma(x) = x^2.
2

%I #14 Dec 05 2020 18:01:33

%S -2,5,-56,382,-3002,25918,-246962,2580478,-29393282,362879997,

%T -4829932803,68976230397,-1052366515203,17086945075197,

%U -294226732800003,5356234211327997,-102793666719744003,2074369080655871997,-43913881247588352003,973160803270655999997

%N Rounded value of 1/(x[n] + n), where x[n] is the n-th negative solution to Gamma(x) = x^2.

%C Consider the equation (x-1)! = x^2. The only solution in the positive integers is x = 1, and (5-1)! ~ 5^2 is a near miss.

%C If (x-1)! is replaced by Gamma(x), then in addition to the positive non-integer solution x = 5.0367... (cf. A264785) there are non-integer solutions increasingly close to each negative integer: x[1] = -1.5259..., x[2] = -1.806544..., x[3] = -3.017901..., x[4] = -3.997382..., x[5] = -5.000333... etc. They are to the left of odd and to the right of even negative integers, and the present sequence gives the reciprocals of the distances 1/(x[n]+n), rounded to nearest integers.

%C We notice an intriguing pattern in the last digits of the terms, which are of the form m*10^k - 3 (resp. -2 for a(4 .. 9)) for increasingly large k. (Positive terms end in ...97, negative terms end in ...03) Is this a coincidence? Will that pattern prevail? Why these values, off by -3 from multiples of 10^k? We would appreciate a simple explanation of this observation.

%e The largest negative solution to Gamma(x) = x^2 is x[1] = -1.525796..., its (signed) distance from -1 is x[1] + 1 = -0.525796..., the reciprocal is 1.901..., which rounded to nearest integer gives a(1) = 2.

%e The next negative solution to Gamma(x) = x^2 is x[2] = -1.806544..., its (signed) distance from -2 is x[2] + 2 = +0.193..., with reciprocal 5.169..., and rounded to nearest integer, a(2) = 5.

%o (PARI) a(n)={1\/(n-solve(x=n+(-1/(n+2))^(n+1),n-(-2/(n+2))^n,gamma(-x)-x^2))}

%Y Cf. A264785, A339161.

%K sign

%O 1,1

%A _M. F. Hasler_, Nov 25 2020.