%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 nth negative solution to Gamma(x) = x^2.
%C Consider the equation (x1)! = x^2. The only solution in the positive integers is x = 1, and (51)! ~ 5^2 is a near miss.
%C If (x1)! is replaced by Gamma(x), then in addition to the positive noninteger solution x = 5.0367... (cf. A264785) there are noninteger 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\/(nsolve(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.
