

A339167


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


2



2, 5, 56, 382, 3002, 25918, 246962, 2580478, 29393282, 362879997, 4829932803, 68976230397, 1052366515203, 17086945075197, 294226732800003, 5356234211327997, 102793666719744003, 2074369080655871997, 43913881247588352003, 973160803270655999997
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OFFSET

1,1


COMMENTS

Consider the equation (x1)! = x^2. The only solution in the positive integers is x = 1, and (51)! ~ 5^2 is a near miss.
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.
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.


LINKS



EXAMPLE

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.
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.


PROG

(PARI) a(n)={1\/(nsolve(x=n+(1/(n+2))^(n+1), n(2/(n+2))^n, gamma(x)x^2))}


CROSSREFS



KEYWORD

sign


AUTHOR



STATUS

approved



