Here is that Gamma infinite product rationalized by your transformation: Product[(3^(-(1/2) + 3*n)*Gamma[2/3 + n + (2*(-3 + x))/(3*(3 + x^2))]* Gamma[1/3 + n - (4*x)/(3*(3 + x^2))]* Gamma[n + (2*(3 + x))/(3*(3 + x^2))])/(2*Pi* Gamma[3*n]), {n, Infinity}] == (BarnesG[4/3]* BarnesG[5/3])/(BarnesG[5/3 + (2*(-3 + x))/(3*(3 + x^2))]* BarnesG[4/3 - (4*x)/(3*(3 + x^2))]* BarnesG[1 + (2*(3 + x))/(3*(3 + x^2))]) =: 1+f[x]. f has extrema at exactly -3-2rt3, -rt3, 3-2rt3, 2rt3-3, rt3, and 2rt3+3 and is visually indistinguishable from an odd, equal-ripple function. But it is neither. The ripples at +-rt3 are .0002 larger, and, instead of 0, f[x]+f[-x] appears to be equal-ripple, varying between 0 and .00023 with eleven turning points! --rwg Puzzle: Recall that In[45]:= Product[Gamma[k], {k, n}] Out[45]= BarnesG[1 + n] What is BarnesG[0]? On 2015-08-02 10:15, Warut Roonguthai wrote:
Nope. Here's how I derived the formula.
Let -8 + 36*a - 27*a^2 = b^2.
So, 3 = (2+b)/(3a-2) * (2-b)/(3a-2).
Let (2+b)/(3a-2) = k, a rational number. So, (2-b)/(3a-2) = 3/k.
Now we can easily solve for a, b, and we have
a = 2/3 + (4/3)*k/(k^2+3), b = 2*(k^2-3)/(k^2+3).
Similarly, this technic can be used to find a parametric solution for Pythagorean triples.
Warut
On Sun, Aug 2, 2015 at 10:51 PM, rwg <rwg@sdf.org> wrote:
Neat! You got that somehow from
In[407]:= Simplify[(2 t/(1 + t^2))^2 + ((1 - t^2)/(1 + t^2))^2]
Out[407]= 1 ? --rwg
On 2015-08-02 05:56, Warut Roonguthai wrote:
I didn't follow math-fun for a long time (i.e., don't know where this equation came from), but the new subject caught my eye.
Using an elementary method, I found that -8 + 36*a - 27*a^2 is a rational square iff
a = 2/3 + (4/3)*k/(k^2+3) for some rational k.
On Sat, Aug 1, 2015 at 11:53 PM, Bill Gosper <billgosper@gmail.com> wrote:
The data are intriguing. Something about 3 and powers of primes 1 mod 6.
What subject is this? BQFs? --rwg