Recent content by RD-BH

[edit 2014.10.11] >>> ERROR <<< The argument, "2014.09.30 for consideration", fails in the superset. The subset, (P!/2)*(2*I+-1)+-2^(J>0)+-K*P!, holds true. ... back to the drawing board 8(

I would not say this is a great math problem as most mathematicians hold it to be true even in absence of a concise proof. Thoughts: ... 1) I know the math is true, I am trying to make it digestible. ... 2) My personal pursuit is a problem I do consider great, The Riemann Hypothesis. ...

@peter grimes I am interested in any/all feedback. Thank you.

[edit 2014.10.11] >>> ERROR <<< The argument, "2014.09.30 for consideration", fails in the superset. The subset, (P!/2)*(2*I+-1)+-2^(J>0)+-K*P!, holds true. [begin 2014.09.30 for consideration] For consideration: I is the set of integers. 2I is the set of integers divisible by 2...

Sadly, I did not like Hyperion or Contact and both were referenced as inspiration for CivBE, and I don't care for CivV. Still, I've preordered.

Just a quick thought on "making the switch" ... I feel Civ IV Beyond the Sword was the spiritual successor to SMAC/X making CivBE just another pretender. ... what am I saying? ... take my money.

As far as I am aware, you can map any powered orbital to any location above the surface of the earth. Perhaps you are thinking of geostationary orbits?

Then I propose the following: ... 1) depleted resources ... 2) limited war over resources ... 3) increased dependence on nuclear power ... 4a) experiment to create new power source (black hole reference) ... 4b) -or- attempt at time travel ... 5) unexpected sub-particle resonance inside...

Clearly "The Great Mistake" was not maintaining intellectual property rights to SMAC/X.

Hmm, maybe a space victory is The Great Mistake?

Thank you 8) === still working on correct wording === Let us examine Integers and Indivisibility. … x/n is an integer when n is a divisor of x … 1i^(4x/n) equals 1 when n is a divisor of x [1i = (-1)^(1/2)] … numberOfDivisors(x) = sum_(n=1)^(x){floor(abs(1i^(4x/n)+1)/2)} …...

Edited original post. Noted TYPO. Added For Consideration: ... Includes: ... ... residue + period (infinite indivisible twin pairs) ... ... period/2 + offset + period (infinite indivisible twin pairs) ... ... odd*period/2 + offset + period (infinite indivisible twin pairs) ... ... residue...

This makes more sense if stated: ... P +- 2^x is not divisible by P for P>2 ... Therefore the product(n_1^x)(P_n) +- (2^I) is not divisible by any element of P for all I != 0 and all x ... Therefore product(n_1^x)(P_n) +- (2^1) and product(n_1^x)(P_n) +- (2^2) are four twin pairs not...

After a lengthy presentation and even longer Q&A, they made no comment on the validity of the proof. This leads me to believe the proof fails. Findings: ... recommended I restructure my paper for educational use ... recommended I publish my work on Indivisibility ... recommended I publish...

Not exactly my statement (though my statement may still be useless). ... original: (x-1)!+x and (x-1)!+x+2 are twin primes. If factors of (x-1)! cannot divide x and x+2 then what about (x-1)!+x and (x-1)!+x+2? ... that is my original point (minus the point of x and x+2 being prime). ... the...