The sum of the prime factors for each number is 6770081261888348088

Continuing with more primes p and this specific version of the algorithm you'll find more pairs as follows:

11458500073142461852512836718403514710
11458500073142461852512836718403514711
3187793133764126630012112493446934643110
3187793133764126630012112493446934643111
41579401268121198749220796887033346345510
41579401268121198749220796887033346345511
25535656878510947884218726327567312941635030
25535656878510947884218726327567312941635031
...
177561956785815018149935855412093504515834133110
177561956785815018149935855412093504515834133111

Step 1: I setup an arbitrary form for a number M in the Ruth-Aaron pair to be a product of two prime factors as follows:
    M = n(n+2v)

Step 2: Since the sum of factors of M above is 2n+2v, I setup a form for a consecutive number to have a large prime factor (2n-d) close to the sum
    M-1 = (2n-d)*Something

Step 3: For this to be satisfied, (2n-d) must divide M-1=n(n+2v)-1. Using some clever substitution I found that choosing n=(d²+(4v+1)d - 4)/2 will always satisfy this condition because (M-1)/(2n-d) will then simplify to 2 v (1+x)+x (2+x)  where d=2x+1. This can then replace the 'Something' above.

Step 4: Since M is odd, M-1 is divisible by 2. Furthermore we need the sum of prime factors to be even and so I added an additional restriction that M-1 be divisible by 5 (arbitrary). This is to balance out the number of odd prime factors in M-1 as we will see. So we express M-1 as:
    M-1 = 10*(2n-d)*[(2 v (1+x)+x (2+x))/10]

Step 5: In order for us to be able to construct a sequence of solutions we need some key variable all the others will be based on. I choose a prime p such that p divides M-1. So:
    M-1 = 10*p*(2n-d)*[(2 v (1+x)+x (2+x))/(10*p)]

Step 6: Now let's lock down the factors involved by assuming the following expressions will simultaneously be prime: {n}, {n+2v}, {2n-d},{(2 v (1+x)+x (2+x))/(10*p)}. We can then setup the Ruth-Aaron equation with sum of factors on each side in terms of p, x, and v as follows:

    2+5+p+(2v(1+x)+x(2+x))/(10*p)+(2n-d) == 2v+2n

This is where adding the additional factor of 5 in Step 4 comes in making this possible. Otherwise the left-side would be odd and right-side even. Solving for v we find that

    v = (10p² - 20p(x-3) + x(2+x)) / (20p - 2(1+x))

Since we are in complete control of the variables involved, I chose x = 10p to simplify v in terms of p. Setting x=10p, v then simplifies to

    v = 5p(9p-8)

Step 7: Now we are all set. We can construct all the variables involved by choosing p. After we choose p,  we calculate v=5p(9p-8), x=10p, d=2x+1, n=(d²+(4v+1)d-4)/2, then M=n(n+2v).

If any of the expressions in Step 6 aren't prime, we just chose another p until they are. :)

The above may make more sense to you by writing down our newly constructed M and M-1 explicitly in terms of p and factoring to give the following:

M = (1800p³ - 1310p² - 50p - 1) (1800p³ - 1220p² - 130p - 1)
M-1 = 2 * 5 * p * (90p² - 61p - 6) (3600p³ - 2620p² - 120p - 3)

And you find that we really have is a Ruth-Aaron polynomial. The sum of the factors of M and the sum of the factors of M-1 are both equal to:

3600p³ - 2530p² - 180p - 2

Obviously we can make different assumptions in the construction above and come up with several variants of this algorithm (and polynomial). For instance using a different prime 'q' instead of 5 in Step 4 and an appropriate v.

Update: After this analysis, I did some further research on what is known about Ruth-Aaron pairs as of April 17, 2002. It turns out that I wasn't the only one who constructed a Ruth-Aaron polynomial (and thus an algorithm for generating RA pairs).  Carl Pomerance, David E. Penney, and Carol Nelson on a joint paper for the journal of Recreational Mathematics in the Spring of 1974 (Vol 7, No. 2), found the following polynomial:

f(x) = 384x³ + 432x² + 112x - 5
f(x) = (8x+5) (48x² + 24x - 1)
f(x)+1 = 4 (1+2x) (48x² + 30x - 1)

FactorSum =48x² + 32x + 4

They probably found the polynomial via inspection as their paper didn't address how they came up with it. It is much simpler than the one I derived, albeit not as adaptable to different forms. Nevertheless, not to be out-done I put together a computerized search and found a simpler Ruth-Aaron cubic as follows:

f(x) = 48x³ - 36x² - 8x + 1
f(x) = (1+4x)(12x² - 12x + 1)
f(x)-1 = 4x(12x² - 9x - 2)

FactorSum = 12x² - 8x + 2

However after translating this requiring x to be odd it ends up being a cousin of the Pomerance/Penney/Nelson cubic after all!

f(x) = 384x³ + 432x² + 128x + 5
f(x) = (5+8x)(48x²+24x+1)
f(x)-1 = 4(1+2x)(48x²+30x+1)

Note: There are several such consecutive polynomials where the irreducible divisors add up to be the same BUT whose polynomial factors cannot all be simultaneously prime. For instance, f(x) = 4x³ - 2x² - 4x is such a case as well as f(x) = 18x³ - 12x² - 3x + 1.

Update 5/30/2003: I put together a generalized pair of RA polynomials for investigational purposes. I'll list it here for posterity:

f(x) = (4kx+5)(12k²x²+12kx-1)
f(x)+1 = 4(kx+1)(12k²x²+15kx-1)

where k is even and NOT divisible by 5. Using odd k will often generate RA pairs too, so keep that in mind for programmatic use.

It's interesting being able to generate a Ruth-Aaron pair that is out of range to be factored by off-the-shelf methods (e.g. 180+ digits). Folks that get just the numbers can't even prove they are Ruth-Aaron. :)  I say "off-the-shelf" because you can actually create a very-fast algorithm to factor Ruth-Aaron numbers.

For your amusement here is a 135-digit Ruth-Aaron pair:

296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475910
296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475911

Links:
    A homepage on Ruth-Aaron numbers (geocites.com)
    Playing around with Ruth-Aaron pairs (maa.org)
    Ruth-Aaron Summary (mathworld.wolfram.com)
    Maris-McGwire-Sosa Numbers (aol.com)
    Ruth-Aaron Triplets (primepuzzles.net)