Never mind, I realized I could experiment using the regular iteration at bases like b = sqrt(2), plus a cheap numerical differentiation procedure, to see how the coeffs. behave.

It seems the terms in

\( b_k = \sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k} \)

for k = 2 (i.e. 2nd coeff of continuum sum of sqrt(2) tetration expanded about 0), f = regular iteration of b = sqrt(2), grow up in magnitude mildly hypergeometrically (but alternate in sign). I get (rounded)

term n = 38: -143.931066425221

term n = 40: 1281.50456722963

term n = 42: -12659.7473479491

term n = 44: 138050.651934470

term n = 46: -1654016.61498652

term n = 48: 21681557.9712844

term n = 50: -309748957.739627

The ratio of magnitudes looks like (mn = magnitude of term n)

m40/m38 = 8.90359947

m42/m40 = 9.87881563

m44/m42 = 10.9046925

m46/m44 = 11.9812300

m48/m46 = 13.1084282

The differences of these are (rn = ratio of term n to term n-2)

r42 - r40 = 0.97521616

r44 - r42 = 1.02587687

r46 - r44 = 1.07653752

r48 - r46 = 1.12719814

In other words, the ratio of the magnitudes of successive coefficients grows slightly faster than linear, but not too much so. I'd venture it is between linear and quadratic.

I heard that the Borel summation is not applicable to sums that grow hypergeometrically like this, but I heard here: (

http://mathworld.wolfram.com/Borel-RegularizedSum.html) that it originates wth the summing of divergent hypergeometric functions, whose term magnitude ratios may grow up even faster than the ones here do. So could the Borel summation be applicable here? It would seem not to have so many fussy parameters, but here:

http://en.wikipedia.org/wiki/Borel_summation I hear it needs an analytic continuation of a certain function to the whole positive real line, but how can you analytically continue a more-or-less arbitrary series like that?