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?