A lower bound of 2[6] (Steinhaus-Moser-Notation) is
$$ M:= (10 \uparrow \uparrow 257) \uparrow \uparrow \uparrow (10 \uparrow \uparrow 257)$$
I would like to bound M in the following way :
$$10 \uparrow^a b < M < 10\uparrow ^a (b+1)$$
Which choice of a and b does the job ?
I would also be content to bound M with conway-chains only differing in the last number such as
$$a\rightarrow b\rightarrow c\rightarrow d < M < a\rightarrow b \rightarrow c \rightarrow (d+1)$$
Any ideas ?
The bounds which strictly adhere to your conditions would not be very tight: $$10↑↑↑↑2<M<10↑↑↑↑3$$
A better pair of bounds would be:
$$10↑↑↑(10↑↑257)<M<10↑↑↑(10↑↑258)$$
Here is a proof for the second statement:
$1.$ Proving the lower bound is trivial:
$$(10↑↑257)>10 ⇒ (10↑↑257)↑↑↑(10↑↑257)>10↑↑↑(10↑↑257)$$
$2.$ The upper bound can be proven by using induction twice:
LEMMA 1: For every natural number $n$, the following holds: $$(10↑↑257)↑↑n<10↑↑(257+2n)$$
PROOF:
For $n=1$ we get:
$$10↑↑257<10↑↑259$$
Which is obviously true.
Now we assume that the lemma is true for some $n=k$, and we show that the inequality holds for $k+1$ as well:
Step 1 (using the recursive definition of Knuth up-arrows): $$(10↑↑257)↑↑(k+1)=(10↑↑257)↑[(10↑↑257)↑↑k]$$
Step 2 (using the induction assumption): $$(10↑↑257)↑[(10↑↑257)↑↑k]<(10↑↑257)↑[10↑↑(257+2k)]$$
Step 3 (using the recursive definition of Knuth up-arrows): $$(10↑↑257)↑[10↑↑(257+2k)]=[10↑(10↑↑256)]↑[10↑↑(257+2k)]$$
Step 4 (replacing every instance of '↑' with ordinary exponential notation): $$[10↑(10↑↑256)]↑[10↑↑(257+2k)]=(10^{10↑↑256}\space)^{10↑↑(257+2k)}=10^{(10↑↑256)\times[10↑↑(257+2k)]}$$
Step 5:
It is easy to see that $(10↑↑256)\times[10↑↑(257+2k)]$ is far smaller than $10↑↑(257+2k+1)$. Therefore: $$10^{(10↑↑256)\times[10↑↑(257+2k)]}<10^{10↑↑(257+2k+1)}=10↑↑(257+2k+2)=10↑↑[257+2(k+1)]$$
$$QED$$
LEMMA 2: For every natural number $n$, the following holds: $$(10↑↑257)↑↑↑n<10↑↑↑(1+2n)$$
PROOF: (quite similar to the proof of Lemma 1)
For $n=1$ we get:
$$10↑↑257<10↑↑(10↑↑10)=10↑↑↑3=10↑↑↑(1+2\times1)$$
Which is obviously true.
Now we assume that the lemma is true for some $n=k$, and we show that the inequality holds for $k+1$ as well:
Step 1 (using the recursive definition of Knuth up-arrows): $$(10↑↑257)↑↑↑(k+1)=(10↑↑257)↑↑[(10↑↑257)↑↑↑k]$$
Step 2 (using the induction assumption): $$(10↑↑257)↑↑[(10↑↑257)↑↑↑k]<(10↑↑257)↑↑[10↑↑↑(1+2k)]$$
Step 3 (using lemma 1): $$(10↑↑257)↑↑[10↑↑↑(1+2k)]<10↑↑[257+2\times(10↑↑↑(1+2k))]$$
Step 4:
It is easy to see that $257+2\times(10↑↑↑(1+2k))$ is far smaller than $10↑↑↑(2+2k)$. Therefore: $$10↑↑[257+2\times(10↑↑↑(1+2k))]<10↑↑[10↑↑↑(2+2k)]=10↑↑↑(3+2k)$$
$$QED$$
Now, using lemma 2 on $n=10↑↑257$ we get:
$$M=(10↑↑257)↑↑↑(10↑↑257)<10↑↑↑[1+2\times(10↑↑257)]<10↑↑↑(10↑↑258)$$
In conclusion:
$$10↑↑↑↑2<10↑↑↑(10↑↑257)<M<10↑↑↑(10↑↑258)<10↑↑↑↑3$$
By the way, the actual value of $2[6]$ is also between these two bounds:
$$2[6]≈10↑↑↑Mega≈10↑↑↑(10↑)^{257}2.79$$ (where $(10↑)^{257}2.79$ is a power tower of $257$ tens and a top exponent of $2.79$)