Infinite Addition Chains

An infinite addition chain is an infinite sequence (a_i)_i of strictly monotonic increasing natural numbers a_i starting with 1, such that each element of this sequence except the first, a₀=1, is the sum of two (not necessarily different) previous elements.
We call such an infinite addition chain minimal, if it contains an infinite subsequence (a_ih)_h such that for each element a_ih of this subsequence ℓ(a_ih) = i_h holds -- i.e. it contains a shortest addition chain for each a_ih.
Furthermore we call a minimal infinite addition chain unbounded minimal if s(a_ih) is unbounded when h tends to infinity. On the other hand, if s(a_ih) is bounded, then exists an index, say i_b, such that all steps in this minimal infinite addition chain with indices j > i_b are double steps and therefore ℓ(a_j)=j.
Finally here are two examples for infinite addition chains; one for a non unbounded minimal: 1,2,4,8,...,2ⁿ,...
and one for a unbounded minimal: 1,2,3,6,12,15,30,60,120,240,255,510,...,2ⁿ(2^2m-1),... with n <= 2^m. The last example is only a conjectural-to-be unbounded minimal addition chain. To the author's knowledge there is no proved-to-be unbounded minimal addition chain known until today (september 2016).
An unbounded minimal addition chain is given by: 1,2,4,5,8,16,32,64,69,128,... such that the next term is either a doublestep or a doubling of the previous-to-previous element or a new smallstep 2^2i-2 for every positive index i. For these numbers n(k)=∑^k_i=1 2^2i-2 must hold ℓ(n)=λ(n)+ν(n)-1 because their bits at index 2ⁱ-2 are spaced sufficient increasingly far apart such that they can't be generated by carries of less than s(n)=ν(n)-1=k-1 smallsteps. Proof idea: s smallsteps can generate at most 2^s bits from a single bit and thus could generate a further bit by their cummulated carries at most 2^s-1 indices higher.
Remark: The same argument/proof works also for numbers n(k)=∑^k_i=0 2^2i-1 respectively the infinite addition chain 1,2,3,4,8,11,16,32,64,128,139,256,....