Arpeggio-Doubling Sequence

[Leroy Quet]

Here is sequence A137293 in its raw unedited
form:

%I A000001
%S A000001
1,1,2,1,2,1,2,3,4,1,2,1,2,3,4,1,2,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,8,1,2,1,2,3,4,1,2,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,8,1,2,1,2,3,4,1,2,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,8,1,2,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,11,12,1,2,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,13,14,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
%N A000001 a(1) = 1. After the initial 1, follow
with runs of integers; the nth run of 2*a(n)
terms is (1,2,3,4,....,2a(n)).
%e A000001 Runs of integers bracketed by
parentheses. (Note that nth run {after the
initial 1} has exactly 2a(n) terms.)
1,(1,2),(1,2),(1,2,3,4),(1,2),(1,2,3,4),(1,2),(1,2,3,4),(1,2,3,4,5,6),(1,2,3,4,5,6,7,8),(1,2),(1,2,3,4),(1,2),(1,2,3,4),(1,2,3,4,5,6),(1,2,3,4,5,6,7,8),(1,2),(1,2,3,4),(1,2),(1,2,3,4),(1,2,3,4,5,6),(1,2,3,4,5,6,7,8),(1,2),(1,2,3,4),(1,2,3,4,5,6),(1,2,3,4,5,6,7,8),(1,2,3,4,5,6,7,8,9,10),(1,2,3,4,5,6,7,8,9,10,11,12),...
%O A000001 1
%K A000001 ,easy,nonn,

The question is, is there a closed form (ie
non-recursive) way of calculating any particular
a(n) directly? (Perhaps something involving the
binary representation of n.)
It seems that there should be.

Also, we can generalize this sequence:

a(1) = 1. After the initial 1, follow with runs
of integers; the nth run of floor(x*a(n)) terms
is (1,2,3,4,....,floor(x*a(n))), where x is any
particular real >= 2.
(For x < 2 and x >= 1, each a(n) = 1.)
For each real x >= 2 there is a unique sequence.
(Maybe someone else wouldn't mind submitting such
sequences for other x's.)
We can also replace each "floor" in the sequence
definition by "ceiling" for another set of
sequences.
Any closed forms for the general cases?

Thanks,
Leroy Quet




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