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In range 0 .. 58784, these are numbers k such that A244158(A239903(n)) = k. (see comments at A244157).

The integer a(n) intervenes as a parameter of an algorithm (Dejter) that generates an ordered tree comprising all Dyck words. The nodes of this ordered tree represent the cyclic classes of vertices, and the cycles of uniform 2-factors (Mutze et al.), of the Odd graphs and Middle-Levels graphs. These factors were used to generate explicit Hamilton cycles in those graphs.

From Antti Karttunen, Jun 22 2014: (Start)

Also called "Greedy Catalan Base" for short.

Note: unlike A239903, this is a true base system, thus A244158(a(n)) = n holds for all n. See also A244159 for another, "less greedy" Catalan Base number system.

No digits larger than 3 will ever appear, because C(n+1)/C(n) approaches 4 from below, but never reaches it. [Where C(n) is the n-th Catalan number, A000108(n)].

3-digits cannot appear earlier than at the fifth digit-position from the right, the first example being a(126) = 30000.

The last digit is always either 0 or 1. (Cf. the sequences A244222 and A244223 which give the corresponding k for "even" and "odd" representations). No term ends as ...21.

No two "odd" terms (ending with 1) may occur consecutively.

A244217 gives the k for which a(k) starts with the digit 1, while A244216 gives the k for which a(k) starts with the digit 2 or 3.

A000108(n+1) gives the position of numeral where 1 is followed by n zeros.

A014138 gives the positions of repunits.

A197433 gives such k that a(k) = A239903(k). [Actually, such k, that the underlying strings of digits/numbers are same].

For the explanations, see the attached notes.

(End)

Algorithm for constructing the sequence: Define a(0) as 0, and for larger values of n, find first the largest Catalan number which is less than or equal to n [which is A081290(n)], and the index k = A244160(n), of that Catalan number. Initialize a vector of k zeros, [0, 0, ..., 0]. Set n_remaining = n - A000108(k) and add 1 to the leftmost element of vector, so that it will become [1, 0, ..., 0]. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the n_remaining, and provided that C(m) <= n_remaining, then set n_remaining = n_remaining - C(m) and increment by one the m-th element of the vector (where the 1st element is the rightmost), otherwise just decrement m by one and keep on doing the same with lesser and lesser Catalan numbers, and whenever it is possible to subtract them from n_remaining (without going less than zero), do so and increment the corresponding m-th element of the vector, as long as either n_remaining becomes zero, or after subtracting C(1) = 1 from n_remaining, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to n_remaining, and start the process again. However, after a finite number of such iterations, n_remaining will finally reach zero, and the result of a(n) is then the vector of numbers constructed, concatenated together and represented as a decimal number.

This shares with "Greedy Catalan Base" (A014418) the property that a simple weighted sum of Sum_{k=1..} digit(k)*C(k) recovers the natural number n, which the given numeral string like A014418(n) or here, a(n), represents. (Here C(k) = the k-th Catalan number, A000108(k), and digit(1) = the digit in the rightmost, least significant digit position.)

In this case, A244158(a(n)) = n holds for only up to 33603, after which comes the first representation containing a "digit" larger than nine, at a(33604), where the underlying string of numbers is [1,2,3,4,5,6,7,8,9,10] but the decimal system used here can no more unambiguously represent them.

On the other hand, with the given Scheme-functions, we always get n back with: (CatBaseSumVec (A244159raw n)).

For n >= 1, A014138(n) gives the positions of repunits: 1, 11, 111, 1111, ...

The "rep-2's": 22222, 222222, 2222222, 22222222, 222222222, ..., etc., occur in positions 128, 392, 1250, 4110, 13834, ... i.e. 2*A014138(n) for n >= 5.

Replace 2^k with A000108(k+1) in binary expansion of n.

From Antti Karttunen, Jun 22 2014: (Start)

On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.

This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.

(End)

After n>0, A000108(n) occurs A000245(n) times.

For all n>0, a(A000108(n)) = A000108(n) [the first occurrence of the n-th Catalan number in this sequence].

Minimal i such that A081289(i) >= A081289(n) [the original definition of the sequence].

In other words, the first position k in A081289 where A081289(n) occurs (the minimal k such that A081289(k) = A081289(n)), and also the first position k in A081288 where A081288(n) occurs (the minimal k such that A081288(k) = A081288(n)). The starting point of the run which contains the n-th term in those sequences.