cp's OEIS Frontend

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)

The first value larger than nine occurs at a(33604) = 10. (33604 = A014143(9)). Note that this sequence is not subject to any corruption by decimal representation as A244159 itself is.

A014143 gives records up to that A014143(9) = 33604, but thereafter, the next record occurs at 57317, for which a(57317) = 11, although A014143(10) = 116103. This is explained by that A244159raw(57317) = [2,3,4,5,6,7,8,9,10,11] and A244159raw(116103) = [1,2,3,4,5,6,7,8,9,10,11] (where A244159raw means the underlying representation, before it is maimed by the decimal representation).

This change is explained by the fact that A014143(n-1) + A014138(n) > A000108(n+1) for n = 1..9, but for n >= 10, A014143(n-1) + A014138(n) < A000108(n+1).

For example, although 16808 = 2*C(9) + 3*C(8) + 4*C(7) + 5*C(6) + 6*C(5) + 7*C(4) + 8*C(3) + 9*C(2) + 10*C(1), its representation in A244159 system is 1000000123, as 16808 = 1*C(10) + 1*C(3) + 2*C(2) + 3*C(1).

Note that a(33604) = 10! = 3628800 because the product is computed from the underlying list (vector) of numbers, and thus is not subject to any corruption by decimal representation as A244159 itself is.