A081290 -id:A081290 - cp's OEIS frontend

A244215 a(0)=0, and for n>=1, if n is the k-th Catalan number (A000108(k)), a(n) = k, otherwise the difference of the indices of the two largest Catalan numbers whose sum is less than or equal to n: a(n) = A244160(n) - A244160(n-A081290(n)).

0, 1, 2, 1, 0, 3, 2, 1, 1, 1, 0, 0, 0, 0, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6

Offset: 0

Antti Karttunen, Jun 23 2014

nonn

After a(0), if the leftmost digit in the Greedy Catalan Base representation of n [= A014418(n)] is larger than 1, then a(n) = 0, otherwise one more than the distance to the next nonzero digit to the right, or to the end of the numeral, if no more nonzero digits are present (i.e., if n is one of the Catalan numbers).

When searching for the two largest Catalan numbers whose sum is less than or equal to n, we first maximize the larger of those two numbers, which is A081290(n) = A000108(A244160(n)), after which we will find the next largest Catalan number that still "fits into" n. - Antti Karttunen, Mar 21 2015

Antti Karttunen, Table of n, a(n) for n = 0..4862

A244216 gives the positions of zeros, A244217 the positions of nonzero terms.

Cf. A000108, A014418, A081290, A244160, A163510.

(define (A244215 n) (- (A244160 n) (A244160 (- n (A081290 n)))))

a(n) = A244160(n) - A244160(n-A081290(n)).

For all n >= 1, a(A000108(n)) = n.

Name improved by Antti Karttunen, Mar 21 2015

A081291 Complement of A072795.

Original entry on oeis.org

0, 3, 6, 7, 8, 14, 15, 16, 17, 18, 19, 20, 21, 22, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128

Offset: 0

Antti Karttunen, Mar 17 2003

nonn

This gives positions of those terms in A063171 which begin with digits 11... Also the elements of table A072764 which do not occur in the leftmost column. See the comment at A081292.

Cf. A072795, A014137, A081288.

Cf. A080300, A081289, A081290, A081292.

A014137(n) = sum(k=0, n, binomial(2*k, k)/(k+1););
A081288(n) = my(i=0); while(binomial(2*i, i)/(i+1) <= n, i++); i;
a(n) = if (n==0, 0,  n + A014137(A081288(n)-1)); \\ Michel Marcus, Apr 28 2020

a(0)=0, a(n) = n + A014137(A081288(n)-1).

a(n) = A080300(A081292(n)) = A081289(n) + n - A081290(n).

A239903 List of Restricted-Growth Strings a_{k-1}a_{k-2}...a_{2}a_{1}, with k=2 and a_1 in {0,1} or k>2, a_{k-1}=1 and a_{j+1}>=1+a_j, for k-1>j>0.

Original entry on oeis.org

0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 120, 121, 122, 123, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1120, 1121, 1122, 1123, 1200, 1201, 1210, 1211, 1212, 1220, 1221, 1222, 1223, 1230, 1231, 1232, 1233, 1234, 10000, 10001, 10010, 10011

Offset: 0

N. J. A. Sloane, Apr 06 2014

We write the nonnegative integers as restricted growth strings (so called by J. Arndt in his book fxtbook.pdf, p. 325) in such a way that the Catalan numbers (cf. A000108) are expressed: 1=1, 10=2, 100=5, 1000=14, etc., 10...0 (with k zeros) = the k-th Catalan number. Once the entries of a restricted-growth string grow above 9, one would need commas or parentheses, say, to separate those entries. See Dejter (2017) for the precise definition.
In the paper "A system of numeration for middle-levels", restricted growth strings (RGSs) are defined as sequences that begin with either 0 or 1, with each successive number to the right being at least zero and at most one greater than its immediate left neighbor. Moreover, apart from case a(0), the RGSs are finite integer sequences of restricted growth which always start with 1 as their first element b_1 in position 1, and from then on, each successive element b_{i+1} in the sequence is restricted to be in range [0,(b_i)+1].
This sequence gives all such finite sequences in size-wise and lexicographic order, represented as decimal numbers by concatenating the integers of such finite sequences (e.g., from [1,2,0,1] we get 1201). The 58784th such sequence is [1, 2, 3, 4, 5, 6, 7, 8, 9, 9], thus a(58784) = 1234567899, after which comes the first RGS, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], where an element larger than 9 is present, which means that the decimal system employed here is unambiguous only up to n=58784. Note that 58785 = A000108(11)-1.
Also, if one considers Stanley's interpretation (u) of Catalan numbers, "sequences of a_1, a_2, ..., a_n of integers such that a_1 = 0 and 0 <= a_{i+1} <= a_{i} + 1" (e.g., 000, 001, 010, 011, 012 for C_3), and discards their initial zero, then one has a bijective correspondence with Dejter's RGSs of one element shorter length, which in turn are in bijective correspondence with the first C_n terms of this sequence (by discarding any leading zeros), from a(0) to a(C_n - 1). From this follows that the k-th Catalan number, A000108(k) (k>0), is represented in this system as 1 followed by k-1 zeros: a(1)=1, a(2)=10, a(5)=100, a(14)=1000, etc., and also that there exist exactly A000245(k) RGSs of length k.
Note how this differs from other number representations utilizing Catalan numbers, A014418 and A244159, in that while the latter are base-systems, where a simple weighted Sum_{k} digit(k)*C(k) recovers the natural number n (which the n-th numeral of such system represents), in contrast here it is the sum of appropriate terms in Catalan's Triangle (A009766, A030237), obtained by unranking a unique instance of a certain combinatorial structure (one of the Catalan interpretations), that gives a correspondence with a unique natural number. (Cf. also A014486.)
This sequence differs from "Semigreedy Catalan Representation", A244159, for the first time at n=10, where a(10) = 120, while A244159(10) = 121. That is also the first position where A244158(a(n)) <> n.
Please see Dejter's preprint for a more formal mathematical definition and how this number system is applied in relation to Havel's Conjecture on the existence of Hamiltonian cycles in the middle-levels graphs.
a(n) is given by the concatenation (with leading zeros removed) of the terms of row n + 23714 of A370222. - Paolo Xausa, Feb 17 2024

			Catalan's Triangle T(row,col) = A009766 begins with row n=0 and 0<=col<=n as:
  Row 0: 1
  Row 1: 1, 1
  Row 2: 1, 2,  2
  Row 3: 1, 3,  5,  5
  Row 4: 1, 4,  9, 14, 14
  Row 5: 1, 5, 14, 28, 42,  42
  Row 6: 1, 6, 20, 48, 90, 132, 132
  (the leftmost diagonal of 1s is "column 0").
  ...
For example, for n=38, we find that A081290(38)=14, which occurs on row A081288(n)-1 = 4, in columns A081288(n)-1 and A081288(n)-2, i.e., as T(4,4) and T(4,3). Thus we subtract 38-14 to get 24, and we see that the next term downward on the same diagonal, 28, is too large to accommodate into the same sum, so we go one diagonal up, starting now from T(3,2) = 5. This fits in, so we now have 24 - 5 = 19, and also the next term on the same diagonal, T(4,2) = 9, fits in, so we now have 19-9 = 10. The next term on the same diagonal, T(5,2) = 14, would not fit in anymore, so we rewind ourselves back to penultimate column, but one step up from where we started on this diagonal, so T(2,1) = 2, which fits in, 10 - 2 = 8, also the next one T(3,1) = 3, 8 - 3 = 5, and the next one T(4,1) = 4, 5 - 4 = 1, after which comes T(5,1) = 5 > 1, thus we jump to T(1,0) = 1, 1-1 = 0, and T(2,0)=1 would not fit anymore, thus next time the row would be zero, and the algorithm is ready with 1 (14), 2 (5+9), 3 (2+3+4) and 1 (1) terms collected, whose total sum 14+5+9+2+3+4+1 = 38, thus a(38) = 1231.
For n=20, the same algorithm results in 1 (14), 1 (5), 0 (not even the first tentative term T(2,1) = 2 from the column 1 would fit, so it is skipped), and from one row higher we get the needed 1 (1), so the total sum of these is 14+5+0+1 = 20, thus a(20) = 1101.

D. E. Knuth, The Art of Computer Programming, Vol. 2: Seminumerical Algorithms, third edition, Addison-Wesley, 1977, p. 192.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999, Exercise 19, interpretation (u).

Antti Karttunen, Table of n, a(n) for n = 0..16796
Joerg Arndt, Matters Computational: Ideas, Algorithm, Source Code, Springer, 2011; freely downloadable here: Fxtbook.
Georg Cantor, Über die einfachen Zahlensysteme, Zeitschrift fur Mathematik und Physik, 14 (1869), 121-128.
Italo J. Dejter, A system of numeration for middle levels, arXiv:1012.0995 [math.CO], (2014).
Italo J. Dejter, The role of restricted growth strings in the two middle levels of the Boolean lattice B_(2k+1), University of Puerto Rico, 2018.
Italo J. Dejter, Reinterpreting Mütze's theorem via natural enumeration of ordered rooted trees, arXiv:1911.02100 [math.CO], (2019).
Italo J. Dejter, A numeral system for the middle-levels graphs, Elec. J. Graph Theory and Applications (2021) Vol. 9, No. 1, 137-156. See p. 138.
Italo J. Dejter, Edge-supplementary arc factorizations of odd graphs, uniform 2-factors and Hamilton cycles, arXiv:2203.05326 [math.CO], 2022.
Italo J. Dejter, Integer sequence of Dyck-path single changes, arXiv:2209.11122 [math.CO], 2022.
Italo J. Dejter, A Dyck-word tree that controls all odd graphs, arXiv:2306.14249 [math.CO], 2023.
Aviezri S. Fraenkel, The Use and Usefulness of Numeration Systems, Inf. Comput., 81(1), (1989), 46-61.
Aviezri S. Fraenkel, Systems of numeration, IEEE Symposium on Computer Arithmetic, (1983), 37-42.
Aviezri S. Fraenkel, Systems of numeration, The American Mathematical Monthly, Vol. 92, No. 2 (Feb., 1985), pp. 105-114.
Antti Karttunen, Catalan ranking and unranking functions, OEIS Wiki.
R. P. Stanley, Exercises on Catalan and Related Numbers, interpretation (u) of Catalan numbers, on page 224 (page 4/27 in the November 1 2000 version of the PDF-file).

Cf. A000108 (Catalan numbers), A000245 (their first differences), A009766 (Catalan's triangle), A236855 (the sum of elements in k-th RGS), A236859 (for n>=1, gives the length of the initial ascent 123... in term a(n)), A244159 (different kinds of Catalan number systems).
Other Catalan combinatorial structures represented as integer sequences: A014486/A063171: Dyck words, parenthesizations, etc., A071156/A071158: Similar restricted words encoded with help of A007623 (Integers written in factorial base), A071153/A079436 (Łukasiewicz words).
Cf. A007001, A007623, A014418, A071154, A071159, A081288, A081290, A081291, A082853, A126307, A244155, A244156, A244157, A244158, A370222.

function CatalanNumerals(z)
    z == 0 && return 0
    f(n) = factorial(n)
    t(j, k) = div(f(k+j)*(k-j+1), f(j)*f(k+1))
    k, i = 2, 0
    while z >= t(i, i + 1) i += 1 end
    dig = fill(0, i); dig[1] = 1
    x = z - t(i - 1, i)
    m = i - 1
    while x > 0
        w, s, p = 0, 0, 0
        while w <= x
            p = w
            w += t(m - 1, m + s)
            s += 1
        end
        dig[k] = s - 1
        m -= 1; k += 1; x -= p
    end
    s = ""; for d in dig s *= string(d) end
    parse(Int, s)
end
[CatalanNumerals(n) for n in 0:42] |> println # Peter Luschny, Nov 10 2019

function [ c ] = catrep(z)
i=0; x=0; y=0; s=0;
while z>=(factorial(2*i+1)*(2))/(factorial(i)*factorial(i+2))
i=i+1;
end
y=(factorial(2*i-1)*(2))/(factorial(i-1)*factorial(i+1));
a=zeros(1,i); a(1,1)=1; k=2; x=z-y; m=1;
while x>0
w=0; s=0; p=0;
while w<=x
p=w;
w=w+(factorial(2*i-2*m+s-1)*(s+2))/(factorial(i-1-m)*factorial(i-m+s+1));
s=s+1;
end
m=m+1; a(1,k)=s-1; k=k+1; x=x-p;
end
a
end

A239903full = With[{r = 2*Range[2, 11]-1}, Reverse[Map[FromDigits[r-#] &, Rest[Select[Subsets[Range[2, 21], {10}, 125477], Min[r-#] >= 0 &]]]]];
A239903full[[;;100]] (* Paolo Xausa, Feb 17 2024 *)

define (t(j,k), (factorial(k+j)*(k-j+1))/(factorial(j)*factorial(k+1)));
i:0;
x:19;
z:0;y:0;s:0;
while x>=t(i,i+1) do (i:i+1);
y:t(i-1,i);a:zeromatrix(1,i);a[1,1]:1;k:2;z:x-y;m:1;
while (z>0) do (
w:0,s:0,p=0,
while (w<=z) do (
p:w,
w:w+t(i-1-m,i-m+s),
s:s+1
),
m:m+1,
a[1,k]:s-1,k:k+1,
z:z-p
);
print(a);

\\ Valid for n<58786 (=A000108(11)).
nxt(w)=if(w[1]==#w, vector(#w+1, i, i>#w), my(k=1); while(w[k]>w[k+1], w[k]=0; k++); w[k]++; w)
seq(n)={my(a=vector(n), w=[1]); a[1]=0; for(i=2, #v, a[i]=fromdigits(Vecrev(w)); w=nxt(w)); a} \\ Andrew Howroyd, Jan 24 2023

(define (A239903_only_upto_16794 n) (if (zero? n) n (A235049 (A071159 (A081291 n))))) ;; Gives correct results only up to 16794.
;; The following gives correct results all the way up to n=58784.
(define (A239903 n) (baselist-as-decimal (A239903raw n)))
(definec (A239903raw n) (if (zero? n) (list) (let loop ((n n) (row (A244160 n)) (col (- (A244160 n) 1)) (srow (- (A244160 n) 1)) (catstring (list 0))) (cond ((or (zero? row) (negative? col)) (reverse! (cdr catstring))) ((> (A009766tr row col) n) (loop n srow (- col 1) (- srow 1) (cons 0 catstring))) (else (loop (- n (A009766tr row col)) (+ row 1) col srow (cons (+ 1 (car catstring)) (cdr catstring))))))))
(define (baselist-as-decimal lista) (baselist->n 10 lista))
(define (baselist->n base bex) (let loop ((bex bex) (n 0)) (cond ((null? bex) n) (else (loop (cdr bex) (+ (* n base) (car bex)))))))
;; From Antti Karttunen, Apr 14-19 2014

To find an RGS corresponding to natural number n, one first finds a maximum row index k such that T(k,k-1) <= n in the Catalan Triangle (A009766) illustrated in the Example section. Note that as the last two columns of this triangle consist of Catalan numbers (that is, T(k,k-1) = T(k,k) = A000108(k)), it means that the first number to be subtracted from n is A081290(n) which occurs as a penultimate element of the row A081288(n)-1, in the column A081288(n)-2. The unranking algorithm then proceeds diagonally downwards, keeping the column index the same, and incrementing the row index, as long as it will encounter terms such that their total sum stays less than or equal to n.
If the total sum of encountered terms on that diagonal would exceed n, the algorithm jumps back to the penultimate column of the triangle, but one row higher from where it started the last time, and again starts summing the terms as long as the total sum stays <= n.
When the algorithm eventually reaches either row zero or column less than zero, the result will be a list of numbers, each element being the number of terms summed from each diagonal, so that the diagonal first traversed appears as the first 1 (as that first diagonal will never allow more than one term), and the number of terms summed from the last traversed diagonal appears the last number in the list. These lists of numbers are then concatenated together as decimal numbers.
These steps can also be played backwards in order to recover the corresponding decimal integer n from such a list of numbers, giving a "ranking function" which will be the inverse to this "unranking function".
For n=1..16794 (where 16794 = A000108(10)-2), a(n) = A235049(A071159(A081291(n))). - Antti Karttunen, Apr 14 2014
Alternative, simpler description of the algorithm from Antti Karttunen, Apr 21 2014: (Start)
Consider the following square array, which is Catalan triangle A009766 without its rightmost, "duplicate" column, appropriately transposed (cf. also tables A030237, A033184 and A054445):
   Row| Terms on that row
   ---+--------------------------
    1 |   1   1   1    1    1 ...
    2 |   2   3   4    5    6 ...
    3 |   5   9  14   20   27 ...
    4 |  14  28  48   75  110 ...
    5 |  42  90 165  275  429 ...
    6 | 132 297 572 1001 1638 ...
To compute the n-th RGS, search first for the greatest Catalan number C_k which is <= n (this is A081290(n), found as the first term of row A081288(n)-1). Then, by a greedy algorithm, select from each successive row (moving towards the top of table) as many terms from the beginning of that row as will still fit into n, subtracting them from n as you go. The number of terms selected from the beginning of each row gives each element of the n-th RGS, so that the number of terms selected from the topmost row (all 1's) appears as its last element.
(End)

Description, formula and examples edited/rewritten by Italo J Dejter, Apr 13 2014 and Antti Karttunen, Apr 18 2014

A244159 Semigreedy Catalan Representation of nonnegative integers.

Original entry on oeis.org

0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 121, 122, 123, 211, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1232, 1233, 1234, 1322, 2111, 2112, 2121, 2122, 2123, 2211, 10000

Offset: 0

Antti Karttunen, Jun 23 2014

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.

			For n = 18, the largest Catalan number <= 18 is C(4) = 14.
Thus we initialize a vector of four zeros [0, 0, 0, 0] and increment the first element to 1: [1, 0, 0, 0] and subtract 14 from 18 to get the remainder 4.
We see that the next smaller Catalan number, C(3) = 5 is greater than 4, so we cannot subtract it without going negative, so the second leftmost element of the vector stays as zero.
We next check C(2) = 2, which is less than 4, thus we increment the zero at that point to 1, and subtract 4 - 2 to get 2.
We compare 2 to C(1) = 1, and as 1 <= 2, it is subtracted 2-1 = 1, and the corresponding element in the vector incremented, thus after the first round, the vector is now [1, 0, 1, 1], and n remaining is 1.
So we start the second round because n has not yet reached the zero, and look for the largest Catalan number <= 1, which in this case is C(1) = 1, so we subtract it from remaining n, and increment the element in the position 1, after which n has reached zero, and the vector is now [1, 0, 1, 2], whose concatenation as decimal numbers thus yields a(18) = 1012.

Antti Karttunen, Table of n, a(n) for n = 0..33603

Cf. A014418 (a classical greedy variant), A244231 (maximum "digit value"), A244232 (sum of digits), A244233 (product of digits), A244314 (positive terms which have at least one zero digit), A244316 (the one-based position of digit incremented last in the described process).
Cf. also A007088, A014138, A176137, A197433, A244230, A244158, A244160.
Differs from A239903 for the first time at n=10, where a(10) = 121, while A239903(10) = 120.

If A176137(n) = 1, a(n) = A007088(A244230(n)), otherwise a(n) = A007088(A244230(n)-1) + a(n-A197433(A244230(n)-1)).
For all n, a(A197433(n)) = A007088(n).
For all n >= 1, a(A000108(n)) = 10^(n-1).
Each a(A014143(n)) has a "triangular" representation [1, 2, 3, ..., n, n+1].

A014420 Minimal number of Catalan numbers that sum to n.

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3, 3, 4, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 3, 4, 4, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 5, 6, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 3, 4, 4, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 5, 6, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 5, 6, 6, 7, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 5, 6, 3

Offset: 0

Olivier Gérard

Olivier Gérard (first 1001 terms) & Antti Karttunen, Table of n, a(n) for n = 0..16796

Cf. A014418, A081290, A244320, A244318, A244232, A236855, A236859.

a(0) = 0, a(n) = 1 + a(n-A081290(n)). - Antti Karttunen, Jul 02 2014

A244160 a(0)=0, and for n >= 1, a(n) = the largest k such that k-th Catalan number <= n.

Original entry on oeis.org

0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6

Offset: 0

Antti Karttunen, Jun 23 2014

nonn

Apart from 0, each n occurs A000245(n) times.

For n >= 1, a(n) gives the largest k such that C(k) <= n, where C(k) stands for the k-th Catalan number, A000108(k).

			For n=1, the largest k such that C(k) <= 1 is 1, thus a(1) = 1.
For n=2, the largest k such that C(k) <= 2 is 2, thus a(2) = 2.
For n=3, the largest k such that C(k) <= 3 is 2, thus a(3) = 2.
For n=4, the largest k such that C(k) <= 4 is 2, thus a(4) = 2.
For n=5, the largest k such that C(k) <= 5 is 3, thus a(5) = 3.

Michael De Vlieger, Table of n, a(n) for n = 0..4861

After zero, one less than A081288.

Cf. A000108, A000245, A081290, A014418, A239903, A244215, A244159, A236859, A126307.

MapIndexed[ConstantArray[First@ #2 - 1, #1] &, Differences@ Array[CatalanNumber, 8, 0]] /. {} -> {0} // Flatten (* Michael De Vlieger, Jun 08 2017 *)
Join[{0},Table[PadRight[{},CatalanNumber[n+1]-CatalanNumber[n],n],{n,6}]// Flatten] (* Harvey P. Dale, Aug 23 2021 *)

from sympy import catalan
def a(n):
    if n==0: return 0
    i=1
    while True:
        if catalan(i)>n: break
        else: i+=1
    return i - 1
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 08 2017

(define (A244160 n) (if (zero? n) n (- (A081288 n) 1)))

a(0) = 0, and for n>=1, a(n) = A081288(n)-1.

For all n>=1, A000108(a(n)) = A081290(n).

A081288 a(n) is the minimal i such that A000108(i) > n.

Original entry on oeis.org

0, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

Antti Karttunen, Mar 17 2003

nonn

Apart from the initial term 0, each n occurs A000245(n-1) times.

Cf. A000108, A000245, A072643, A081289, A081290. Used to compute A081291.

A081288(n) = my(i=0); while(binomial(2*i, i)/(i+1) <= n, i++); i; \\ Michel Marcus, Apr 28 2020

from sympy import catalan
def a(n):
    if n==0: return 0
    i=1
    while True:
        if catalan(i)>n: return i
        else: i+=1
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 08 2017

A081289 a(0) = 0, a(n) = A081293(A081288(n)-1).

Original entry on oeis.org

0, 3, 6, 6, 6, 14, 14, 14, 14, 14, 14, 14, 14, 14, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107

Offset: 0

Antti Karttunen, Mar 17 2003

nonn

After n>0, A081293(n) occurs A000245(n) times.

Cf. A000108, A000245, A081288, A081290, A081293.

A000108(n) = binomial(2*n, n)/(n+1);
A081293(n) = A000108(n) + sum(k=0, n, A000108(k));
A081288(n) = my(i=0); while(binomial(2*i, i)/(i+1) <= n, i++); i;
a(n) = if (n==0, 0, A081293(A081288(n)-1)); \\ Michel Marcus, Apr 28 2020

A244217 Numbers n less than twice the largest Catalan number that is less than or equal to n.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 9, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 132, 133, 134

Offset: 1

Antti Karttunen, Jun 23 2014

nonn

Equally: Numbers k such that if m is the largest Catalan number <= k [= A081290(k)], then k < 2*m.

Numbers k such that the Greedy Catalan Base representation of k (A014418(k)) starts with digit 1.

			41 is not a member, because the largest Catalan number less than or equal to 41 is C(4) = 14, and 41 is not less than 2*14 = 28.
42 is a member, because the largest Catalan number less than or equal to 42 is C(5) = 42 itself, and 42 certainly is less than 2*42 = 84.
See also the examples given for the complement of this sequence: A244216.

Antti Karttunen, Table of n, a(n) for n = 1..6918

Positions of nonzeros in A244215.
Complement of A244216.
Cf. A000108 (a subsequence), A014418, A081290, A244314.

A244316 a(0) = 0, after which, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

Original entry on oeis.org

0, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 6

Offset: 0

Antti Karttunen, Jun 25 2014

nonn

For n >= 1, a(n) tells the one-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.
Algorithm for constructing the sequence: Find the largest Catalan number which is less than or equal to n (this is A081290(n) = A000108(k), where k = A244160(n), that is, the corresponding index of that Catalan number), and subtract that from n. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the remaining n, and if it does not, then subtract that also from n, and keep on doing the same for lesser and lesser Catalan numbers, comparing and also subtracting them (whenever it is possible without going less than zero) from n, until either n becomes zero, or after subtracting C(1) = 1 from n, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to remaining n, and start the process again. However, when at some point n finally reaches zero, then the index k of the last Catalan number, A000108(k) which was subtracted from n before it reached zero, is our result, a(n) = k. [Here n = the original value of n, from which we started subtracting initially from].
If n is one of the terms of A197433, meaning that if it can be represented as a sum of distinct Catalan numbers as n = C(i) + C(j) + ... + C(k) (which representation then necessarily is unique), then a(n) = min(i,j,...,k).

Antti Karttunen, Table of n, a(n) for n = 0..16796

One more than A244315.

Cf. A000108, A001511, A176137, A197433, A244230, A244159, A244160, A081290.

a(0) = 0, and for n >= 1, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).
For n >= 1, a(n) = A244315(n)+1.
For n >= 1, a(A000108(n)) = n and a(A014138(n)) = a(A014143(n)) = 1.

cp's OEIS Frontend

A244215 a(0)=0, and for n>=1, if n is the k-th Catalan number (A000108(k)), a(n) = k, otherwise the difference of the indices of the two largest Catalan numbers whose sum is less than or equal to n: a(n) = A244160(n) - A244160(n-A081290(n)).

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A081291 Complement of A072795.

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A239903 List of Restricted-Growth Strings a_{k-1}a_{k-2}...a_{2}a_{1}, with k=2 and a_1 in {0,1} or k>2, a_{k-1}=1 and a_{j+1}>=1+a_j, for k-1>j>0.

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Julia

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Maxima

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A244159 Semigreedy Catalan Representation of nonnegative integers.

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A014420 Minimal number of Catalan numbers that sum to n.

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A244160 a(0)=0, and for n >= 1, a(n) = the largest k such that k-th Catalan number <= n.

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A081288 a(n) is the minimal i such that A000108(i) > n.

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A081289 a(0) = 0, a(n) = A081293(A081288(n)-1).

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