cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A126301 A071158-codes for the fixed points of Vaillé's 1997 bijection on Dyck paths.

Original entry on oeis.org

0, 1, 23211, 2432211, 2354543212221, 335465432122211
Offset: 0

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Author

Antti Karttunen, Jan 02 2007

Keywords

Comments

Vaille gives the terms a(2)-a(4) on the last page of the 1997 paper. Note that this sequence migh be finite, for two reasons: (a) there are no more fixed points after some limit (the next one after a(5) must have at least 19 digits. All the terms must be of odd length). (b) some of the fixed points would require digits higher than "9", in which case the factorial expansion can nomore presented unambiguously in decimal. However, the sequence A126311 can accommodate also those cases.

Examples

			This sequence consists of those terms of A071158 for which the first factorial digit is equal to the number of 1's in the term and the following algorithm results the remaining factorial digits of the same term: First, extract all maximal subsequences from the term (for d ranging from 1 to the largest digit present) that consist of digits d and d+1 and place them next to each other, from left to right. E.g. for the term 2354543212221 this yields the sequence: 2212221,2332222,3443,5454,55. After discarding the last digit (here 5) and replacing in each batch the smaller number with +1 and larger number with -1, we get:
-1,-1,+1,-1,-1,-1,+1,+1,-1,-1,+1,+1,+1,+1,+1,-1,-1,+1,-1,+1,-1,+1,+1.
and summing these from RIGHT, we get the following partial sums:
1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1.
Retaining only the partial sums under the +1's (i.e. the rightmost one and all the partial sums that are larger than the preceding partial sum one step to the right) we obtain: 3,5,4,5,4,3,2,1,2,2,2 and 1. These, after appended to the number of 1's in the original term (2), yields the same term 2354543212221 from which we started from, which thus is a member of this sequence. Similarly, the term 2432211 belongs to this sequence, because the same procedure yields:
22211,2322,43,4 and after discarding the last 4:
-1,-1,-1,+1,+1,+1,-1,+1,+1,-1,+1 and summing from the right:
1, 2, 3, 4, 3, 2, 1, 2, 1, 0, 1.
collecting all the partial sums larger than their right neighbor (those under +1's), which appended after the number of 1's (2), results the same term 2432211.
		

Crossrefs

a(n) = A071158(A126300(n)) = A007623(A126311(n)). Subset of A126299. Cf. A126295.

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

Views

Author

N. J. A. Sloane, Apr 06 2014

Keywords

Comments

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

Examples

			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.
		

References

  • 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).

Crossrefs

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).

Programs

  • Julia
    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
    
  • MATLAB
    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
    
  • Mathematica
    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 *)
  • Maxima
    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);
    
  • PARI
    \\ 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
  • Scheme
    (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
    

Formula

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)

Extensions

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

A071156 Apart from the initial term (0), lists all integers whose factorial expansion ends with 1 (i.e., are odd numbers), do not contain a digit zero and each successive digit to the left is at most one greater than the preceding digit.

Original entry on oeis.org

0, 1, 3, 5, 9, 11, 15, 17, 23, 33, 35, 39, 41, 47, 57, 59, 63, 65, 71, 87, 89, 95, 119, 153, 155, 159, 161, 167, 177, 179, 183, 185, 191, 207, 209, 215, 239, 273, 275, 279, 281, 287, 297, 299, 303, 305, 311, 327, 329, 335, 359, 417, 419, 423, 425, 431, 447, 449
Offset: 0

Views

Author

Antti Karttunen, May 14 2002

Keywords

Comments

a(n) = A085198(A014486(n)) = A071155(A057164(n)). Catalan numbers A000108(n) gives the number of terms whose factorial expansion contain n digits.
0 is included by considering it to have the empty string as its factorial base representation. - Franklin T. Adams-Watters, Jun 28 2006

Crossrefs

The beginning of this sequence expanded in the factorial number system: A071158. Inverse function: A085199. First differences: A085191.
Cf. A000108 (row lengths), A071155, A120696.

A125986 Signature-permutation of the inverse of Vaillé's 1997 bijection on Dyck paths.

Original entry on oeis.org

0, 1, 3, 2, 8, 6, 7, 5, 4, 22, 19, 20, 15, 14, 21, 17, 18, 13, 11, 16, 12, 10, 9, 64, 60, 61, 52, 51, 62, 54, 55, 41, 39, 53, 40, 38, 37, 63, 57, 58, 46, 44, 59, 49, 50, 36, 33, 47, 34, 29, 28, 56, 45, 48, 35, 31, 43, 32, 27, 25, 42, 30, 26, 24, 23, 196, 191, 192, 178, 177
Offset: 0

Views

Author

Antti Karttunen, Jan 02 2007

Keywords

Crossrefs

Inverse: A125985. Cf. A057515, A071158. Algorithm is partially described in A126301.

A071157 The zero-free, right-to-left factorial walk encoding for each rooted plane tree encoded by A014486. Sequence A071155 shown with factorial expansion (A007623).

Original entry on oeis.org

0, 1, 11, 21, 111, 211, 121, 221, 321, 1111, 2111, 1211, 2211, 3211, 1121, 2121, 1221, 2221, 3221, 1321, 2321, 3321, 4321, 11111, 21111, 12111, 22111, 32111, 11211, 21211, 12211, 22211, 32211, 13211, 23211, 33211, 43211, 11121, 21121, 12121
Offset: 0

Views

Author

Antti Karttunen, May 14 2002

Keywords

Comments

Apart from the initial term (0, which encodes the null tree), if we scan the digits from the right (the least significant digit which is always 1) to the left (the most significant), then each successive digit to the left is at most one greater than the previous and never less than one.
Note: this finite decimal representation works only up to the 23712nd term, as the 23713rd such walk is already (10,9,8,7,6,5,4,3,2,1). The sequence A071158 shows the initial portion of this sequence sorted.

Crossrefs

Corresponding Łukasiewicz words: A071153.
Essentially the same as A071159 but with digits reversed.

Formula

a(n) = A007623(A071155(n)).

A126299 Factorial codes for the fixed points of the square of Vaillé's 1997 bijection on Dyck paths.

Original entry on oeis.org

0, 1, 11, 21, 111, 321, 1111, 4321, 11111, 23211, 54321, 111111, 122211, 324321, 654321, 1111111, 1222121, 2432211, 3244321, 7654321, 11111111, 22121321, 34443211, 87654321, 111111111, 122322121, 325543321, 987654321, 1111111111
Offset: 0

Views

Author

Antti Karttunen, Jan 02 2007

Keywords

Comments

From n=31 the terms cannot anymore be presented unambiguously with decimal numbers, as A126298(31)=23713, A071156(23713)=39916799 and A007623(39916799) would result the factorial expansion "T987654321", where T stands for digit "ten".

Crossrefs

Superset of A126301. Number of terms of length n is given by A126295(n).

Formula

a(n) = A071158(A126298(n)).
Showing 1-6 of 6 results.