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 1-10 of 13 results. Next

A129593 Prime-factorization encoded partition codes for the Łukasiewicz-words in A071153.

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 9, 9, 5, 16, 27, 27, 6, 25, 27, 25, 6, 27, 25, 25, 25, 25, 7, 32, 81, 81, 18, 125, 81, 125, 18, 18, 15, 125, 15, 15, 49, 81, 125, 125, 15, 49, 18, 15, 18, 81, 125, 15, 125, 15, 49, 125, 49, 15, 125, 49, 15, 15, 125, 49, 49, 49, 49, 49, 11, 64, 243, 243, 54
Offset: 0

Views

Author

Antti Karttunen, May 01 2007

Keywords

Comments

If the signature-permutation of a Catalan automorphism SP satisfies the condition A129593(SP(n)) = A129593(n) for all n, then it is called a Łukasiewicz-word permuting automorphism. In addition to all the automorphisms whose signature permutation satisfies the more restricted condition A127301(SP(n)) = A127301(n) for all n, this includes also certain automorphisms like *A072797 that do not preserve the non-oriented form of the general tree. A000041(n) distinct values occur in each range [A014137(n-1)..A014138(n-1)]. All natural numbers occur. Cf. A129599.

Examples

			The terms A071153(5..7) are 201, 210 and 120. After discarding zero and sorting, each produces partition 1+2. Converting it to prime-exponents like explained in A129595, we get 2^0 * 3^2 = 9, thus a(5) = a(6) = a(7) = 9.
		

Crossrefs

a(n) = a(A072797(n)).
Variant: A129599. To be computed: the position of the first and the last occurrence of n, the number of occurrences of each n.

Formula

Construction: remove zeros from the Łukasiewicz-word of a general plane tree encoded by A014486(n) (i.e. A071153(n)), sort the numbers into ascending order and interpreting it as a partition of a natural number, encode it in the manner explained in A129595.

A063171 Dyck language interpreted as binary numbers in ascending order.

Original entry on oeis.org

0, 10, 1010, 1100, 101010, 101100, 110010, 110100, 111000, 10101010, 10101100, 10110010, 10110100, 10111000, 11001010, 11001100, 11010010, 11010100, 11011000, 11100010, 11100100, 11101000, 11110000, 1010101010, 1010101100, 1010110010, 1010110100, 1010111000
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 09 2001

Keywords

Comments

a(n) is the binary expansion of A014486(n). - Joerg Arndt, Feb 27 2013
Replacing "1" by "(" and "0" by ")" yields well-formed bracket expressions (the first term being the empty string)
, (), ()(), (()), ()()(), ()(()), (())(), (()()), ((())), ()()()(), ()()(()), ()(())(), ()(()()), ()((())), (())()(), (())(()), (()())(), (()()()), (()(())), ((()))(), ((())()), ((()())), (((()))), ()()()()(), ()()()(()), ()()(())(), ()()(()()), ()()((())), ()(())()(), ()(())(()), ()(()())(), ()(()()()), ()(()(())), ()((()))(), ()((())()), ()((()())), ()(((()))), (())()()(), (())()(()), (())(())(), (())(()()), (())((())), (()())()(), (()())(()), (()()())(), (()()()()), (()()(())), (()(()))(), (()(())()), (()(()())), (()((()))), ((()))()(), ((()))(()), ((())())(), ((())()()), ((())(())), ((()()))(), ((()())()), ((()()())), ((()(()))), (((())))(), (((()))()), (((())())), (((()()))), ((((()))))
The term a(0)=0 stands for the empty string. - Joerg Arndt, Feb 27 2013

Examples

			s -> ss -> 1s0s -> 11s00s -> 111000s -> 11100010
		

References

  • Donald E. Knuth, The Art of Computer Programming, Vol. 4A: Combinatorial Algorithms, Part 1, Addison-Wesley, 2011, Section 7.2.1.6, pp. 443 (Algorithm P).

Crossrefs

A014486 gives these terms as converted from decimal to binary system.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, union, fromList)
    newtype Word = Word String deriving (Eq, Show, Read)
    instance Ord Word where
       Word us <= Word vs | length us == length vs = us <= vs
                          | otherwise              = length us <= length vs
    a063171 n = a063171_list !! (n-1)
    a063171_list = dyck $ singleton (Word "S") where
       dyck s | null ws   = (read w :: Integer) : dyck s'
              | otherwise = dyck $ union s' (fromList $ concatMap gen ws)
              where ws = filter ((== 'S') . head . snd) $
                                map (`splitAt` w) [0..length w - 1]
                    (Word w, s') = deleteFindMin s
       gen (us,vs) = map (Word . (us ++) . (++ tail vs)) ["10", "1S0", "SS"]
    -- Reinhard Zumkeller, Mar 09 2011
    
  • Maple
    seq(convert(d, binary), d in select(isA014486, [seq(0..640)]));  # Peter Luschny, Mar 13 2024
  • Mathematica
    balancedQ[0] = True; balancedQ[n_] := (s = 0; Do[s += If[b == 1, 1, -1]; If[s < 0, Return[False]], {b, IntegerDigits[n, 2]}]; Return[s == 0]); FromDigits /@ IntegerDigits[ Select[Range[0, 684], balancedQ], 2] (* Jean-François Alcover, Jul 24 2013 *)
    (* Uses Algorithm P from Knuth's TAOCP section 7.2.1.6 - see References and Links. *)
    alist[n_] := Block[{a = Flatten[Table[{1, 0}, n]], m = 2*n - 1, j, k},
        FromDigits /@ Reap[
        While[True,
            Sow[a]; a[[m]] = 0;
            If[a[[m - 1]] == 0,
                a[[--m]] = 1, j = m - 1; k = 2*n - 1;
                While[j > 1 && a[[j]] == 1, a[[j--]] = 0; a[[k]] = 1; k -= 2];
                If[j == 1, Break[]];
                a[[j]] = 1; m = 2*n - 1]
        ]][[2, 1]]];
    Join[{{0}, {10}}, Array[alist, 4, 2]] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a_rows(N) = my(a=Vec([[0]], N)); for(r=1, N-1, my(b=a[r], c=List()); foreach(b, t, for(i=1, if(t, valuation(t, 10), 0)+1, listput(~c, t*100+10^i))); a[r+1]=Vec(c)); a; \\ Ruud H.G. van Tol, Jun 02 2024
  • Python
    def A063171_list(limit):
        return [0] + [int(bin(k)[2::]) for k in range(1, limit) if is_A014486(k)]
    print(A063171_list(700))  # Peter Luschny, Jul 30 2022
    
  • Python
    from itertools import count, islice
    from sympy.utilities.iterables import multiset_permutations
    def A063171_gen(): # generator of terms
        yield 0
        for l in count(1):
            for s in multiset_permutations('0'*l+'1'*(l-1)):
                c, m = 0, (l<<1)-1
                for i in range(m):
                    if s[i] == '1':
                        c += 2
                    if cA063171_list = list(islice(A063171_gen(),30)) # Chai Wah Wu, Nov 28 2023
    

Formula

Chomsky-2 grammar with axiom s, terminal alphabet {0, 1} and three rules s -> ss, s -> 1s0, s -> 10.
a(n) = A071152(n)/2.
a(n) = A007088(A014486(n)).

Extensions

a(0)=0 prepended by Joerg Arndt, Feb 27 2013

A057515 Number of separate "mountains" in mountain ranges encoded by A014486, number of bottom branches (trunks) in the corresponding rooted plane trees, i.e., the degree of the root node.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Sep 03 2000

Keywords

Comments

This sequence is produced when the function 'length' (present in programming languages like Lisp, Scheme, Prolog and Haskell) acts on symbolless S-expressions encoded by A014486/A063171.

Crossrefs

a(n) gives the first digit of A071153(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

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

A057514 Number of peaks in mountain ranges encoded by A014486, number of leaves in the corresponding rooted plane trees (the root node is never counted as a leaf).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Sep 03 2000

Keywords

Comments

Sum_{i=A014137(n)..(A014137(n+1)-1)} a(i) = A001700(n), i.e., A001700(n) gives the total number of leaves in all ordered trees with n + 1 edges.

Crossrefs

a(n)-1 gives the number of zeros in A071153(n) (for n>=1).

Programs

  • Python
    def a005811(n): return bin(n^(n>>1))[2:].count("1")
    def ok(n): # This function after Peter Luschny
        B=bin(n)[2:] if n!=0 else 0
        s=0
        for b in B:
            s+=1 if b=="1" else -1
            if s<0: return 0
        return s==0
    def A(n): return [0] + [i for i in range(1, n + 1) if ok(i)]
    l=A(200)
    print([a005811(l[i])//2 for i in range(len(l))]) # Indranil Ghosh, May 21 2017

Formula

a(n) = A005811(A014486(n))/2 = A000120(A003188(A014486(n)))/2.

A071159 Integers whose decimal expansion start with 1, do not contain zeros and each successive digit to the right is at most one greater than the previous digit.

Original entry on oeis.org

1, 11, 12, 111, 112, 121, 122, 123, 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1231, 1232, 1233, 1234, 11111, 11112, 11121, 11122, 11123, 11211, 11212, 11221, 11222, 11223, 11231, 11232, 11233, 11234, 12111, 12112, 12121
Offset: 1

Views

Author

Antti Karttunen, May 14 2002

Keywords

Crossrefs

Essentially the same as A071157 but with digits reversed.
Corresponding Łukasiewicz words: A071153.

Programs

  • Maple
    R[1]:= [1]:
    for d from 2 to 6 do
    R[d]:= map(t -> seq(10*t+j,j=1..min((t mod 10)+1,9)), R[d-1])
    od:
    A:= map(op, [seq(R[d],d=1..6)]); # Robert Israel, Jan 31 2017
  • Mathematica
    desQ[n_]:=Module[{idn=IntegerDigits[n]},idn[[1]]==1&&FreeQ[idn,0]&&Max[ Differences[ idn]]<2]; Select[Range[13000],desQ] (* Harvey P. Dale, Feb 19 2017 *)

A071155 The Catalan factorial walks (for each rooted plane tree encoded by A014486) encoded as zero-free numbers in factorial base (A007623).

Original entry on oeis.org

0, 1, 3, 5, 9, 15, 11, 17, 23, 33, 57, 39, 63, 87, 35, 59, 41, 65, 89, 47, 71, 95, 119, 153, 273, 177, 297, 417, 159, 279, 183, 303, 423, 207, 327, 447, 567, 155, 275, 179, 299, 419, 161, 281, 185, 305, 425, 209, 329, 449, 569, 167, 287, 191, 311, 431, 215, 335
Offset: 0

Views

Author

Antti Karttunen, May 14 2002

Keywords

Crossrefs

Same sequence sorted: A071156, expanded in the factorial number system: A071157. Corresponding Łukasiewicz words: A071153.
Cf. A000108 (row lengths), A120695.

A071154 Totally balanced decimal numbers: if we assign the weight w(d) = d-1 to each digit d (i.e., w(0) = -1, w(1) = 0, ..., w(9) = 8) and then read the digits of the term from left to right, the partial sum of the weights is never negative and the total weighted sum is zero.

Original entry on oeis.org

1, 11, 20, 111, 120, 201, 210, 300, 1111, 1120, 1201, 1210, 1300, 2011, 2020, 2101, 2110, 2200, 3001, 3010, 3100, 4000, 11111, 11120, 11201, 11210, 11300, 12011, 12020, 12101, 12110, 12200, 13001, 13010, 13100, 14000, 20111, 20120, 20201
Offset: 1

Views

Author

Antti Karttunen, May 14 2002

Keywords

Comments

The initial portion of this sequence (up to the 6917th term) is equal to A071153 (Łukasiewicz words for rooted plane trees) sorted in ascending order.

Crossrefs

Subset of A061384. Superset of A071161.
Cf. A014486 (totally balanced binary numbers), A071153.

Programs

  • PARI
    isok(n) = {my(s = 0); my(d = digits(n)); for (k=1, #d, s += d[k]-1; if (s<0, return (0));); if (s, 0, 1);} \\ Michel Marcus, Oct 16 2015

A079436 Full Łukasiewicz word for each rooted plane tree (interpretation e in Stanley's exercise 19) encoded by A014486 (or A063171).

Original entry on oeis.org

0, 10, 200, 110, 3000, 2010, 2100, 1200, 1110, 40000, 30010, 30100, 20200, 20110, 31000, 21010, 22000, 13000, 12010, 21100, 12100, 11200, 11110, 500000, 400010, 400100, 300200, 300110, 401000, 301010, 302000, 203000, 202010, 301100, 202100
Offset: 0

Views

Author

Antti Karttunen, Jan 09 2003

Keywords

Comments

Note: Here the last leaf is explicit, i.e. the terms are obtained from those of A071153 by multiplying them by 10.
Note: this finite decimal representation works only up to the 6917th term, as the 6918th such word is already "x0000000000" (where x stands for digit "ten").

Crossrefs

a(n) = 10*A071153(n).
For n > 1, the number of zeros in the term a(n) is given by A057514(n).
The first digit of each term is given by A057515.

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)).
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