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 16 results. Next

A144218 Equals product A*B, where A is an infinite lower triangular matrix with A086246 in every column and B is the diagonal matrix with A001006 as diagonal.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 4, 4, 2, 2, 4, 9, 9, 4, 4, 4, 9, 21, 21, 9, 8, 8, 9, 21, 51, 51, 21, 18, 16, 18, 21, 51, 127, 127, 51, 42, 36, 36, 42, 51, 127, 323, 323, 127, 102, 84, 81, 84, 102, 127, 323, 835, 835, 323, 254, 204, 189, 189, 204, 254, 323, 835, 2188
Offset: 0

Views

Author

Gary W. Adamson, Sep 14 2008

Keywords

Comments

Right border is A001006.
Row sums give A001006 without the initial 1.
Left border is A086246 (A001006 with an additional leading 1).
Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle:
    1;
    1,   1;
    1,   1,   2;
    2,   1,   2,   4;
    4,   2,   2,   4,   9;
    9,   4,   4,   4,   9,  21;
   21,   9,   8,   8,   9,  21,  51;
   51,  21,  18,  16,  18,  21,  51, 127;
  127,  51,  42,  36,  36,  42,  51, 127, 323;
  323, 127, 102,  84,  81,  84, 102, 127, 323, 835;
  835, 323, 254, 204, 189, 189, 204, 254, 323, 835, 2188;
  ...
Row 4 = (4, 2, 2, 4, 9) = termwise products of (4, 2, 1, 1, 1) and (1, 1, 2, 4, 9) = (4*1, 2*1, 1*2, 1*4, 1*9).
		

Crossrefs

Programs

  • Mathematica
    nmax = 10;
    T[0, 0] = T[1, 0] = 1;
    T[n_, 0]  := Hypergeometric2F1[3/2, 1-n, 3, 4] // Abs;
    T[n_, n_] := Hypergeometric2F1[(1-n)/2, -n/2, 2, 4];
    row[n_] := row[n] = Table[T[m, 0], {m, n, 0, -1}]*Table[T[m, m], {m, 0, n} ];
    T[n_, k_] /; 0Jean-François Alcover, Aug 07 2018 *)

Extensions

Edited by Joerg Arndt, Jan 26 2024

A348197 Composition of the g.f. of A086246 with itself.

Original entry on oeis.org

0, 1, 2, 4, 10, 28, 84, 264, 860, 2880, 9862, 34392, 121770, 436688, 1583146, 5793216, 21370806, 79391536, 296760222, 1115327844, 4212125662, 15976390684, 60833679424, 232452408632, 891060970152, 3425639505624, 13204738280326, 51024408662932, 197607503526934
Offset: 0

Views

Author

Alexander Burstein, Oct 06 2021

Keywords

Comments

G.f.: A(x) is the pseudo-involutory Riordan companion of 2*M(x)-1, where M(x) is the g.f. of A001006.
For 1 <= n <= 7, a(n) coincides with A068875(n-1).
Conjecture: a(n) > A068875(n-1) for n > 7 (equivalently, a(n) > 2*A000108(n-1) for n > 7).

Crossrefs

Programs

  • Maple
    gf:= (f-> f(f(x)))(x->(1+x-sqrt(1-2*x-3*x^2))/2):
    a:= n-> coeff(series(gf,x,n+1),x,n):
    seq(a(n), n=0..40);  # Alois P. Heinz, Oct 06 2021
  • Mathematica
    f[x_] := (1 + x - Sqrt[1 - 2*x - 3*x^2])/2; a[n_] := SeriesCoefficient[f[f[x]], {x, 0, n}]; Array[a, 25, 0] (* Amiram Eldar, Oct 06 2021 *)
  • PARI
    f(x) = (1+x-sqrt(1-2*x-3*x^2))/2;
    my(x='x+O('x^30)); concat(0, Vec(f(f(x)))) \\ Michel Marcus, Oct 06 2021

Formula

G.f.: A(x) = F(F(x)), where F(x) is the g.f. of A086246.
Let G(x) = 2*M(x) - 1, where M(x) is the g.f. of A001006 (equivalently, x*G(x) is the g.f. of A007971). Then G(-A(x)) = 1/G(x).
A(-A(x)) = -x.
a(n) ~ ((1 + sqrt(3))^(n - 1/2) * 3^(n - 1/2)) / (sqrt(Pi) * n^(3/2) * 2^n). - Vaclav Kotesovec, Oct 07 2021

A001006 Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n (labeled) points on a circle.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211
Offset: 0

Views

Author

Keywords

Comments

Number of 4321-, (3412,2413)-, (3412,3142)- and 3412-avoiding involutions in S_n.
Number of sequences of length n-1 consisting of positive integers such that the first and last elements are 1 or 2 and the absolute difference between any 2 consecutive elements is 0 or 1. - Jon Perry, Sep 04 2003
From David Callan, Jul 15 2004: (Start)
Also number of Motzkin n-paths: paths from (0,0) to (n,0) in an n X n grid using only steps U = (1,1), F = (1,0) and D = (1,-1).
Number of Dyck n-paths with no UUU. (Given such a Dyck n-path, change each UUD to U, then change each remaining UD to F. This is a bijection to Motzkin n-paths. Example with n=5: U U D U D U U D D D -> U F U D D.)
Number of Dyck (n+1)-paths with no UDU. (Given such a Dyck (n+1)-path, mark each U that is followed by a D and each D that is not followed by a U. Then change each unmarked U whose matching D is marked to an F. Lastly, delete all the marked steps. This is a bijection to Motzkin n-paths. Example with n=6 and marked steps in small type: U U u d D U U u d d d D u d -> U U u d D F F u d d d D u d -> U U D F F D.) (End)
a(n) is the number of strings of length 2n+2 from the following recursively defined set: L contains the empty string and, for any strings a and b in L, we also find (ab) in L. The first few elements of L are e, (), (()), ((())), (()()), (((()))), ((()())), ((())()), (()(())) and so on. This proves that a(n) is less than or equal to C(n), the n-th Catalan number. See Orrick link (2024). - Saul Schleimer (saulsch(AT)math.rutgers.edu), Feb 23 2006 (Additional linked comment added by William P. Orrick, Jun 13 2024.)
a(n) = number of Dyck n-paths all of whose valleys have even x-coordinate (when path starts at origin). For example, T(4,2)=3 counts UDUDUUDD, UDUUDDUD, UUDDUDUD. Given such a path, split it into n subpaths of length 2 and transform UU->U, DD->D, UD->F (there will be no DUs for that would entail a valley with odd x-coordinate). This is a bijection to Motzkin n-paths. - David Callan, Jun 07 2006
Also the number of standard Young tableaux of height <= 3. - Mike Zabrocki, Mar 24 2007
a(n) is the number of RNA shapes of size 2n+2. RNA Shapes are essentially Dyck words without "directly nested" motifs of the form A[[B]]C, for A, B and C Dyck words. The first RNA Shapes are []; [][]; [][][], [[][]]; [][][][], [][[][]], [[][][]], [[][]][]; ... - Yann Ponty (ponty(AT)lri.fr), May 30 2007
The sequence is self-generated from top row A going to the left starting (1,1) and bottom row = B, the same sequence but starting (0,1) and going to the right. Take dot product of A and B and add the result to n-th term of A to get the (n+1)-th term of A. Example: a(5) = 21 as follows: Take dot product of A = (9, 4, 2, 1, 1) and (0, 1, 1, 2, 4) = (0, + 4 + 2 + 2 + 4) = 12; which is added to 9 = 21. - Gary W. Adamson, Oct 27 2008
Equals A005773 / A005773 shifted (i.e., (1,2,5,13,35,96,...) / (1,1,2,5,13,35,96,...)). - Gary W. Adamson, Dec 21 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
a(n) is the number of involutions of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(4)=9; indeed, p=3412=(13)(24) is the only involution of {1,2,3,4} with genus > 0. This follows easily from the fact that a permutation p of {1,2,...,n} has genus 0 if and only if the cycle decomposition of p gives a noncrossing partition of {1,2,...,n} and each cycle of p is increasing (see Lemma 2.1 of the Dulucq-Simion reference). [Also, redundantly, for p=3412=(13)(24) we have cp'=2341*3412=4123=(1432) and so g(p)=(1/2)(4+1-2-1)=1.] - Emeric Deutsch, May 29 2010
Let w(i,j,n) denote walks in N^2 which satisfy the multivariate recurrence w(i,j,n) = w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Then a(n) = Sum_{i = 0..n, j = 0..n} w(i,j,n) is the number of such walks of length n. - Peter Luschny, May 21 2011
a(n)/a(n-1) tends to 3.0 as N->infinity: (1+2*cos(2*Pi/N)) relating to longest odd N regular polygon diagonals, by way of example, N=7: Using the tridiagonal generator [cf. comment of Jan 07 2009], for polygon N=7, we extract an (N-1)/2 = 3 X 3 matrix, [0,1,0; 1,1,1; 0,1,1] with an e-val of 2.24697...; the longest Heptagon diagonal with edge = 1. As N tends to infinity, the diagonal lengths tend to 3.0, the convergent of the sequence. - Gary W. Adamson, Jun 08 2011
Number of (n+1)-length permutations avoiding the pattern 132 and the dotted pattern 23\dot{1}. - Jean-Luc Baril, Mar 07 2012
Number of n-length words w over alphabet {a,b,c} such that for every prefix z of w we have #(z,a) >= #(z,b) >= #(z,c), where #(z,x) counts the letters x in word z. The a(4) = 9 words are: aaaa, aaab, aaba, abaa, aabb, abab, aabc, abac, abca. - Alois P. Heinz, May 26 2012
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=1, r(k)<=k, and r(k)!=r(k-1); for example, the 9 RGS for n=4 are 1010, 1012, 1201, 1210, 1212, 1230, 1231, 1232, 1234. - Joerg Arndt, Apr 16 2013
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=0, r(k)<=k and r(k)-r(k-1) != 1; for example, the 9 RGS for n=4 are 0000, 0002, 0003, 0004, 0022, 0024, 0033, 0222, 0224. - Joerg Arndt, Apr 17 2013
Number of (4231,5276143)-avoiding involutions in S_n. - Alexander Burstein, Mar 05 2014
a(n) is the number of increasing unary-binary trees with n nodes that have an associated permutation that avoids 132. For more information about unary-binary trees with associated permutations, see A245888. - Manda Riehl, Aug 07 2014
a(n) is the number of involutions on [n] avoiding the single pattern p, where p is any one of the 8 (classical) patterns 1234, 1243, 1432, 2134, 2143, 3214, 3412, 4321. Also, number of (3412,2413)-, (3412,3142)-, (3412,2413,3142)-avoiding involutions on [n] because each of these 3 sets actually coincides with the 3412-avoiding involutions on [n]. This is a complete list of the 8 singles, 2 pairs, and 1 triple of 4-letter classical patterns whose involution avoiders are counted by the Motzkin numbers. (See Barnabei et al. 2011 reference.) - David Callan, Aug 27 2014
From Tony Foster III, Jul 28 2016: (Start)
A series created using 2*a(n) + a(n+1) has Hankel transform of F(2n), offset 3, F being the Fibonacci bisection, A001906 (empirical observation).
A series created using 2*a(n) + 3*a(n+1) + a(n+2) gives the Hankel transform of Sum_{k=0..n} k*Fibonacci(2*k), offset 3, A197649 (empirical observation). (End)
Conjecture: (2/n)*Sum_{k=1..n} (2k+1)*a(k)^2 is an integer for each positive integer n. - Zhi-Wei Sun, Nov 16 2017
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n." - Eric M. Schmidt, Dec 16 2017
Number of U_{k}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018
If tau_1 and tau_2 are two distinct permutation patterns chosen from the set {132,231,312}, then a(n) is the number of valid hook configurations of permutations of [n+1] that avoid the patterns tau_1 and tau_2. - Colin Defant, Apr 28 2019
Number of permutations of length n that are sorted to the identity by a consecutive-321-avoiding stack followed by a classical-21-avoiding stack. - Colin Defant, Aug 29 2020
From Helmut Prodinger, Dec 13 2020: (Start)
a(n) is the number of paths in the first quadrant starting at (0,0) and consisting of n steps from the infinite set {(1,1), (1,-1), (1,-2), (1,-3), ...}.
For example, denoting U=(1,1), D=(1,-1), D_ j=(1,-j) for j >= 2, a(4) counts UUUU, UUUD, UUUD_2, UUUD_3, UUDU, UUDD, UUD_2U, UDUU, UDUD.
This step set is inspired by {(1,1), (1,-1), (1,-3), (1,-5), ...}, suggested by Emeric Deutsch around 2000.
See Prodinger link that contains a bijection to Motzkin paths. (End)
Named by Donaghey (1977) after the Israeli-American mathematician Theodore Motzkin (1908-1970). In Sloane's "A Handbook of Integer Sequences" (1973) they were called "generalized ballot numbers". - Amiram Eldar, Apr 15 2021
Number of Motzkin n-paths a(n) is split into A107587(n), number of even Motzkin n-paths, and A343386(n), number of odd Motzkin n-paths. The value A107587(n) - A343386(n) can be called the "shadow" of a(n) (see A343773). - Gennady Eremin, May 17 2021
Conjecture: If p is a prime of the form 6m+1 (A002476), then a(p-2) is divisible by p. Currently, no counterexample exists for p < 10^7. Personal communication from Robert Gerbicz: mod such p this is equivalent to A066796 with comment: "Every A066796(n) from A066796((p-1)/2) to A066796(p-1) is divisible by prime p of form 6m+1". - Serge Batalov, Feb 08 2022
From Rob Burns, Nov 11 2024: (Start)
The conjecture is proved in the 2017 paper by Rob Burns in the Links below. The result is contained in Tables 4 and 5 of the paper, which show that a(p-2) == 0 (mod p) when p == 1 (mod 6) and a(p-2) == -1 (mod p) when p == -1 (mod 6).
In fact, the 2017 paper by Burns establishes more general congruences for a(p^k - 2) where k >= 1.
If p == 1 (mod 6) then a(p^k - 2) == 0 (mod p) for k >= 1.
If p == -1 (mod 6) then a(p^k - 2) == -1 (mod p) when k is odd and a(p^k - 2) == 0 (mod p) when k is even.
These are consequences of the transitions provided in Tables 4, 5 and 6 of the paper.
The 2024 paper by Nadav Kohen also proves the conjecture. Proposition 6 of the paper states that a prime p divides a(p-2) if and only if p = (1 mod 3). (End)
From Peter Bala, Feb 10 2022: (Start)
Conjectures:
(1) For prime p == 1 (mod 6) and n, r >= 1, a(n*p^r - 2) == -A005717(n-1) (mod p), where we take A005717(0) = 0 to match Batalov's conjecture above.
(2) For prime p == 5 (mod 6) and n >= 1, a(n*p - 2) == -A005773(n) (mod p).
(3) For prime p >= 3 and k >= 1, a(n + p^k) == a(n) (mod p) for 0 <= n <= (p^k - 3).
(4) For prime p >= 5 and k >= 2, a(n + p^k) == a(n) (mod p^2) for 0 <= n <= (p^(k-1) - 3). (End)
The Hankel transform of this sequence with a(0) omitted gives the period-6 sequence [1, 0, -1, -1, 0, 1, ...] which is A010892 with its first term omitted, while the Hankel transform of the current sequence is the all-ones sequence A000012, and also it is the unique sequence with this property which is similar to the unique Hankel transform property of the Catalan numbers. - Michael Somos, Apr 17 2022
The number of terms in which the exponent of any variable x_i is not greater than 2 in the expansion of Product_{j=1..n} Sum_{i=1..j} x_i. E.g.: a(4) = 9: 3*x1^2*x2^2, 4*x1^2*x2*x3, 2*x1^2*x2*x4, x1^2*x3^2, x1^2*x3*x4, 2*x1*x2^2*x3, x1*x2^2*x4, x1*x2*x3^2, x1*x2*x3*x4. - Elif Baser, Dec 20 2024

Examples

			G.f.: 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 21*x^5 + 51*x^6 + 127*x^7 + 323*x^8 + ...
.
The 21 Motzkin-paths of length 5: UUDDF, UUDFD, UUFDD, UDUDF, UDUFD, UDFUD, UDFFF, UFUDD, UFDUD, UFDFF, UFFDF, UFFFD, FUUDD, FUDUD, FUDFF, FUFDF, FUFFD, FFUDF, FFUFD, FFFUD, FFFFF.
		

References

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  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pp. 24, 298, 618, 912.
  • A. J. Bu, Automated counting of restricted Motzkin paths, Enumerative Combinatorics and Applications, ECA 1:2 (2021) Article S2R12.
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Crossrefs

Bisections: A026945, A099250.
Sequences related to chords in a circle: A001006, A054726, A006533, A006561, A006600, A007569, A007678. See also entries for chord diagrams in Index file.
a(n) = A005043(n)+A005043(n+1).
A086246 is another version, although this is the main entry. Column k=3 of A182172.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.
Cf. A004148, A004149, A023421, A023422, A023423, A290277 (inv. Euler Transf.).

Programs

  • Haskell
    a001006 n = a001006_list !! n
    a001006_list = zipWith (+) a005043_list $ tail a005043_list
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    # Three different Maple scripts for this sequence:
    A001006 := proc(n)
        add(binomial(n,2*k)*A000108(k),k=0..floor(n/2)) ;
    end proc:
    A001006 := proc(n) option remember; local k; if n <= 1 then 1 else procname(n-1) + add(procname(k)*procname(n-k-2),k=0..n-2); end if; end proc:
    # n -> [a(0),a(1),..,a(n)]
    A001006_list := proc(n) local w, m, j, i; w := proc(i,j,n) option remember;
    if min(i,j,n) < 0 or max(i,j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi else
    w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) fi end:
    [seq( add( add( w(i, j, m), i = 0..m), j = 0..m), m = 0..n)] end:
    A001006_list(29); # Peter Luschny, May 21 2011
  • Mathematica
    a[0] = 1; a[n_Integer] := a[n] = a[n - 1] + Sum[a[k] * a[n - 2 - k], {k, 0, n - 2}]; Array[a, 30]
    (* Second program: *)
    CoefficientList[Series[(1 - x - (1 - 2x - 3x^2)^(1/2))/(2x^2), {x, 0, 29}], x] (* Jean-François Alcover, Nov 29 2011 *)
    Table[Hypergeometric2F1[(1-n)/2, -n/2, 2, 4], {n,0,29}] (* Peter Luschny, May 15 2016 *)
    Table[GegenbauerC[n,-n-1,-1/2]/(n+1),{n,0,100}] (* Emanuele Munarini, Oct 20 2016 *)
    MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
    Table[MotzkinNumber[n], {n, 0, 29}] (* Jean-François Alcover, Oct 27 2021 *)
  • Maxima
    a[0]:1$
    a[1]:1$
    a[n]:=((2*n+1)*a[n-1]+(3*n-3)*a[n-2])/(n+2)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • Maxima
    M(n) := coeff(expand((1+x+x^2)^(n+1)),x^n)/(n+1);
    makelist(M(n),n,0,60); /* Emanuele Munarini, Apr 04 2012 */
    
  • Maxima
    makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */
    
  • PARI
    {a(n) = polcoeff( ( 1 - x - sqrt((1 - x)^2 - 4 * x^2 + x^3 * O(x^n))) / (2 * x^2), n)}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp(x + x * O(x^n)) * besseli(1, 2 * x + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • Python
    from gmpy2 import divexact
    A001006 = [1, 1]
    for n in range(2, 10**3):
        A001006.append(divexact(A001006[-1]*(2*n+1)+(3*n-3)*A001006[-2],n+2))
    # Chai Wah Wu, Sep 01 2014
    
  • Python
    def mot():
        a, b, n = 0, 1, 1
        while True:
            yield b//n
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
    A001006 = mot()
    print([next(A001006) for n in range(30)]) # Peter Luschny, May 16 2016
    
  • Python
    # A simple generator of Motzkin-paths (see the first comment of David Callan).
    C = str.count
    def aGen(n: int):
        a = [""]
        for w in a:
            if len(w) == n:
                if C(w, "U") == C(w, "D"): yield w
            else:
                for j in "UDF":
                    u = w + j
                    if C(u, "U") >= C(u, "D"): a += [u]
        return a
    for n in range(6):
        MP = [w for w in aGen(n)];
        print(len(MP), ":", MP)  # Peter Luschny, Dec 03 2024

Formula

G.f.: A(x) = ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
G.f. A(x) satisfies A(x) = 1 + x*A(x) + x^2*A(x)^2.
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2). - Joerg Arndt, Oct 23 2012
a(n) = (-1/2) Sum_{i+j = n+2, i >= 0, j >= 0} (-3)^i*C(1/2, i)*C(1/2, j).
a(n) = (3/2)^(n+2) * Sum_{k >= 1} 3^(-k) * Catalan(k-1) * binomial(k, n+2-k). [Doslic et al.]
a(n) ~ 3^(n+1)*sqrt(3)*(1 + 1/(16*n))/((2*n+3)*sqrt((n+2)*Pi)). [Barcucci, Pinzani and Sprugnoli]
Limit_{n->infinity} a(n)/a(n-1) = 3. [Aigner]
a(n+2) - a(n+1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). [Bernhart]
a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!). [Bernhart]
From Len Smiley: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k+1), inv. Binomial Transform of A000108.
a(n) = (1/(n+1))*Sum_{k=0..ceiling((n+1)/2)} binomial(n+1, k)*binomial(n+1-k, k-1);
D-finite with recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + (3*n-3)*a(n-2). (End)
a(n) = Sum_{k=0..n} C(n, 2k)*A000108(k). - Paul Barry, Jul 18 2003
E.g.f.: exp(x)*BesselI(1, 2*x)/x. - Vladeta Jovovic, Aug 20 2003
a(n) = A005043(n) + A005043(n+1).
The Hankel transform of this sequence gives A000012 = [1, 1, 1, 1, 1, 1, ...]. E.g., Det([1, 1, 2, 4; 1, 2, 4, 9; 2, 4, 9, 21; 4, 9, 21, 51]) = 1. - Philippe Deléham, Feb 23 2004
a(m+n) = Sum_{k>=0} A064189(m, k)*A064189(n, k). - Philippe Deléham, Mar 05 2004
a(n) = (1/(n+1))*Sum_{j=0..floor(n/3)} (-1)^j*binomial(n+1, j)*binomial(2*n-3*j, n). - Emeric Deutsch, Mar 13 2004
a(n) = A086615(n) - A086615(n-1) (n >= 1). - Emeric Deutsch, Jul 12 2004
G.f.: A(x)=(1-y+y^2)/(1-y)^2 where (1+x)*(y^2-y)+x=0; A(x)=4*(1+x)/(1+x+sqrt(1-2*x-3*x^2))^2; a(n)=(3/4)*(1/2)^n*Sum_(k=0..2*n, 3^(n-k)*C(k)*C(k+1, n+1-k) ) + 0^n/4 [after Doslic et al.]. - Paul Barry, Feb 22 2005
G.f.: c(x^2/(1-x)^2)/(1-x), c(x) the g.f. of A000108. - Paul Barry, May 31 2006
Asymptotic formula: a(n) ~ sqrt(3/4/Pi)*3^(n+1)/n^(3/2). - Benoit Cloitre, Jan 25 2007
a(n) = A007971(n+2)/2. - Zerinvary Lajos, Feb 28 2007
a(n) = (1/(2*Pi))*Integral_{x=-1..3} x^n*sqrt((3-x)*(1+x)) is the moment representation. - Paul Barry, Sep 10 2007
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([0,1,1]), see the 6th formula. - Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
G.f.: 1/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 08 2009
a(n) = (-3)^(1/2)/(6*(n+2)) * (-1)^n*(3*hypergeom([1/2, n+1],[1],4/3) - hypergeom([1/2, n+2],[1],4/3)). - Mark van Hoeij, Nov 12 2009
G.f.: 1/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Mar 02 2010
G.f.: 1/(1-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 [Adds apparently a third '1' in front. - R. J. Mathar, Jan 29 2011]
Let A(x) be the g.f., then B(x)=1+x*A(x) = 1 + 1*x + 1*x^2 + 2*x^3 + 4*x^4 + 9*x^5 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x) (continued fraction); more generally B(x)=C(x/(1+x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^n*sqrt(1-x^2). - Peter Luschny, Sep 11 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = 1/2/(x^2)-1/2/x-1/2/(x^2)*G(0); G(k) = 1+(4*k-1)*x*(2+3*x)/(4*k+2-x*(2+3*x)*(4*k+1)*(4*k+2) /(x*(2+3*x)*(4*k+1)+(4*k+4)/G(k+1))), if -1 < x < 1/3; (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = (-1 + 1/G(0))/(2*x); G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * ( -3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) unless n=-2. - Michael Somos, Mar 23 2012
a(n) = (-1)^n*hypergeometric([-n,3/2],[3],4). - Peter Luschny, Aug 15 2012
Representation in terms of special values of Jacobi polynomials P(n,alpha,beta,x), in Maple notation: a(n)= 2*(-1)^n*n!*JacobiP(n,2,-3/2-n,-7)/(n+2)!, n>=0. - Karol A. Penson, Jun 24 2013
G.f.: Q(0)/x - 1/x, where Q(k) = 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
Catalan(n+1) = Sum_{k=0..n} binomial(n,k)*a(k). E.g.: 42 = 1*1 + 4*1 + 6*2 + 4*4 + 1*9. - Doron Zeilberger, Mar 12 2015
G.f. A(x) with offset 1 satisfies: A(x)^2 = A( x^2/(1-2*x) ). - Paul D. Hanna, Nov 08 2015
a(n) = GegenbauerPoly(n,-n-1,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = a(n-1) + A002026(n-1). Number of Motzkin paths that start with an F step plus number of Motzkin paths that start with an U step. - R. J. Mathar, Jul 25 2017
G.f. A(x) satisfies A(x)*A(-x) = F(x^2), where F(x) is the g.f. of A168592. - Alexander Burstein, Oct 04 2017
G.f.: A(x) = exp(int((E(x)-1)/x dx)), where E(x) is the g.f. of A002426. Equivalently, E(x) = 1 + x*A'(x)/A(x). - Alexander Burstein, Oct 05 2017
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^k. - Ilya Gutkovskiy, Apr 11 2019
From Gennady Eremin, May 08 2021: (Start)
G.f.: 2/(1 - x + sqrt(1-2*x-3*x^2)).
a(n) = A107587(n) + A343386(n) = 2*A107587(n) - A343773(n) = 2*A343386(n) + A343773(n). (End)
Revert transform of A049347 (after Michael Somos). - Gennady Eremin, Jun 11 2021
Sum_{n>=0} 1/a(n) = 2.941237337631025604300320152921013604885956025483079699366681494505960039781389... - Vaclav Kotesovec, Jun 17 2021
Let a(-1) = (1 - sqrt(-3))/2 and a(n) = a(-3-n)*(-3)^(n+3/2) for all n in Z. Then a(n) satisfies my previous formula relation from Mar 23 2012 now for all n in Z. - Michael Somos, Apr 17 2022
Let b(n) = 1 for n <= 1, otherwise b(n) = Sum_{k=2..n} b(k-1) * b(n-k), then a(n) = b(n+1) (conjecture). - Joerg Arndt, Jan 16 2023
From Peter Bala, Feb 03 2024: (Start)
G.f.: A(x) = 1/(1 + x)*c(x/(1 + x))^2 = 1 + x/(1 + x)*c(x/(1 + x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
A(x) = 1/(1 - 3*x)*c(-x/(1 -3*x))^2.
a(n+1) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n, k)*A000245(k+1).
a(n) = 3^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], 4/3). (End)
G.f. A(x) satisfies A(x) = exp( x*A(x) + Integral x*A(x)/(1 - x^2*A(x)) dx ). - Paul D. Hanna, Mar 04 2024
a(n) = hypergeom([-n/2,1/2-n/2],[2],4). - Karol A. Penson, May 18 2025

A014137 Partial sums of Catalan numbers (A000108).

Original entry on oeis.org

1, 2, 4, 9, 23, 65, 197, 626, 2056, 6918, 23714, 82500, 290512, 1033412, 3707852, 13402697, 48760367, 178405157, 656043857, 2423307047, 8987427467, 33453694487, 124936258127, 467995871777, 1757900019101, 6619846420553, 24987199492705, 94520750408709, 358268702159069
Offset: 0

Views

Author

Keywords

Comments

This is also the result of applying the transformation on generating functions A -> 1/((1 - x)*(1 - x*A)) to the g.f. for the Catalan numbers.
p divides a(p) - 3 for prime p = 3 and p = {7, 13, 19, 31, 37, 43, ...} = A002476 (Primes of the form 6*n + 1). p^2 divides a(p^2) - 3 for prime p > 3. - Alexander Adamchuk, Jul 11 2006
Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 (Primes congruent to {0, 2} mod 3); and A045309 (Primes p such that x^3 = n (integer) has only one solution mod p). Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk, Feb 23 2007
For p prime >= 5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan, Nov 29 2007
Hankel transform is A010892(n+1). - Paul Barry, Apr 24 2009
Equals INVERTi transform of A000245: (1, 3, 9, 28, ...). - Gary W. Adamson, May 15 2009
The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367; see A121852. - Jonathan Vos Post, Feb 10 2010
Number of lattice paths from (0,0) to (n,n) which do not go above the diagonal x=y using steps (1,k), (k,1) with k >= 1 including two kinds of (1,1). - Alois P. Heinz, Oct 14 2015
Binomial transform of A086246(n+1) = [1, 1, 1, 2, 4, 9, ...], or, equivalently, of A001006 (Motzkin numbers) with 1 prepended.

Examples

			G.f. = 1 + 2*x + 4*x^2 + 9*x^3 + 23*x^4 + 65*x^5 + 197*x^6 + 626*x^7 + 2056*x^8 + ...
		

Crossrefs

Programs

  • Magma
    [(&+[Catalan(k): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jun 30 2024
  • Maple
    a:= proc(n) option remember; `if`(n<2, n+1,
          ((5*n-1)*a(n-1)-(4*n-2)*a(n-2))/(n+1))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, May 18 2013
    A014137List := proc(m) local A, P, n; A := [1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-n]]);
    A := [op(A), P[-1]] od; A end: A014137List(30); # Peter Luschny, Mar 26 2022
  • Mathematica
    Table[Sum[(2k)!/(k!)^2/(k+1),{k,0,n}],{n,0,30}] (* Alexander Adamchuk, Jul 11 2006 *)
    Accumulate[CatalanNumber[Range[0,30]]] (* Harvey P. Dale, May 08 2012 *)
    a[ n_] := SeriesCoefficient[ (1 - (1 - 4 x)^(1/2)) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, Oct 24 2015 *)
    Table[(1 + CatalanNumber[n] (3 (n + 1) Hypergeometric2F1[1, -n, 1/2 - n, 1/4] - 4 n - 2))/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)
  • PARI
    Vec((1-(1-4*x)^(1/2))/(2*x*(1-x))+O(x^99)) \\ Charles R Greathouse IV, Feb 11 2011
    
  • PARI
    sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }
    C(n)=binomial(2*n, n)/(n+1);
    sm(vector(66, n, C(n-1)))
    /* Joerg Arndt, May 04 2013 */
    
  • Python
    from _future_ import division
    A014137_list, b, s = [], 1, 0
    for n in range(10**2):
        s += b
        A014137_list.append(s)
        b = b*(4*n+2)//(n+2) # Chai Wah Wu, Jan 28 2016
    
  • Sage
    def A014137():
        f, c, n = 1, 1, 1
        while True:
            yield f
            n += 1
            c = c * (4*n - 6) // n
            f = c + f
    a = A014137()
    print([next(a) for  in range(29)]) # _Peter Luschny, Nov 30 2016
    

Formula

a(n) = A014138(n-1) + 1.
G.f.: (1 - (1 - 4*x)^(1/2))/(2*x*(1 - x)).
a(n) = Sum_{k=0..n} (2k)!/(k!)^2/(k+1). - Alexander Adamchuk, Jul 11 2006
D-finite with recurrence: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
Mathar's formula reduces to 2*(2*n-1)*C(n-1) = (n+1)*C(n), which is a known recurrence of the Catalan numbers, so the conjecture is true. - Peter J. Taylor, Mar 23 2015
Let C(n+1) = binomial(2*n+2,n+1)/(n+2) and H(n) = hypergeometric([1,n+3/2],[n+3],4) then A014137(n) = -(-1)^(2/3) - C(n+1)*H(n) and A014138(n) = -I^(2/3) - C(n+1)*H(n). - Peter Luschny, Aug 09 2012
G.f. (conjecture): Q(0)/(1-x), where Q(k)= 1 + (4*k + 1)*x/(k + 1 - 2*x*(k + 1)*(4*k + 3)/(2*x*(4*k + 3) + (2*k + 3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) ~ 2^(2*n + 2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 10 2013
0 = a(n)*(16*a(n+1) - 26*a(n+2) + 10*a(n+3)) + a(n+1)*(-14*a(n+1) + 23*a(n+2) - 11*a(n+3)) + a(n+2)*(a(n+2) + a(n+3)) if n >= 0. - Michael Somos, Oct 24 2015
a(n) = (1 + A000108(n)*(3*(n+1)*hypergeom([1,-n], [1/2-n], 1/4) - 4*n - 2))/2. - Vladimir Reshetnikov, Oct 03 2016
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^2. - Ilya Gutkovskiy, Jul 25 2021
From Peter Luschny, Nov 16 2022: (Start)
a(n) = C(n)*hypergeom([1, -n - 1], [1/2 - n], 1/4) + 1/2.
a(n) = A358436(n) / C(n). (End)
E.g.f.: exp(2*x)*(BesselI(0, 2*x)/2 - BesselI(1, 2*x)) + exp(x)/2*(3*Integral_{x=-oo..oo} BesselI(0,2*x)*exp(x) dx + 1). - Mélika Tebni, Sep 01 2024

A106228 G.f. satisfies A(x) = 1 + x*A(x)/(1 - x*A(x)^2).

Original entry on oeis.org

1, 1, 2, 6, 21, 80, 322, 1347, 5798, 25512, 114236, 518848, 2384538, 11068567, 51817118, 244370806, 1159883685, 5536508864, 26560581688, 127993221140, 619280193640, 3007251366000, 14651743202152, 71601107803904, 350873710447210, 1723795243004223
Offset: 0

Views

Author

Paul D. Hanna, May 19 2005

Keywords

Comments

Number of paths from (0,0) to (3n-3,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and have no tripledescents (ddd). Example: a(3)=6 because we have udud, Uddud, udUdd, UddUdd, uudd and Ududd (the remaining four paths contain the string ddd: uUddd, UdUddd, Uuddd and UUdddd; see A027307). - Emeric Deutsch, Jun 08 2005
a(n) = number of node-labeled ordered trees (A000108) on n vertices, each node labeled with a positive integer <= its outdegree. A node is a non-root non-leaf vertex. Example. a(3)=6 counts the 5 ordered trees on 4 vertices with all labels 1 and the tree
.|.
/ \
with its (one and only) node labeled 2. - David Callan, Jul 14 2006
a(n) = number of Schroeder (n-1)-paths with no triple descents. Example: a(4)=21 counts all 22 Schroeder 3-paths (A006318) except UUUDDD. - David Callan, Jul 14 2006
(1 + 2x + 6x^2 + ...)*(1 + x + 2x^2 + 6x^3) = (1 + 3x + 10x^2 + 37x^3 + ...), where A109081 = (1, 1, 3, 10, 37, ...). - Gary W. Adamson, Nov 15 2011
a(n) = number of Motzkin paths of length 2n-1 with no downsteps in odd position. Example: a(3)=6 counts FFFFF, FFUDF, FUFDF, UDFFF, UDUDF, UFFDF with U an upstep (1,1), F a flatstep (1,0), and D a downstep (1,-1). - David Callan, May 20 2015
Number of permutations of length n that avoid 4123, 4132, and 4213. - Jay Pantone, Oct 01 2015
Conjecturally, the number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j) and e(i) <= e(k). [Martinez and Savage, 2.21] - Eric M. Schmidt, Jul 17 2017
a(n) is the number of permutations of length n avoiding the partially ordered pattern (POP) {1>3, 1>4, 4>2} of length 4. That is, the number of length n permutations having no subsequences of length 4 in which the first element is the largest and the fourth element is larger than the second element. - Sergey Kitaev, Dec 10 2020
a(n) is the number of peakless Motzkin paths of length 2n that do not start with an up edge and where every pair of matching up and down edges occupies positions of the same parity. Equivalently, the number of RNA secondary structures on 2n vertices where the leftmost vertex is not matched and only vertices of the same parity can be matched. - Alexander Burstein, May 17 2021

Examples

			A = 1 + x*A + x^2*A^3 + x^3*A^5 + x^4*A^7 + x^5*A^9 + ...
a(4) = 21 since the top row terms of Q^3 = (10, 7, 3, 1). - _Gary W. Adamson_, Nov 15 2011
G.f. = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 80*x^5 + 322*x^6 + 1347*x^7 + ...
		

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n<2 then 1 elif n=2 then 2 else ((380*n^3-840*n^2+496*n-72)*a(n-1)+(76*n^3-282*n^2+302*n-84)*a(n-2)+(57*n^3-297*n^2+402*n-72)*a(n-3))/(76*n^3-54*n^2-46*n) fi end: seq(a(i),i=0..23); # Peter Luschny, Aug 03 2012
  • Mathematica
    Flatten[{1,Table[1/n*Sum[Binomial[n,k]*Binomial[n+k,n-k-1],{k,0,n-1}],{n,1,20}]}] (* Vaclav Kotesovec, Sep 16 2013 *)
    a[ n_] := If[ n < 0, 0, HypergeometricPFQ[{-n, 1 - n, n + 1}, {1, 3/2}, 1/4]]; (* Michael Somos, May 27 2014 *)
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(k=1,n,A=1+x*A/(1-x*A^2)); polcoeff(A,n)}
    
  • PARI
    {a(n) = local(A); if( n<0, 0, n++; A = (1 + x - sqrt(1 - 2*x - 3*x^2 + x * O(x^n))) / 2; polcoeff( serreverse( x^2 / A), n))}; /* Michael Somos, Jun 18 2005 */
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x /  (1 + 2*x + 2*x^2 + x^3) + x * O(x^n)), n))}; /* Michael Somos, Dec 31 2014 */
  • Sage
    from mpmath import mp
    mp.dps = 32; mp.pretty = True
    def A106228(n) : return int(mp.hyper([-n, 1-n, n+1], [1, 3/2], 1/4))
    [A106228(n) for n in (0..23)] # Peter Luschny, Aug 02 2012
    

Formula

G.f.: A(x) = (1/x)*series_reversion[x/(1 + x*G001006(x))] and thus G.f. satisfies: A(x) = 1 + x*A(x)*G001006(x*A(x)) where G001006(x) is the g.f. of Motzkin numbers A001006.
G.f.: 1 + x*exp( Sum_{n>=1} A082759(n)*x^n/n ), where A082759(n) = Sum_{k=0..n} binomial(n,k)*trinomial(n,k). - Paul D. Hanna, Nov 02 2012
a(n) = (1/n)sum(binomial(n, j+1)*b(n, j), j=0..n-1), where b(n, j) are the trinomial coefficients [b(n, j)=A027907(n, j)=coefficient of x^j in (1+x+x^2)^n]. - Emeric Deutsch, Jun 08 2005
Given g.f. A(x), then B(x) = x*A(x) satisfies 0 = f(x, B(x)) where f(x, y) = y^3 - (1+y)*x*(y-x). - Michael Somos, Jun 18 2005
a(n+1) = Sum[binomial(2n-2k,n-k)*binomial(n+k,n)/(n+1),{k,0,n}]. - David Callan, Aug 16 2006
For n>0: a(n) = 1/n*sum(binomial(n,j)*sum(binomial(j,i)*binomial(n-j,2*j-n-i-1)*2^(2*n-3*j+2*i+1),i=0..n-1), j=0..n); - Vladimir Kruchinin, Dec 26 2010
a(n) = 1/(n+1)*sum(binomial(n+1,k)*binomial(n+k+1,n-k),k,0,n); - Vladimir Kruchinin, Feb 28 2010
a(n) = upper left term in M^n, M = the production matrix:
1, 1
1, 1, 1
2, 2, 1, 1
3, 3, 2, 1, 1
4, 4, 3, 2, 1, 1
5, 5, 4, 3, 2, 1, 1
...
- Gary W. Adamson, Jul 08 2011
D-finite with recurrence: 4*n*(2*n+1)*a(n) + 2*(6-5*n-10*n^2)*a(n-1) + 12*(-9*n^2+35*n-33)*a(n-2) - 2*(n-3)*(13*n-28)*a(n-3) - 15*(n-3)*(n-4)*a(n-4) = 0. - R. J. Mathar, Nov 14 2011
From Gary W. Adamson, Nov 15 2011: (Start)
a(n) is the sum of top row terms of Q^(n-1), where Q = the following infinite square production matrix:
1, 1, 0, 0, 0, ...
2, 1, 1, 0, 0, ...
3, 2, 1, 1, 0, ...
4, 3, 2, 1, 1, ...
5, 4, 3, 2, 1, ...
... (End)
a(n) = 3_F_2([-n, 1-n, n+1], [1, 3/2], 1/4). - Peter Luschny, Aug 02 2012
A four-term recurrence equation is given in the Maple program. Peter Luschny, Aug 03 2012
a(n) ~ 1/228*sqrt(114)*sqrt((32129+3933*sqrt(57))^(1/3) * ((32129+3933*sqrt(57))^(2/3) + 532 + 38*(32129+3933*sqrt(57))^(1/3))) / ((32129+3933*sqrt(57))^(1/3)) * (((1261+57*sqrt(57))^(2/3) + 112 + 10*(1261+57*sqrt(57))^(1/3)) / (6*(1261+57*sqrt(57))^(1/3)))^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Sep 16 2013
G.f. satisfies x*F(x)^3 - x*F(x)^2 + (x-1)*F(x) + 1 = 0. - Jay Pantone, Oct 01 2015
G.f. satisfies A(-x*A(x)^3) = 1/A(x). - Alexander Burstein, Dec 05 2019
G.f.: A(x) = sqrt(B(x)) where B(x) is the g.f. of A262441. - Seiichi Manyama, Mar 31 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n/2+3*k/2+1/2,n)/(n+3*k+1). - Seiichi Manyama, Apr 04 2024

A130777 Coefficients of first difference of Chebyshev S polynomials.

Original entry on oeis.org

1, -1, 1, -1, -1, 1, 1, -2, -1, 1, 1, 2, -3, -1, 1, -1, 3, 3, -4, -1, 1, -1, -3, 6, 4, -5, -1, 1, 1, -4, -6, 10, 5, -6, -1, 1, 1, 4, -10, -10, 15, 6, -7, -1, 1, -1, 5, 10, -20, -15, 21, 7, -8, -1, 1, -1, -5, 15, 20, -35, -21, 28, 8, -9, -1, 1, 1, -6, -15, 35, 35, -56, -28, 36, 9, -10, -1, 1
Offset: 0

Views

Author

Philippe Deléham, Jul 14 2007

Keywords

Comments

Inverse of triangle in A061554.
Signed version of A046854.
From Paul Barry, May 21 2009: (Start)
Riordan array ((1-x)/(1+x^2),x/(1+x^2)).
This triangle is the coefficient triangle for the Hankel transforms of the family of generalized Catalan numbers that satisfy a(n;r)=r*a(n-1;r)+sum{k=1..n-2, a(k)*a(n-1-k;r)}, a(0;r)=a(1;r)=1. The Hankel transform of a(n;r) is h(n)=sum{k=0..n, T(n,k)*r^k} with g.f. (1-x)/(1-r*x+x^2). These sequences include A086246, A000108, A002212. (End)
From Wolfdieter Lang, Jun 11 2011: (Start)
The Riordan array ((1+x)/(1+x^2),x/(1+x^2)) with entries Phat(n,k)= ((-1)^(n-k))*T(n,k) and o.g.f. Phat(x,z)=(1+z)/(1-x*z+z^2) for the row polynomials Phat(n,x) is related to Chebyshev C and S polynomials as follows.
Phat(n,x) = (R(n+1,x)-R(n,x))/(x+2) = S(2*n,sqrt(2+x))
with R(n,x)=C_n(x) in the Abramowitz and Stegun notation, p. 778, 22.5.11. See A049310 for the S polynomials. Proof from the o.g.f.s.
Recurrence for the row polynomials Phat(n,x):
Phat(n,x) = x*Phat(n-1,x) - Phat(n-2,x) for n>=1; Phat(-1,x)=-1, Phat(0,x)=1.
The A-sequence for this Riordan array Phat (see the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices) is given by 1, 0, -1, 0, -1, 0, -2, 0, -5,.., starting with 1 and interlacing the negated A000108 with zeros (o.g.f. 1/c(x^2) = 1-c(x^2)*x^2, with the o.g.f. c(x) of A000108).
The Z-sequence has o.g.f. sqrt((1-2*x)/(1+2*x)), and it is given by A063886(n)*(-1)^n.
The A-sequence of the Riordan array T(n,k) is identical with the one for the Riordan array Phat, and the Z-sequence is -A063886(n).
(End)
The row polynomials P(n,x) are the characteristic polynomials of the adjacency matrices of the graphs which look like P_n (n vertices (nodes), n-1 lines (edges)), but vertex no. 1 has a loop. - Wolfdieter Lang, Nov 17 2011
From Wolfdieter Lang, Dec 14 2013: (Start)
The zeros of P(n,x) are x(n,j) = -2*cos(2*Pi*j/(2*n+1)), j=1..n. From P(n,x) = (-1)^n*S(2*n,sqrt(2-x)) (see, e.g., the Lemma 6 of the W. Lang link).
The discriminants of the P-polynomials are given in A052750. (End)

Examples

			The triangle T(n,k) begins:
n\k  0   1   1   3    4    5    6    7    8    9  10  11  12  13 14 15 ...
0:   1
1:  -1   1
2:  -1  -1   1
3:   1  -2  -1   1
4:   1   2  -3  -1    1
5:  -1   3   3  -4   -1    1
6:  -1  -3   6   4   -5   -1    1
7:   1  -4  -6  10    5   -6   -1    1
8:   1   4 -10 -10   15    6   -7   -1    1
9:  -1   5  10 -20  -15   21    7   -8   -1    1
10: -1  -5  15  20  -35  -21   28    8   -9   -1   1
11:  1  -6 -15  35   35  -56  -28   36    9  -10  -1   1
12:  1   6 -21 -35   70   56  -84  -36   45   10 -11  -1   1
13: -1   7  21 -56  -70  126   84 -120  -45   55  11 -12  -1   1
14: -1  -7  28  56 -126 -126  210  120 -165  -55  66  12 -13  -1  1
15:  1  -8 -28  84  126 -252 -210  330  165 -220 -66  78  13 -14 -1  1
...  reformatted and extended - _Wolfdieter Lang_, Jul 31 2014.
---------------------------------------------------------------------------
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
-1, 1,
-2, 0, 1,
-2, -1, 0, 1,
-4, 0, -1, 0, 1,
-6, -1, 0, -1, 0, 1,
-12, 0, -1, 0, -1, 0, 1,
-20, -2, 0, -1, 0, -1, 0, 1,
-40, 0, -2, 0, -1, 0, -1, 0, 1,
-70, -5, 0, -2, 0, -1, 0, -1, 0, 1 (End)
Row polynomials as first difference of S polynomials:
P(3,x) = S(3,x) - S(2,x) = (x^3 - 2*x) - (x^2 -1) = 1 - 2*x - x^2 +x^3.
Alternative triangle recurrence (see a comment above): T(6,2) = T(5,2) + T(5,1) = 3 + 3 = 6. T(6,3) = -T(5,3) + 0*T(5,1) = -(-4) = 4. - _Wolfdieter Lang_, Jul 31 2014
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964. Tenth printing, Wiley, 2002 (also electronically available).

Crossrefs

Cf. A066170, A046854, A057077 (first column).
Row sums: A010892(n+1); repeat(1,0,-1,-1,0,1). Alternating row sums: A061347(n+2); repeat(1,-2,1).

Programs

  • Maple
    A130777 := proc(n,k): (-1)^binomial(n-k+1,2)*binomial(floor((n+k)/2),k) end: seq(seq(A130777(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    T[n_, k_] := (-1)^Binomial[n - k + 1, 2]*Binomial[Floor[(n + k)/2], k];
    Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017, from Maple *)
  • Sage
    @CachedFunction
    def A130777(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        h = A130777(n-1,k) if n==1 else 0
        return A130777(n-1,k-1) - A130777(n-2,k) - h
    for n in (0..9): [A130777(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

Number triangle T(n,k) = (-1)^C(n-k+1,2)*C(floor((n+k)/2),k). - Paul Barry, May 21 2009
From Wolfdieter Lang, Jun 11 2011: (Start)
Row polynomials: P(n,x) = sum(k=0..n, T(n,k)*x^k) = R(2*n+1,sqrt(2+x)) / sqrt(2+x), with Chebyshev polynomials R with coefficients given in A127672 (scaled T-polynomials).
R(n,x) is called C_n(x) in Abramowitz and Stegun's handbook, p. 778, 22.5.11.
P(n,x) = S(n,x)-S(n-1,x), n>=0, S(-1,x)=0, with the Chebyshev S-polynomials (see the coefficient triangle A049310).
O.g.f. for row polynomials: P(x,z):= sum(n>=0, P(n,x)*z^n ) = (1-z)/(1-x*z+z^2).
(from the o.g.f. for R(2*n+1,x), n>=0, computed from the o.g.f. for the R-polynomials (2-x*z)/(1-x*z+z^2) (see A127672))
Proof of the Chebyshev connection from the o.g.f. for Riordan array property of this triangle (see the P. Barry comment above).
For the A- and Z-sequences of this Riordan array see a comment above. (End)
abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) T(n,n-k) = A108299(n,k); abs(T(n,n-k)) = A065941(n,k). - Johannes W. Meijer, Aug 08 2011
From Wolfdieter Lang, Jul 31 2014: (Start)
Similar to the triangles A157751, A244419 and A180070 one can give for the row polynomials P(n,x) besides the usual three term recurrence another one needing only one recurrence step. This uses also a negative argument, namely P(n,x) = (-1)^(n-1)*(-1 + x/2)*P(n-1,-x) + (x/2)*P(n-1,x), n >= 1, P(0,x) = 1. Proof by computing the o.g.f. and comparing with the known one. This entails the alternative triangle recurrence T(n,k) = (-1)^(n-k)*T(n-1,k) + (1/2)*(1 + (-1)^(n-k))*T(n-1,k-1), n >= m >= 1, T(n,k) = 0 if n < k and T(n,0) = (-1)^floor((n+1)/2) = A057077(n+1). [P(n,x) recurrence corrected Aug 03 2014]
(End)

Extensions

New name and Chebyshev comments by Wolfdieter Lang, Jun 11 2010

A168049 Expansion of (3 -x -sqrt(1-2*x-3*x^2))/2.

Original entry on oeis.org

1, 0, 1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209
Offset: 0

Views

Author

Paul Barry, Nov 17 2009

Keywords

Comments

A variant of the Motzkin numbers A001006. Hankel transform is A168050.
Essentially the same as A086246. - R. J. Mathar, Dec 20 2011
Alternatively, this sequence corresponds to the number of positive walks with n steps {-1,0,1} starting at the origin, ending at altitude 1, and staying strictly above the x-axis. - David Nguyen, Dec 01 2016

Examples

			G.f. = 1 + x^2 + x^3 + 2*x^4 + 4*x^5 + 9*x^6 + 21*x^7 + 51*x^8 + ... - _Michael Somos_, Sep 26 2018
		

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((3 -x - Sqrt(1-2*x-3*x^2))/2)); // G. C. Greubel, Sep 25 2018
  • Mathematica
    CoefficientList[Series[(3-x-Sqrt[1-2*x-3*x^2])/2, {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
  • PARI
    Vec((3-x-sqrt(1-2*x-3*x^2))/2) \\ Charles R Greathouse IV, Dec 01 2016
    

Formula

D-finite with recurrence: n*a(n) +(3-2n)*a(n-1) +3(3-n)*a(n-2)=0. - R. J. Mathar, Dec 20 2011
0 = a(n)*(+9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1)*(-3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2)*(+a(n+2) + a(n+3)) if n>0. - Michael Somos, Jan 31 2014
a(n) ~ 3^(n+1/2) / (6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Feb 12 2014
G.f.: 1 + x^2/(1 - x - x^2/(1 - x - x^2/(1 - x - x^2/(1 - ...)))), a continued fraction. - Ilya Gutkovskiy, Sep 23 2017

A327871 Number of self-avoiding planar walks starting at (0,0), ending at (n,n), remaining in the first quadrant and using steps (0,1), (-1,1), and (1,-1) with the restriction that (-1,1) and (1,-1) are always immediately followed by (0,1).

Original entry on oeis.org

1, 1, 3, 14, 70, 369, 2002, 11076, 62127, 352070, 2010998, 11559030, 66780155, 387444085, 2255875650, 13174629240, 77143234950, 452738296890, 2662359410158, 15683996769460, 92540962166016, 546799192200261, 3235027635603828, 19161631961190036, 113617798289197650
Offset: 0

Views

Author

Alois P. Heinz, Sep 28 2019

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(x, y, t) option remember; `if`(min(x, y)<0, 0,
          `if`(max(x, y)=0, 1, b(x-1, y, 1)+
          `if`(t=1, b(x-1, y+1, 0)+b(x+1, y-1, 0), 0)))
        end:
    a:= n-> b(n$2, 0):
    seq(a(n), n=0..25);
  • Mathematica
    b[x_, y_, t_] := b[x, y, t] = If[Min[x, y] < 0, 0, If[Max[x, y]==0, 1, b[x - 1, y, 1] + If[t==1, b[x - 1, y + 1, 0] + b[x + 1, y - 1, 0], 0]]];
    a[n_] := b[n, n, 0];
    a /@ Range[0, 25] (* Jean-François Alcover, May 13 2020, after Maple *)
    a[n_] := Binomial[2n - 1, n - 1] Hypergeometric2F1[(2 - n)/2, (1 - n)/2, n + 2, 4];
    a[0] := 1; Table[a[n], {n, 0, 24}] (* Peter Luschny, May 19 2021 *)

Formula

a(n) ~ sqrt(5 + 1/sqrt(13)) * (70 + 26*sqrt(13))^n / (2^(3/2) * sqrt(Pi*n) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Oct 12 2019
From Peter Luschny, May 19 2021: (Start)
Next three formulas for n >= 1:
a(n) = A026300(2*n - 1, n - 1).
a(n) = Sum_{j=0..floor((n-1)/2)} C(2*n-1, 2*j + n)*(C(2*j + n, j) - C(2*j +n, j-1)).
a(n) = binomial(2*n - 1, n - 1)*hypergeom([(2 - n)/2, (1 - n)/2], [n + 2], 4). (End)

A091836 A triangle of Motzkin ballot numbers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 6, 6, 4, 1, 9, 13, 13, 10, 5, 1, 21, 30, 30, 24, 15, 6, 1, 51, 72, 72, 59, 40, 21, 7, 1, 127, 178, 178, 148, 105, 62, 28, 8, 1, 323, 450, 450, 378, 276, 174, 91, 36, 9, 1, 835, 1158, 1158, 980, 730, 480, 273, 128, 45, 10, 1, 2188, 3023, 3023
Offset: 0

Views

Author

Emeric Deutsch, Mar 09 2004

Keywords

Comments

T(n-1,k) is the number of Motzkin paths of length n that have k points on the horizontal axis (besides the first and last point). For example T(1,0)=1 counts the path UD with 2 steps and no intermediate interception with the y=0 axis, and T(1,1)=1 counts the path FF with 2 steps, staying on the y=0 axis. - R. J. Mathar, Jul 23 2017
Riordan matrix A=(g(t),t*g(t)), where g(t)=1+t*M(t)=C(t/(1-t)), where M(t) and C(t) are the g.f. of Motzkin and Catalan numbers. A is a pseudo-involution. - Emanuele Munarini, Jul 03 2024

Examples

			Triangle begins:
   1;
   1,  1;
   1,  2,  1;
   2,  3,  3,  1;
   4,  6,  6,  4,  1;
   9, 13, 13, 10,  5,  1;
  21, 30, 30, 24, 15,  6,  1;
  ...
		

Crossrefs

Mirror image of A034929.
T(n, 0) = A086246(n+1) = A001006(n-1).
T(n, 1) = A005554(n).
Row sums are the Motzkin numbers (A001006).

Programs

  • Mathematica
    T[n_, m_] := If[n == m, 1, (-1)^m (m Sum[k (-1)^(n+k) Binomial[n+k-1, n-1] Sum[Binomial[j, -n+m-k+2j] Binomial[n-m, j], {j, 0, n-m}], {k, 1, n-m}])/ (n(n-m))];
    Table[T[n, m], {n, 1, 11}, {m, 1, n}] // Flatten (* Jean-François Alcover, Jul 27 2018, after Vladimir Kruchinin *)
  • Maxima
    T(n,m):=if n=m then 1 else (-1)^m*(m*sum(k*(-1)^(n+k)*binomial(n+k-1,n-1)*sum(binomial(j,-n+m-k+2*j)*binomial(n-m,j),j,0,n-m),k,1,n-m))/(n*(n-m)); /* Vladimir Kruchinin, Aug 20 2012 */

Formula

Column k has g.f.: z^k(1+zM)^(k+1).
G.f.: (1+zM)/(1-tz(1+zM)), where M = 1 + zM+ z ^2M^2 is the g.f. of the Motzkin numbers (A001006).
T(n,m) = (m*(Sum_{k=1..n-m} k*(-1)^(n+m+k)*binomial(n+k-1,n-1) * Sum_{j=0..n-m} binomial(j,-n+m-k+2*j)*binomial(n-m,j)))/(n*(n-m)), n>m, T(n,n)=1. - Vladimir Kruchinin, Aug 20 2012
From Emanuele Munarini, Jul 03 2024: (Start)
T(n,k) = Sum_{i=0..n-k} (-1)^(n-k-i)*binomial(n-k-1,n-k-i) * binomial(2*i+k,i+k) * (k+1) / (i+k+1).
T(n,k) = Sum_{i=0..n-k} binomial(n-k-1,n-k-i)*binomial(n-i+1,i)*(k+1)/(n-i+1) for k < n.
T(n,k) = Sum_{i=0..n-k} trinomial(n-k,n-k-i)*binomial(k+1,i)*i/(n-k) for k < n, where trinomial(n,k) = A027907(n,k).
Recurrence: T(n+2,k+2) = T(n+2,k+1) + T(n+1,k+1) - T(n+1,k) - T(n,k). (End)

A292440 Expansion of (1 - x + sqrt(1 - 2*x - 3*x^2))/2 in powers of x.

Original entry on oeis.org

1, -1, -1, -1, -2, -4, -9, -21, -51, -127, -323, -835, -2188, -5798, -15511, -41835, -113634, -310572, -853467, -2356779, -6536382, -18199284, -50852019, -142547559, -400763223, -1129760415, -3192727797, -9043402501, -25669818476, -73007772802
Offset: 0

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Author

Seiichi Manyama, Sep 16 2017

Keywords

Comments

Apart from a(1) the same as A168051. - R. J. Mathar, Sep 18 2017

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-x +Sqrt(1-2*x-3*x^2))/2)); // G. C. Greubel, Aug 13 2018
  • Mathematica
    CoefficientList[Series[(1-x +Sqrt[1-2*x-3*x^2])/2, {x, 0, 50}], x] (* G. C. Greubel, Aug 13 2018 *)
  • PARI
    x='x+O('x^50); Vec((1 - x + sqrt(1 - 2*x - 3*x^2))/2) \\ G. C. Greubel, Aug 13 2018
    

Formula

Convolution inverse of A001006.
Let f(x) = (1 - x - sqrt(1 - 2*x - 3*x^2))/(2*x^2).
G.f.: 1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(... (continued fraction).
G.f.: 1/f(x) = 1 - x - x^2*f(x).
a(n) = -A001006(n-2) for n > 1.
a(n) ~ -3^(n - 1/2) / (2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 14 2018
D-finite with recurrence: n*a(n) +(-2*n+3)*a(n-1) +3*(-n+3)*a(n-2)=0. - R. J. Mathar, Jan 23 2020
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