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 301 results. Next

A006319 Royal paths in a lattice (convolution of A006318).

Original entry on oeis.org

1, 1, 4, 16, 68, 304, 1412, 6752, 33028, 164512, 831620, 4255728, 22004292, 114781008, 603308292, 3192216000, 16989553668, 90890869312, 488500827908, 2636405463248, 14281895003716, 77631035881072, 423282220216964, 2314491475510816, 12688544297945348
Offset: 0

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Author

Keywords

Comments

Number of peaks at level 1 in all Schröder paths of semilength n (n>=1). Example: a(2)=4 because in the six Schröder paths of semilength two, HH, H(UD), (UD)H, (UD)(UD), UHD and UUDD (where H=(2,0), U=(1,1), D=(1,-1)), we have four peaks at level 1 (shown between parentheses). - Emeric Deutsch, Dec 27 2003
a(n) = number of Schroder n-paths (subdiagonal paths of steps E = (1,0), N = (0,1), and D = (1,1) from the origin to (n,n) ) that start with an E step. For example, a(2) = 4 counts END, ENEN, EDN, EENN. - David Callan, May 15 2022

Examples

			a(4) = 68 since the top row of M^3 = (22, 22, 16, 8, 0, 0, 0, ...); where 68 = (22 + 22 + 16 + 8).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First differences of A006318. Second diagonal of A033877.

Programs

  • Magma
    [1] cat [&+[2/(k+2)*Binomial(n+k,k+1)*Binomial(n-1,k): k in [0..n]]: n in [1..25]]; // Vincenzo Librandi, Jul 22 2017
    
  • Mathematica
    d[n_] := d[n] = Sum[(n - j)*Sum[d[i]d[j - i], {i, 0, j}], {j, 0, n-1}]; d[0] = 1; Table[d[n], {n, 0, 26}]
    a[0] := 1; a[n_] := n Hypergeometric2F1[1 - n, n + 1, 3, -1];
    Array[a, 25, 0] (* Peter Luschny, Jan 31 2020 *)
  • PARI
    apply( {A006319(n)=!n+sum(k=0,n-1,binomial(n+k,k+1)*binomial(n-1,k)*2/(k+2))}, [0..30]) \\ M. F. Hasler, Jan 29 2020
  • Sage
    def A006319_list(n) :
        D = [0]*(n+1); D[1] = 1
        b = True; h = 2; R = [1]
        for i in range(2*n-2) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1;
            else :
                for k in range(1,h, 1) : D[k] += D[k-1]
                R.append(D[h-2]);
            b = not b
        return R
    A006319_list(25) # Peter Luschny, Jun 03 2012
    

Formula

All listed terms satisfy the recurrence a(1) = 1 and, for n > 1, a(n) = 4*a(n-1) + Sum_{k=2..n-2} a(k)*a(n-k-1). - John W. Layman, Feb 23 2001
From Mario Catalani (mario.catalani(AT)unito.it), Jun 19 2003: (Start)
a(n) = Sum_{j=0..n} (n-j)*(Sum_{i=0..j} a(i)*a(j-i)) for n > 0, a(0)=1.
G.f.: A(x) = (1/(2x))((1-x)^2 - sqrt((1-x)^4 - 4*x*(1-x)^2)) (End)
a(n) = 0^n + Sum_{k=0..n-1} binomial(n+k, 2*k+1)*A000108(k+1). - Paul Barry, Feb 01 2009
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z = x/(1-x)^2 (continued fraction); more generally g.f. C(x/(1-x)^2) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) is the sum of top row terms in M^(n-1), M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
1, 1, 2, 0, 0, 0, ...
1, 1, 1, 2, 0, 0, ...
1, 1, 1, 1, 2, 0, ...
1, 1, 1, 1, 1, 2, ...
... - Gary W. Adamson, Aug 23 2011
a(n) ~ 2^(1/4)*(3+2*sqrt(2))^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 09 2013
D-finite with recurrence: (n+1)*a(n) + (-7*n+4)*a(n-1) + (7*n-17)*a(n-2) + (-n+4)*a(n-3) = 0. - R. J. Mathar, Oct 16 2013
a(n) = Sum_{k=0..n} (2/(k+2))*binomial(n+k,k+1)*binomial(n-1,k) for n >= 1. - David Callan, Jul 21 2017
G.f. A(x) satisfies: A(x) = 1/(1 - Sum_{k>=1} k*x^k*A(x)). - Ilya Gutkovskiy, Apr 10 2018
From Peter Bala, Jan 28 2020: (Start)
a(n) = A006318(n) - A006318(n-1) for n >= 1.
(2*n-3)*(n+1)*a(n) = 12*(n-1)^2*a(n-1) - (2*n-1)*(n-3)*a(n-2) with a(1) = 1, a(2) = 4.
O.g.f. A(x) = (1 - x)*( (1 - x) - sqrt(1 - 6*x + x^2) )/(2*x) = (1 - x)*S(x) = 1 + x*S(x)^2, where S(x) is the o.g.f. for the large Schröder numbers A006318. (End)
a(n) = 0^n + n*hypergeom([1 - n, n + 1], [3], -1). - Peter Luschny, Jan 31 2020

A086616 Partial sums of the large Schroeder numbers (A006318).

Original entry on oeis.org

1, 3, 9, 31, 121, 515, 2321, 10879, 52465, 258563, 1296281, 6589727, 33887465, 175966211, 921353249, 4858956287, 25786112993, 137604139011, 737922992937, 3974647310111, 21493266631001, 116642921832963, 635074797251889, 3467998148181631, 18989465797056721, 104239408386028035
Offset: 0

Views

Author

Paul D. Hanna, Jul 24 2003

Keywords

Comments

Row sums of triangle A086614. - Paul D. Hanna, Jul 24 2003
Hankel transform is A136577(n+1). - Paul Barry, Jun 03 2009

Examples

			a(1) = 2 + 1 = 3;
a(2) = 3 + 4 + 2 = 9;
a(3) = 4 + 10 + 12 + 5 = 31;
a(4) = 5 + 20 + 42 + 40 + 14 = 121.
		

Crossrefs

Cf. A086614 (triangle), A086615 (antidiagonal sums).
Cf. A006318.

Programs

  • Mathematica
    Table[SeriesCoefficient[(1-x-Sqrt[1-6*x+x^2])/(2*x*(1-x)),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
  • PARI
    x='x+O('x^66); Vec((1-x-sqrt(1-6*x+x^2))/(2*x*(1-x))) \\ Joerg Arndt, May 10 2013
  • Sage
    # Generalized algorithm of L. Seidel
    def A086616_list(n) :
        D = [0]*(n+2); D[1] = 1
        b = True; h = 2; R = []
        for i in range(2*n) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
            else :
                for k in range(1,h, 1) : D[k] += D[k-1]
                R.append(D[h-1]); h += 1;
            b = not b
        return R
    A086616_list(23) # Peter Luschny, Jun 02 2012
    

Formula

G.f.: A(x) = 1/(1 - x)^2 + x*A(x)^2.
a(1) = 1 and a(n) = n + Sum_{i=1..n-1} a(i)*a(n-i) for n >= 2. - Benoit Cloitre, Mar 16 2004
G.f.: (1 - x - sqrt(1 - 6*x + x^2))/(2*x*(1 - x)). Cf. A001003. - Ralf Stephan, Mar 23 2004
a(n) = Sum_{k=0..n} C(n+k+1, 2*k+1) * A000108(k). - Paul Barry, Jun 03 2009
Recurrence: (n+1)*a(n) = (7*n-2)*a(n-1) - (7*n-5)*a(n-2) + (n-2)*a(n-3). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(24 + 17*sqrt(2))*(3 + 2*sqrt(2))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
A(x) = 1/(1 - x)^2 * c(x/(1-x^2)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. - Peter Bala, Aug 29 2024

Extensions

Name changed using a comment of Emeric Deutsch from Dec 20 2004. - Peter Luschny, Jun 03 2012

A333090 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of S(x)^n evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 3, 21, 183, 1729, 17003, 171237, 1752047, 18130433, 189218451, 1987916021, 20996253479, 222730436161, 2371369720827, 25325636818629, 271189884041183, 2910628489408513, 31302328583021091, 337241582882175189, 3639109029230457751, 39324814984207649729
Offset: 0

Views

Author

Peter Bala, Mar 22 2020

Keywords

Comments

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. See A333091 for m = 2 and A333092 for m = 3. For similarly defined sequences see A333093 through A333097.

Examples

			n-th order Taylor polynomial of S(x)^n:
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^1 = 1 + 2*x + O(x^2)
  n = 2: S(x)^2 = 1 + 4*x + 16*x^2 + O(x^3)
  n = 3: S(x)^3 = 1 + 6*x + 30*x^2 + 146*x^3 + O(x^4)
  n = 4: S(x)^4 = 1 + 8*x + 48*x^2 + 264*x^3 + 1408*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 16 = 21, a(3) = 1 + 6 + 30 + 146 = 183 and a(4) = 1 + 8 + 48 + 264 + 1408 = 1729.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^n, n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    2    1                           3
  n = 2 |   16    4    1                     21
  n = 3 |  146   30    6   1                183
  n = 4 | 1408  264   48   8   1           1729
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence is [x^n]S(x)^n = A103885(n).
Examples of supercongruences:
a(13) - a(1) = 2371369720827 - 3 = (2^3)*(3^2)*(13^3)*83*180617 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 425495386400395896971 - 183 = (2^2)*(7^3*)*19*47* 347287606554703 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 5894174066435445232142003 - 17003 = (2^3)*(3^4)*(5^6)*17* 41*101*5081*1627513421 == 0 ( mod 5^6 ).
		

Crossrefs

Programs

  • Maple
    S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
    G := (x,n) -> series(S(x)^n, x, 51):
    seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);
  • Mathematica
    Table[SeriesCoefficient[((1 + x)*(1 - Sqrt[1 - 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = [x^n] ( (1 + x)*S(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 10*x^2 + 66*x^3 + 498*x^4 + ... = (1/x)*Revert( x/S(x) ) is the o.g.f. of A027307.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ phi^(5*n+2) / (2*5^(3/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 28 2020

A186826 Riordan array (s(x),x*S(x)) where s(x) is the g.f. of the little Schroeder numbers A001003, and S(x) is the g.f. of the large Schroeder numbers A006318.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 11, 11, 5, 1, 45, 45, 23, 7, 1, 197, 197, 107, 39, 9, 1, 903, 903, 509, 205, 59, 11, 1, 4279, 4279, 2473, 1061, 347, 83, 13, 1, 20793, 20793, 12235, 5483, 1949, 541, 111, 15, 1, 103049, 103049, 61463, 28435, 10717, 3285, 795, 143, 17, 1, 518859, 518859, 312761, 148249, 58351, 19199, 5197, 1117, 179, 19, 1
Offset: 0

Views

Author

Paul Barry, Feb 27 2011

Keywords

Comments

Reverse of A144944. Inverse of A186827.

Examples

			Triangle begins
       1;
       1,      1;
       3,      3,      1;
      11,     11,      5,      1;
      45,     45,     23,      7,     1;
     197,    197,    107,     39,     9,     1;
     903,    903,    509,    205,    59,    11,    1;
    4279,   4279,   2473,   1061,   347,    83,   13,    1;
   20793,  20793,  12235,   5483,  1949,   541,  111,   15,   1;
  103049, 103049,  61463,  28435, 10717,  3285,  795,  143,  17,  1;
  518859, 518859, 312761, 148249, 58351, 19199, 5197, 1117, 179, 19, 1;
Production matrix of this triangle begins
  1, 1;
  2, 2, 1;
  2, 2, 2, 1;
  2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 2, 2, 1;
For instance, 107=1*45+2*23+2*7+2*1.
		

Crossrefs

Cf. A001003, A006318, A010683 (row sums), A144944 (row reverse), A186827 (inverse), A186828 (diagonal sums), A239204.

Programs

  • Haskell
    a186826 n k = a186826_tabl !! n !! k
    a186826_row n = a186826_tabl !! n
    a186826_tabl = map reverse a144944_tabl
    -- Reinhard Zumkeller, May 11 2013
    
  • Mathematica
    t[, 0]=1; t[p, p_]:= t[p, p]= t[p, p-1]; t[p_, q_]:= t[p, q]= t[p, q -1] + t[p-1, q] + t[p-1, q-1];
    Table[t[p, q], {p,0,10}, {q,p,0,-1}]//Flatten (* Jean-François Alcover, Jul 16 2019 *)
  • SageMath
    @CachedFunction
    def t(n,k):
        if (k<0 or k>n): return 0
        elif (k==0): return 1
        elif (kA186826(n,k): return t(n+2,n-k)
    flatten([[A186826(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 11 2023

Formula

Riordan array ((1+x+sqrt(1-6*x+x^2))/(4*x), (1-x-sqrt(1-6*x+x^2))/2).
Sum_{k=0..n} T(n,k) = A010683(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A186828(n).
R(n,k) = k*Sum_{i=0..n-k} (A001003(i)/(n-i))*Sum_{m=0..n-k-i} binomial(n-i,m)*binomial(2*(n-i)-m-k-1, n-i-1), k>0, R(n,0) = A001003(n). - Vladimir Kruchinin, Mar 09 2011
Sum_{k=0..n} (-1)^k*T(n, k) = A239204(n-2). - G. C. Greubel, Mar 11 2023

A227505 Schroeder triangle sums: a(n) = A006603(n+3) - A006318(n+3) - A006319(n+2).

Original entry on oeis.org

1, 6, 31, 154, 763, 3808, 19197, 97772, 502749, 2607658, 13630635, 71743478, 379949431, 2023314980, 10828048409, 58206726936, 314157742457, 1701817879214, 9249717805207, 50427858276754, 275695956722547, 1511164724634440, 8302888160922965
Offset: 1

Views

Author

Johannes W. Meijer, Jul 15 2013

Keywords

Comments

The terms of this sequence equal the Kn23 sums, see A180662, of the Schroeder triangle A033877 (with offset 1 and n for columns and k for rows).

Crossrefs

Programs

  • Maple
    A227505 := proc(n) local k, T; T := proc(n, k) option remember; if n=1 then return(1) fi; if kA227505(n), n = 1..23);
    A227505 := proc(n): A006603(n+3) - A006318(n+3) - A006319(n+2) end: A006603 := n ->  add((k*add(binomial(n-k+2, i)*binomial(2*n-3*k-i+3, n-k+1), i= 0.. n-2*k+2))/(n-k+2), k= 1.. n/2+1): A006318 := n -> add(binomial(n+k, n-k) * binomial(2*k, k)/(k+1), k=0..n): A006319 := proc(n): if n=0 then 1 else A006318(n) - A006318(n-1) fi: end: seq(A227505(n), n=1..23);

Formula

a(n) = sum(A033877(n-2*k+2,n-k+3), k=1..floor((n+1)/2)).
a(n) = A006603(n+3) - A006318(n+3) - A006319(n+2).

A035011 A006318(n) - 1.

Original entry on oeis.org

0, 1, 5, 21, 89, 393, 1805, 8557, 41585, 206097, 1037717, 5293445, 27297737, 142078745, 745387037, 3937603037, 20927156705, 111818026017, 600318853925, 3236724317173
Offset: 0

Views

Author

Keywords

Comments

Number of occurrences of UD, UHD, UHHD, UHHHD, ... starting at level zero in all Schroeder paths of semilength n (i.e., lattice paths from (0,0) to (2n,0), with steps H=(2,0), U=(1,1) and D=(1,-1) and not going below the x-axis). Example: a(2) = 5 because in the six paths of semilength 2, namely HH, H(UD), (UD)H, (UHD), (UD)(UD), UUDD, we have 5 required occurrences (shown between parentheses). - Emeric Deutsch, Dec 28 2003

Crossrefs

Cf. A006318.

Programs

  • Mathematica
    CoefficientList[Series[(1-4*x+x^2)/(2*x*(1-x))-Sqrt[1-6*x+x^2]/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 31 2014 *)
  • Sage
    def A035011_list(n) :
        D = [0]*(n+2); D[1] = 1
        b = True; h = 2; R = []
        for i in range(2*n) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
            else :
                for k in range(1,h, 1) : D[k] += D[k-1]
                R.append(D[h-2]); h += 1;
            b = not b
        return R
    A035011_list(20) # Peter Luschny, Jun 03 2012

Formula

G.f.: (1-4*z+z^2)/(2*z*(1-z))-sqrt(1-6*z+z^2)/(2*z). - Emeric Deutsch, Dec 28 2003
Recurrence: (2*n^2 + 11*n + 12)*a(n+3) = (14*n^2 + 59*n + 60)*a(n+2) - (14*n^2 + 53*n + 48)*a(n+1) + (2*n^2 + 5*n)*a(n). - Ralf Stephan, Feb 11 2014
Asymptotics: a(n) ~ (3+2*sqrt(2))^n*(2^(1/4)+1/2^(1/4))/sqrt(2*Pi*n^3). - Fung Lam, Mar 31 2014
From Vaclav Kotesovec, Mar 31 2014: (Start)
Recurrence: (n+1)*a(n) = (8*n-3)*a(n-1) - 7*(2*n-3)*a(n-2) + (8*n-21)*a(n-3) - (n-4)*a(n-4).
Recurrence: (n+1)*(2*n-3)*a(n) = (2*n-1)*(7*n-9)*a(n-1) - (2*n-3)*(7*n-5)*a(n-2) + (n-3)*(2*n-1)*a(n-3).
(End)

A227504 Schroeder triangle sums: a(n) = A006603(n+1) - A006318(n+1).

Original entry on oeis.org

1, 4, 17, 74, 335, 1566, 7515, 36836, 183709, 929392, 4758477, 24611950, 128411643, 675051770, 3572165431, 19012868648, 101718917721, 546707554844, 2950563205705, 15983712882930, 86880753686279, 473710078493718, 2590187432233363, 14199709022579788
Offset: 1

Views

Author

Johannes W. Meijer, Jul 15 2013

Keywords

Comments

The terms of this sequence equal the Kn22 sums, see A180662, of the Schroeder triangle A033877 (with offset 1 and n for columns and k for rows).

Crossrefs

Programs

  • Maple
    A227504 := proc(n) local k, T; T := proc(n, k) option remember; if n=1 then return(1) fi; if kA227504(n), n = 1..24);
    A227504 := proc(n): A006603(n+1) - A006318(n+1) end: A006603 := n -> add((k*add(binomial(n-k+2, i)*binomial(2*n-3*k-i+3, n-k+1), i= 0.. n-2*k+2)) / (n-k+2), k= 1.. n/2+1): A006318 := n -> add(binomial(n+k, n-k) * binomial(2*k, k)/(k+1), k=0..n): seq(A227504(n), n=1..24);

Formula

a(n) = sum(A033877(n-2*k+2, n-k+2), k=1..floor((n+1)/2)).
a(n) = A006603(n+1) - A006318(n+1).

A333473 a(n) = [x^n] ( S(x/(1 + x)) )^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 2, 12, 92, 752, 6352, 54768, 478928, 4231424, 37680320, 337622912, 3040354176, 27492359936, 249463806464, 2270319909632, 20714443816192, 189418898063360, 1735482632719360, 15928224355854336, 146414296847992832, 1347721096376573952, 12421053168197722112
Offset: 0

Views

Author

Peter Bala, Mar 23 2020

Keywords

Comments

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence s(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: s(n*p^k) == s(n*p^(k-1)) ( mod p^(k) ) for any prime p and positive integers n and k. For certain power series F(x) we may get stronger congruences. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = S(x/(1 + x)) = 1 + 2*x + 4*x^2 + 12*x^3 + 40*x^4 + 154*x^5 + 544*x^6 + ...(see A025227), where S(x) is the o.g.f. of the large Schröder numbers A006318. We conjecture that the associated sequence a(n) = [x^n] ( S(x/(1 + x)) )^n satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333472.
More generally, we conjecture that for a positive integer r and integer s, the sequence a(r,s;n) := [x^(r*n)] ( S(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note the sequence b(n) := [x^n] ( S(x) )^n = A103885(n) appears to satisfy the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*S(x/(1 + x)) )^n = A333090(n) also appears to satisfy the same congruences.

Examples

			Examples of congruences:
a(11) - a(1) = 3040354176 - 2 = 2*(11^2)*13*966419 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 12421053168197722112 - 92 = (2^2)*(3^7)*5*(7^2)* 5795401942927 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 90551762251592215396352 - 6352 = (2^4)*(5^4)*293* 30905038311123623 == 0 ( mod 5^4 ).
		

Crossrefs

Main diagonal of A378317.

Programs

  • Maple
    Sch := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
    G := x → Sch(x/(1+x));
    H := (x, n) -> series(G(x)^n, x, 51):
    seq(coeff(H(x, n), x, n), n = 0..25)
  • Mathematica
    Table[SeriesCoefficient[((1 - Sqrt[1- 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(n+2*k-1, 2*k)); \\ Seiichi Manyama, Nov 24 2024

Formula

a(n) = [x^n] ( (1 - sqrt(1- 4*x - 4*x^2))/(2*x) )^n.
a(n) ~ sqrt(((sqrt(2) + 1)^(2/3) + (sqrt(2) - 1)^(2/3) - 1)/3) * ((3*(71 + 8*sqrt(2))^(1/3) + 3*(71 - 8*sqrt(2))^(1/3) + 13))^n / (sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+2*k-1,2*k). - Seiichi Manyama, Nov 24 2024

A108891 Triangle read by rows: T(n,k) = number of Schroeder (or royal) n-paths (A006318) containing k returns to the diagonal y=x. (A northeast step lying on y=x contributes a return.)

Original entry on oeis.org

2, 2, 4, 6, 8, 8, 22, 28, 24, 16, 90, 112, 96, 64, 32, 394, 484, 416, 288, 160, 64, 1806, 2200, 1896, 1344, 800, 384, 128, 8558, 10364, 8952, 6448, 4000, 2112, 896, 256, 41586, 50144, 43392, 31616, 20160, 11264, 5376, 2048, 512
Offset: 1

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Author

David Callan, Jul 25 2005

Keywords

Examples

			Table begins
  n\k  1    2    3    4    5    6
  -------------------------------
  1 |  2
  2 |  2    4
  3 |  6    8    8
  4 | 22   28   24   16
  5 | 90  112   96   64   32
  6 |394  484  416  288  160   64
The paths DD, END, DEN, ENEN each have 2 returns (E=east, N=north, D=northeast); so T(2,2)=4.
From _Philippe Deléham_, Nov 02 2013: (Start)
Triangle (0, 1, 2, 1, 2, 1, 2, ...) DELTA (1, 0, 0, 0, ...) begins:
  1;
  0,   2;
  0,   2,   4;
  0,   6,   8,   8;
  0,  22,  28,  24,  16;
  0,  90, 112,  96,  64,  32;
  0, 394, 484, 416, 288, 160,  64; (End)
		

Crossrefs

Row sums are the large Schroeder numbers A006318. Column k=1 is twice the little Schroeder numbers A001003. The main diagonal consists of powers of 2, A000079. The first subdiagonal is A036289. The analogous Catalan triangle is A009766 (with rows reversed).

Programs

  • Mathematica
    T[n_, k_] := (-1)^(n - k) Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, 2]; Table[T[n - 1, k - 1]*2^k, {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Sep 21 2022, after Peter Luschny at A104219 *)

Formula

Column k is the k-fold convolution of column 1.
T(n, k) = A104219(n-1, k-1)*2^k. - Philippe Deléham, Jul 31 2005
Triangle T(n,k), 1 <= k <= n, read by rows given by (0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 02 2013

A138462 A bisection of A006318.

Original entry on oeis.org

1, 6, 90, 1806, 41586, 1037718, 27297738, 745387038, 20927156706, 600318853926, 17518619320890, 518431875418926, 15521467648875090, 469286147871837366, 14308406109097843626, 439442782615614361662, 13582285614213903189954, 422158527806921249683014
Offset: 0

Views

Author

N. J. A. Sloane, May 08 2008

Keywords

Programs

  • Maxima
    a(n):=sum(binomial(2*n+1,i+1)*binomial(2*n+i,i),i,0,2*n)/(2*n+1); /* Vladimir Kruchinin, Apr 02 2017 */
    
  • PARI
    a(n) = sum(i=0, 2*n, binomial(2*n+1,i+1)*binomial(2*n+i,i))/(2*n+1) \\ Colin Barker, Dec 28 2017

Formula

a(n) = A006318(2n).
a(n) = Sum_{i=0..2n} (C(2*n+1,i+1)*C(2*n+i,i))/(2*n+1). - Vladimir Kruchinin, Apr 02 2017
D-finite with recurrence +n*(2*n+1)*a(n) +(-70*n^2+73*n-15)*a(n-1) +(70*n^2-277*n+270)*a(n-2) -(2*n-5)*(n-3)*a(n-3)=0. - R. J. Mathar, Mar 25 2024
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