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.

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A027307 Number of paths from (0,0) to (3n,0) that stay in first quadrant (but may touch horizontal axis) and where each step is (2,1), (1,2) or (1,-1).

Original entry on oeis.org

1, 2, 10, 66, 498, 4066, 34970, 312066, 2862562, 26824386, 255680170, 2471150402, 24161357010, 238552980386, 2375085745978, 23818652359682, 240382621607874, 2439561132029314, 24881261270812490, 254892699352950850
Offset: 0

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Author

Keywords

Comments

These are the 3-Schroeder numbers according to Yang-Jiang (2021). - N. J. A. Sloane, Mar 28 2021
Equals row sums of triangle A104978 which has g.f. F(x,y) that satisfies: F = 1 + x*F^2 + x*y*F^3. - Paul D. Hanna, Mar 30 2005
a(n) counts ordered complete ternary trees with 2*n + 1 leaves, where the internal vertices come in two colors and such that each vertex and its rightmost child have different colors. See [Drake, Example 1.6.9]. An example is given below. - Peter Bala, Sep 29 2011
a(n) for n >= 1 is the number of compact coalescent histories for matching lodgepole gene trees and species trees with n cherries and 2n+1 leaves. - Noah A Rosenberg, Jun 21 2022
a(n) is the maximum number of distinct sets that can be obtained as complete parenthesizations of “S_1 union S_2 intersect S_3 union S_4 intersect S_5 union ... union S_{2*n} intersect S_{2*n+1}”, where n union and n intersection operations alternate, starting with a union, and S_1, S_2, ... , S_{2*n+1} are sets. - Alexander Burstein, Nov 22 2023

Examples

			a(2) = 10. Internal vertices colored either b(lack) or w(hite); 5 uncolored leaf vertices shown as o.
........b...........b.............w...........w.....
......./|\........./|\.........../|\........./|\....
....../.|.\......./.|.\........./.|.\......./.|.\...
.....b..o..o.....o..b..o.......w..o..o.....o..w..o..
..../|\............/|\......../|\............/|\....
.../.|.\........../.|.\....../.|.\........../.|.\...
..o..o..o........o..o..o....o..o..o........o..o..o..
....................................................
........b...........b.............w...........w.....
......./|\........./|\.........../|\........./|\....
....../.|.\......./.|.\........./.|.\......./.|.\...
.....w..o..o.....o..w..o.......b..o..o.....o..b..o..
..../|\............/|\......../|\............/|\....
.../.|.\........../.|.\....../.|.\........../.|.\...
..o..o..o........o..o..o....o..o..o........o..o..o..
....................................................
........b...........w..........
......./|\........./|\.........
....../.|.\......./.|.\........
.....o..o..w.....o..o..b.......
........../|\........./|\......
........./.|.\......./.|.\.....
........o..o..o.....o..o..o....
...............................
From _Alexander Burstein_, Feb 14 2025: (Start)
a(2) = 10 as the maximum number of distinct sets obtained as complete parenthesizations of S_1 u(nion) S_2 (i)n(tersect) S_3 u(nion) S_4 (i)n(tersect) S_5:
S_1 u (S_2 n (S_3 u (S_4 n S_5))),
S_1 u (S_2 n ((S_3 u S_4) n S_5)) = S_1 u ((S_2 n (S_3 u S_4)) n S_5),
S_1 u ((S_2 n S_3) u (S_4 n S_5)) = (S_1 u (S_2 n S_3)) u (S_4 n S_5),
S_1 u (((S_2 n S_3) u S_4) n S_5),
(S_1 u S_2) n (S_3 u (S_4 n S_5)),
(S_1 u S_2) n ((S_3 u S_4) n S_5) = ((S_1 u S_2) n (S_3 u S_4)) n S_5,
((S_1 u S_2) n S_3) u (S_4 n S_5),
(S_1 u (S_2 n (S_3 u S_4))) n S_5,
(S_1 u ((S_2 n S_3) u S_4)) n S_5 = ((S_1 u (S_2 n S_3)) u S_4) n S_5,
(((S_1 u S_2) n S_3) u S_4) n S_5. (End)
		

References

  • Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.

Crossrefs

The sequences listed in Yang-Jiang's Table 1 appear to be A006318, A001003, A027307, A034015, A144097, A243675, A260332, A243676. - N. J. A. Sloane, Mar 28 2021
Apart from first term, this is 2*A034015. - N. J. A. Sloane, Mar 28 2021

Programs

  • Mathematica
    a[n_] := ((n+1)*(2n)!*Hypergeometric2F1[-n, 2n+1, n+2, -1]) / (n+1)!^2;
    Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Nov 14 2011, after Pari *)
    a[n_] := If[n == 0, 1, 2*Hypergeometric2F1[1 - n, -2 n, 2, 2]];
    Table[a[n], {n, 0, 19}]  (* Peter Luschny, Nov 08 2021 *)
  • PARI
    a(n)=if(n<1,n==0,sum(i=0,n-1,2^(i+1)*binomial(2*n,i)*binomial(n,i+1))/n)
    
  • PARI
    a(n)=sum(k=0,n,binomial(2*n+k,n+2*k)*binomial(n+2*k,k)/(n+k+1)) \\ Paul D. Hanna
    
  • PARI
    a(n)=sum(k=0,n, binomial(n,k)*binomial(2*n+k+1,n)/(2*n+k+1) ) /* Michael Somos, May 23 2005 */

Formula

G.f.: (2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3.
a(n) = (1/n) * Sum_{i=0..n-1} 2^(i+1)*binomial(2*n, i)*binomial(n, i+1), n>0.
a(n) = 2*A034015(n-1), n>0.
a(n) = Sum_{k=0..n} C(2*n+k, n+2*k)*C(n+2*k, k)/(n+k+1). - Paul D. Hanna, Mar 30 2005
Given g.f. A(x), y=A(x)x satisfies 0=f(x, y) where f(x, y)=x(x-y)+(x+y)y^2 . - Michael Somos, May 23 2005
Series reversion of x(Sum_{k>=0} a(k)x^k) is x(Sum_{k>=0} A085403(k)x^k).
G.f. A(x) satisfies A(x)=A006318(x*A(x)). - Vladimir Kruchinin, Apr 18 2011
The function B(x) = x*A(x^2) satisfies B(x) = x+x*B(x)^2+B(x)^3 and hence B(x) = compositional inverse of x*(1-x^2)/(1+x^2) = x+2*x^3+10*x^5+66*x^7+.... Let f(x) = (1+x^2)^2/(1-4*x^2+x^4) and let D be the operator f(x)*d/dx. Then a(n) equals 1/(2*n+1)!*D^(2*n)(f(x)) evaluated at x = 0. For a refinement of this sequence see A196201. - Peter Bala, Sep 29 2011
D-finite with recurrence: 2*n*(2*n+1)*a(n) = (46*n^2-49*n+12)*a(n-1) - 3*(6*n^2-26*n+27)*a(n-2) - (n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(50+30*sqrt(5))*((11+5*sqrt(5))/2)^n/(20*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 08 2012. Equivalently, a(n) ~ phi^(5*n + 1) / (2 * 5^(1/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
a(n) = 2*hypergeom([1 - n, -2*n], [2], 2) for n >= 1. - Peter Luschny, Nov 08 2021
From Peter Bala, Jun 16 2023: (Start)
P-recursive: n*(2*n + 1)*(5*n - 7)*a(n) = (110*n^3 - 264*n^2 + 181*n - 36)*a(n-1) + (n - 2)*(2*n - 3)*(5*n - 2)*a(n-2) with a(0) = 1 and a(1) = 2.
The g.f. A(x) = 1 + 2*x + 10*x^2 + 66*x^3 + ... satisfies A(x)^2 = (1/x) * the series reversion of x*((1 - x)/(1 + x))^2.
Define b(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = (1/2) * [x^n] ((1 + x)/(1 - x))^(2*n) = A103885(n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ). (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 09 2023

A035607 Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 8, 2, 1, 8, 18, 12, 2, 1, 10, 32, 38, 16, 2, 1, 12, 50, 88, 66, 20, 2, 1, 14, 72, 170, 192, 102, 24, 2, 1, 16, 98, 292, 450, 360, 146, 28, 2, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 1, 20, 162, 688, 1666, 2364, 1970, 952, 258, 36, 2, 1, 22, 200, 978, 2816
Offset: 0

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Keywords

Comments

Table also gives coordination sequences of same lattices.
Rows sums are given by A001333. Rising and falling diagonals are the tribonacci numbers A000213, A001590. - Paul Barry, Feb 13 2003
a(d,m) also gives the number of ways to choose m squares from a 2 X (d-1) grid so that no two squares in the selection are (horizontally or vertically) adjacent. - Jacob A. Siehler, May 13 2006
Mirror image of triangle A113413. - Philippe Deléham, Oct 15 2006
The Ca1 sums lead to A126116 and the Ca2 sums lead to A070550, see A180662 for the definitions of these triangle sums. - Johannes W. Meijer, Aug 05 2011
A035607 is jointly generated with the Delannoy triangle A008288 as an array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 05 2012
Also, the polynomial v(n,x) above is x + (x + 1)*f(n-1,x), where f(0,x) = 1. - Clark Kimberling, Oct 24 2014
Rows also give the coefficients of the independence polynomial of the n-ladder graph. - Eric W. Weisstein, Dec 29 2017
Considering both sequences as square arrays (offset by one row), the rows of A035607 are the first differences of the rows of A008288, and the rows of A008288 are the partial sums of the rows of A035607. - Shel Kaphan, Feb 23 2023
Considering only points with nonnegative coordinates, the number of points at L1 distance = m in d dimensions is the same as the number of ways of putting m indistinguishable balls into d distinguishable urns, binomial(m+d-1, d-1). This is one facet of the cross-polytope. Allowing for + and - coordinates, there are binomial(d,i)*2^i facets containing points with up to i nonzero coordinates. Eliminating double counting of points with any coordinates = 0, there are Sum_{i=1..d} (-1)^(d-i)*binomial(m+i-1,i-1)*binomial(d,i)*2^i points at distance m in d dimensions. One may avoid the alternating sum by using binomial(m-1,i-1) to count only the points per facet with exactly i nonzero coordinates, avoiding any double counting, but the result is the same. - Shel Kaphan, Mar 04 2023

Examples

			From _Clark Kimberling_, Oct 24 2014: (Start)
As a triangle of coefficients in polynomials v(n,x) in Comments, the first 6 rows are
  1
  1   2
  1   4   2
  1   6   8   2
  1   8  18  12   2
  1  10  32  38  16   2
  ... (End)
From _Shel Kaphan_, Mar 04 2023: (Start)
For d=3, m=4:
There are binomial(3,1)*2^1 = 6 facets (vertices) of binomial(4+1-1,1-1) = 1 point with <= one nonzero coordinate.
There are binomial(3,2)*2^2 = 12 facets (edges) of binomial(4+2-1,2-1) = 5 points with <= two nonzero coordinates.
There are binomial(3,3)*2^3 = 8 facets (faces) of binomial(4+3-1,3-1) = 15 points with <= three nonzero coordinates.
a(3,4) = 8*15 - 12*5 + 6*1 = 120 - 60 + 6 = 66. (End)
		

Crossrefs

Other versions: A113413, A119800, A122542, A266213.
Cf. A008288, which has g.f. 1/(1-x-x*y-x^2*y).
Cf. A078057 (row sums), A050146 (central terms).
Cf. A050146.

Programs

  • Haskell
    a035607 n k = a035607_tabl !! n !! k
    a035607_row n = a035607_tabl !! n
    a035607_tabl = map fst $ iterate
       (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                          zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 2])
    -- Reinhard Zumkeller, Jul 20 2013
    
  • Maple
    A035607 := proc(d,m) local j: add(binomial(floor((d-1+j)/2),d-m-1)*binomial(d-m-1, floor((d-1-j)/2)),j=0..d-1) end: seq(seq(A035607(d,m),m=0..d-1),d=1..11); # d=dimension, m=norm # Johannes W. Meijer, Aug 05 2011
  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
    v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A008288 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A035607 *)
    (* Clark Kimberling, Mar 09 2012 *)
    Reverse /@ CoefficientList[CoefficientList[Series[(1 + x)/(1 - x - x y - x^2 y), {x, 0, 10}], x], y] // Flatten (* Eric W. Weisstein, Dec 29 2017 *)
  • PARI
    T(n, k) = if (k==0, 1, sum(i=0, k-1, binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1)));
    tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(T(n, k), ", ")); print); \\ as a triangle; Michel Marcus, Feb 27 2018
  • Sage
    def A035607_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return prec(n-1,k-1)+2*sum(prec(n-i,k-1) for i in (2..n-k+1))
        return [prec(n, n-k) for k in (0..n-1)]
    for n in (1..10): print(A035607_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

From Johannes W. Meijer, Aug 05 2011: (Start)
f(d,m) = Sum_{j=0..d-1} binomial(floor((d-1+j)/2), d-m-1)*binomial(d-m-1, floor((d-1-j)/2)), d >= 1 and 0 <= m <= d-1.
f(d,m) = f(d-1,m-1) + f(d-1,m) + f(d-2,m-1) (d >= 3 and 1 <= m <= d-1) with f(d,0) = 1 (d >= 1) and f(d,d-1) = 2 (d>=2). (End)
From Roger Cuculière, Apr 10 2006: (Start)
The generating function G(x,y) of this double sequence is the sum of a(n,p)*x^n*y^p, n=1..oo, p=0..oo, which is G(x,y) = x*(1+y)/(1-x-y-x*y).
The horizontal generating function H_n(y), which generates the rows of the table: (1, 2, 2, 2, 2, ...), (1, 4, 8, 12, 16, ...), (1, 6, 18, 38, 66, ...), is the sum of a(n,p)*y^p, p=0..oo, for each fixed n. This is H_n(y) = ((1+y)^n)/((1-y)^n).
The vertical generating function V_p(x), which generates the columns of the table: (1, 1, 1, 1, 1, ...), (2, 4, 6, 8, 10, ...), (2, 8, 18, 32, 50, ...), is the sum of a(n,p)*x^n, n=1..oo, for each fixed p. This is V_p(x) = 2*((1+x)^(p-1))/((1-x)^(p+1)) for p >= 1 and V_0(x) = x/(1-x). (End)
G.f.: (1+x)/(1-x-x*y-x^2*y). - Vladeta Jovovic, Apr 02 2002 (But see previous lines!)
T(2*n,n) = A050146(n+1). - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle read by rows: T(n,0) = 1, for n > 1: T(n,n-1) = 2, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-1), 0 < k < n. - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle T(n,k) with 0 <= k < n read by rows: T(n,0)=1 for n > 0 and T(n,k) = Sum_{i=0..k-1} binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1) for k > 0. - Werner Schulte, Feb 22 2018
With p >= 1 and q >= 0, as a square array a(p,q) = T(p+q-1,q) = 2*p*Hypergeometric2F1[1-p, 1-q, 2, 2] for q >= 1. Consequently, a(p,q) = a(q,p)*p/q. - Shel Kaphan, Feb 14 2023
For n >= 1, T(2*n,n) = A002003(n), T(3*n,2*n) = A103885(n) and T(4*n,3*n) = A333715(n). - Peter Bala, Jun 15 2023

Extensions

More terms from David W. Wilson
Maple program corrected and information added by Johannes W. Meijer, Aug 05 2011

A033877 Triangular array read by rows associated with Schroeder numbers: T(1,k) = 1; T(n,k) = 0 if k < n; T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k).

Original entry on oeis.org

1, 1, 2, 1, 4, 6, 1, 6, 16, 22, 1, 8, 30, 68, 90, 1, 10, 48, 146, 304, 394, 1, 12, 70, 264, 714, 1412, 1806, 1, 14, 96, 430, 1408, 3534, 6752, 8558, 1, 16, 126, 652, 2490, 7432, 17718, 33028, 41586, 1, 18, 160, 938, 4080, 14002, 39152, 89898, 164512, 206098
Offset: 1

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Author

Keywords

Comments

A106579 is in some ways a better version of this sequence, but since this was entered first it will be the main entry for this triangle.
The diagonals of this triangle are self-convolutions of the main diagonal A006318: 1, 2, 6, 22, 90, 394, 1806, ... . - Philippe Deléham, May 15 2005
From Johannes W. Meijer, Sep 22 2010, Jul 15 2013: (Start)
Note that for the terms T(n,k) of this triangle n indicates the column and k the row.
The triangle sums, see A180662, link Schroeder's triangle with several sequences, see the crossrefs. The mirror of this triangle is A080247.
Quite surprisingly the Kn1p sums, p >= 1, are all related to A026003 and crystal ball sequences for n-dimensional cubic lattices (triangle offset is 0): Kn11(n) = A026003(n), Kn12(n) = A026003(n+2) - 1, Kn13(n) = A026003(n+4) - A005408(n+3), Kn14(n) = A026003(n+6) - A001844(n+4), Kn15(n) = A026003(n+8) - A001845(n+5), Kn16(n) = A026003(n+10) - A001846(n+6), Kn17(n) = A026003(n+12) - A001847(n+7), Kn18(n) = A026003(n+14) - A001848(n+8), Kn19(n) = A026003(n+16) - A001849(n+9), Kn110(n) = A026003(n+18) - A008417(n+10), Kn111(n) = A026003(n+20) - A008419(n+11), Kn112(n) = A026003(n+22) - A008421(n+12). (End)
T(n,k) is the number of normal semistandard Young tableaux with two columns, one of height k and one of height n. The recursion can be seen by performing jeu de taquin deletion on all instances of the smallest value. (If there are two instances of the smallest value, jeu de taquin deletion will always shorten the right column first and the left column second.) - Jacob Post, Jun 19 2018

Examples

			Triangle starts:
  1;
  1,    2;
  1,    4,    6;
  1,    6,   16,   22;
  1,    8,   30,   68,   90;
  1,   10,   48,  146,  304,  394;
  1,   12,   70,  264,  714, 1412, 1806;
  ... - _Joerg Arndt_, Sep 29 2013
		

Crossrefs

Essentially same triangle as A080247 and A080245 but with rows read in reversed order. Also essentially the same triangle as A106579.
Cf. A001003 (row sums), A026003 (antidiagonal sums).
Triangle sums (see the comments): A001003 (Row1, Row2), A026003 (Kn1p, p >= 1), A006603 (Kn21), A227504 (Kn22), A227505 (Kn23), A006603(2*n) (Kn3), A001850 (Kn4), A227506 (Fi1), A010683 (Fi2).

Programs

  • Haskell
    a033877 n k = a033877_tabl !! n !! k
    a033877_row n = a033877_tabl !! n
    a033877_tabl = iterate
       (\row -> scanl1 (+) $ zipWith (+) ([0] ++ row) (row ++ [0])) [1]
    -- Reinhard Zumkeller, Apr 17 2013
    
  • Magma
    function t(n,k)
      if k le 0 or k gt n then return 0;
      elif k eq 1 then return 1;
      else return t(n,k-1) + t(n-1,k-1) + t(n-1,k);
      end if;
    end function;
    [t(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Mar 23 2023
  • Maple
    T := proc(n, k) option remember; if n=1 then return(1) fi; if kJohannes W. Meijer, Sep 22 2010, revised Jul 17 2013
  • Mathematica
    T[1, ]:= 1; T[n, k_]/;(k
    				
  • Sage
    def A033877_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return prec(n-1,k-1)-2*sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^k*prec(n, n-k) for k in (0..n-1)]
    for n in (1..10): print(A033877_row(n)) # Peter Luschny, Mar 16 2016
    
  • SageMath
    @CachedFunction
    def t(n, k): # t = A033847
        if (k<0 or k>n): return 0
        elif (k==1): return 1
        else: return t(n, k-1) + t(n-1, k-1) + t(n-1, k)
    flatten([[t(n,k) for k in range(1,n+1)] for n in range(1, 16)]) # G. C. Greubel, Mar 23 2023
    

Formula

As an upper right triangle: a(n, k) = a(n, k-1) + a(n-1, k-1) + a(n-1, k) if k >= n >= 0 and a(n, k) = 0 otherwise.
G.f.: Sum T(n, k)*x^n*y^k = (1-x*y-(x^2*y^2-6*x*y+1)^(1/2)) / (x*(2*y+x*y-1+(x^2*y^2-6*x*y+1)^(1/2))). - Vladeta Jovovic, Feb 16 2003
Another version of A000007 DELTA [0, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] = 1, 1, 0, 1, 2, 0, 1, 4, 6, 0, 1, 6, 16, 22, 0, 1, ..., where DELTA is Deléham's operator defined in A084938.
Sum_{n=1..floor((k+1)/2)} T(n+p-1, k-n+p) = A026003(2*p+k-3) - A008288(2*p+k-3, p-2), p >= 2, k >= 1. - Johannes W. Meijer, Sep 28 2013
From G. C. Greubel, Mar 23 2023: (Start)
(t(n, k) as a lower triangle)
t(n, k) = t(n, k-1) + t(n-1, k-1) + t(n-1, k) with t(n, 1) = 1.
t(n, n) = A006318(n-1).
t(2*n-1, n) = A330801(n-1).
t(2*n-2, n) = A103885(n-1), n > 1.
Sum_{k=1..n-1} t(n, k) = A238112(n), n > 1.
Sum_{k=1..n} t(n, k) = A001003(n).
Sum_{k=1..n-1} (-1)^(k-1)*t(n, k) = (-1)^n*A001003(n-1), n > 1.
Sum_{k=1..n} (-1)^(k-1)*t(n, k) = A080243(n-1).
Sum_{k=1..floor((n+1)/2)} t(n-k+1, k) = A026003(n-1). (End)

Extensions

More terms from David W. Wilson

A002003 a(n) = 2 * Sum_{k=0..n-1} binomial(n-1, k)*binomial(n+k, k).

Original entry on oeis.org

0, 2, 8, 38, 192, 1002, 5336, 28814, 157184, 864146, 4780008, 26572086, 148321344, 830764794, 4666890936, 26283115038, 148348809216, 838944980514, 4752575891144, 26964373486406, 153196621856192, 871460014012682, 4962895187697048, 28292329581548718
Offset: 0

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Author

Keywords

Comments

a(n) is the number of order-preserving partial self maps of {1,...,n}. For example, a(2) = 8 because there are 8 order-preserving partial self maps of {1,2}: (1 2), (1 1), (2 2), (1 -), (2 -), (- 1), (- 2), (- -). Here for example (2 -) represents the partial map which maps 1 to 2 but does not include 2 in its domain. - James East, Oct 25 2005
From Peter Bala, Mar 02 2020: (Start)
For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(2*m,n)*b(n+1) + P(2*m,-n)*b(n-1) = Q(2*m,n)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial Q(2*m,n) is an even function of n; its 2*m zeros seem to belong to the interval [-1, 1] and 2*m - 2 of these zeros appear to lie close to the rational numbers of the form +-(2*k + 1)/(2*m), where 0 <= k <= m - 2. Cf. A103885. (End)
a(n), n>0, is the number of points at L1 distance = n from any given point in Z^n. The sequence is also the difference between the central diagonal (A001850) and +-1 diagonal (A002002) of the Delannoy number triangle (A008288). - Shel Kaphan, Feb 15 2023

Examples

			G.f. = 2*x + 8*x^2 + 38*x^3 + 192*x^4 + 1002*x^5 + 5336*x^6 + 28814*x^7 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A064861 := proc(n,k) option remember; if n = 1 then 1; elif k = 0 then 0; else A064861(n,k-1)+(3/2-1/2*(-1)^(n+k))*A064861(n-1,k); fi; end; seq(A064861(i,i-1),i=1..40);
  • Mathematica
    Flatten[{0,Table[SeriesCoefficient[((1+x)/Sqrt[1-6*x+x^2]-1)/2,{x,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Oct 04 2012 *)
    a[ n_] := If[ n < 1, 0, Hypergeometric2F1[ n, -n, 1, -1]]; (* Michael Somos, Aug 24 2014 *)
    Table[2*Sum[Binomial[n-1,k]Binomial[n+k,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Sep 18 2024 *)
  • PARI
    {a(n) = if( n<1, 0, polcoeff( ((1 - x^2) / (1 - x)^2 + x * O(x^n))^n, n))} /* Michael Somos, Sep 24 2003 */
    
  • Python
    from math import comb
    def A002003(n): return sum(comb(n,k)**2*k<Chai Wah Wu, Mar 22 2023

Formula

a(n) = 2*A047781(n).
From Vladeta Jovovic, Mar 28 2004: (Start)
G.f.: ((1+x)/sqrt(1-6*x+x^2)-1)/2.
E.g.f.: exp(3*x)*(2*BesselI(0, 2*sqrt(2)*x)+sqrt(2)*BesselI(1, 2*sqrt(2)*x)). (End)
a(n) = T(n, n-1), array T as in A064861.
a(n) = T(n, n-2), array T as in A049600.
a(n+1) = A110110(2n+1). - Tilman Neumann, Feb 05 2009
a(n) = 2 * JacobiP(n-1,0,1,3) = ((7*n+3)*LegendreP(n,3) - (n+1)*LegendreP(n+1,3)) /(2*n) for n > 0. - Mark van Hoeij, Jul 12 2010
Logarithmic derivative of A006318, the large Schroeder numbers. - Paul D. Hanna, Oct 25 2010
D-finite with recurrence: 4*(3*n^2-6*n+2)*a(n-1) - (n-2)*(2*n-1)*a(n-2) - n*(2*n-3)*a(n)=0. - Vaclav Kotesovec, Oct 04 2012
a(n) ~ (3+2*sqrt(2))^n/(2^(3/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 04 2012
Recurrence (an alternative): n*a(n) = (6-n)*a(n-6) + 2*(5*n-27)*a(n-5) + (84-15*n)*a(n-4) + 52*(3-n)*a(n-3) + 3*(2-5*n)*a(n-2) + 2*(5*n-3)*a(n-1), n>=7. - Fung Lam, Feb 05 2014
a(n) = Hyper2F1([-n, n], [1], -1) for n > 0. - Peter Luschny, Aug 02 2014
a(n) = [x^n] ((1+x)/(1-x))^n for n > 0. - Seiichi Manyama, Jun 07 2018
From Peter Bala, Mar 13 2020: (Start)
a(n) = 2 * Sum_{k = 0..n-1} 2^k*C(n,k+1)*C(n-1,k).
a(n) = 2 * (-1)^(n+1) * Sum_{k = 0..n-1} (-2)^k*C(n+k,n-1)*C(n-1,k).
a(n) = Sum_{k = 0..n} C(n,k)*C(2*n-k-1,n-1).
Conjecture: a(n) = - [x^n] (1 - F(x))^n, where F(x) = 2*x + 6*x^2 + 34*x^3 + 238*x^4 + ... is the o.g.f. of A108424. Equivalently, a(n) = -[x^n](G(x))^(-n), where G(x) = 1 + 2*x + 10*x^2 + 66*x^3 + 498*x^4 + ... is the o.g.f. of A027307.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
a(n) = Sum_{k = 1..n} C(n, k) * C(n-1, k-1) * 2^k. - Michael Somos, May 23 2021
a(n) = A001850(n) - A002002(n), for n > 0. - Shel Kaphan, Feb 15 2023

Extensions

More terms from Barbara Haas Margolius (b.margolius(AT)csuohio.edu), Oct 10 2001

A119259 Central terms of the triangle in A119258.

Original entry on oeis.org

1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889
Offset: 0

Views

Author

Reinhard Zumkeller, May 11 2006

Keywords

Comments

The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 06 2022

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Haskell
    a119259 n = a119258 (2 * n) n  -- Reinhard Zumkeller, Aug 06 2014
    
  • Mathematica
    Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
  • Python
    from itertools import count, islice
    def A119259_gen(): # generator of terms
        yield from (1,3)
        a, c = 2, 1
        for n in count(1):
            yield (a<>1
    A119259_list = list(islice(A119259_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n,k). - Seiichi Manyama, Jul 31 2025

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

A080247 Formal inverse of triangle A080246. Unsigned version of A080245.

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 22, 16, 6, 1, 90, 68, 30, 8, 1, 394, 304, 146, 48, 10, 1, 1806, 1412, 714, 264, 70, 12, 1, 8558, 6752, 3534, 1408, 430, 96, 14, 1, 41586, 33028, 17718, 7432, 2490, 652, 126, 16, 1, 206098
Offset: 0

Views

Author

Paul Barry, Feb 15 2003

Keywords

Comments

Row sums are little Schroeder numbers A001003. Diagonal sums are generalized Fibonacci numbers A006603. Columns include A006318, A006319, A006320, A006321.
T(n,k) is the number of dissections of a convex (n+3)-gon by nonintersecting diagonals with exactly k diagonals emanating from a fixed vertex. Example: T(2,1)=4 because the dissections of the convex pentagon ABCDE having exactly one diagonal emanating from the vertex A are: {AC}, {AD}, {AC,EC} and {AD,BD}. - Emeric Deutsch, May 31 2004
For more triangle sums, see A180662, see the Schroeder triangle A033877 which is the mirror of this triangle. - Johannes W. Meijer, Jul 15 2013

Examples

			Triangle starts:
[0]    1
[1]    2,    1
[2]    6,    4,   1
[3]   22,   16,   6,   1
[4]   90,   68,  30,   8,  1
[5]  394,  304, 146,  48, 10,  1
[6] 1806, 1412, 714, 264, 70, 12, 1
...
From _Gary W. Adamson_, Jul 25 2011: (Start)
n-th row = top row of M^n, M = the following infinite square production matrix:
  2, 1, 0, 0, 0, ...
  2, 2, 1, 0, 0, ...
  2, 2, 2, 1, 0, ...
  2, 2, 2, 2, 1, ...
  ... (End)
		

Crossrefs

Cf. A000007, A033877 (mirror), A084938.

Programs

  • Maple
    A080247:=(n,k)->(k+1)*add(binomial(n+1,k+j+1)*binomial(n+j,j),j=0..n-k)/(n+1):
    seq(seq(A080247(n,k),k=0..n),n=0..9);
  • Mathematica
    Clear[w] w[n_, k_] /; k < 0 || k > n := 0 w[0,0]=1 ; w[n_, k_] /; 0 <= k <= n && !n == k == 0 := w[n, k] = w[n-1,k-1] + w[n-1,k] + w[n,k+1] Table[w[n,k],{n,0,10},{k,0,n}] (* David Callan, Jul 03 2006 *)
    T[n_, k_] := Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, -1];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Peter Luschny, Jan 08 2018 *)
  • Maxima
    T(n,k):=((k+1)*sum(2^m*binomial(n+1,m)*binomial(n-k-1,n-k-m),m,0,n-k))/(n+1); /* Vladimir Kruchinin, Jan 10 2022 */
  • Sage
    def A080247_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return prec(n-1,k-1)-2*sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^(n-k)*prec(n, k) for k in (1..n)]
    for n in (1..10): print(A080247_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 2/(2+y*x-y+y*(x^2-6*x+1)^(1/2))/y/x. - Vladeta Jovovic, Feb 16 2003
Essentially same triangle as triangle T(n,k), n > 0 and k > 0, read by rows; given by [0, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] DELTA A000007 where DELTA is Deléham's operator defined in A084938.
T(n, k) = T(n-1, k-1) + 2*Sum_{j>=0} T(n-1, k+j) with T(0, 0) = 1 and T(n, k)=0 if k < 0. - Philippe Deléham, Jan 19 2004
T(n, k) = (k+1)*Sum_{j=0..n-k} (binomial(n+1, k+j+1)*binomial(n+j, j))/(n+1). - Emeric Deutsch, May 31 2004
Recurrence: T(0,0)=1; T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n,k+1). - David Callan, Jul 03 2006
T(n, k) = binomial(n, k)*hypergeom([k - n, n + 1], [k + 2], -1). - Peter Luschny, Jan 08 2018
T(n,k) = (k+1)/(n+1)*Sum_{m=0..n-k} 2^m*binomial(n+1,m)*binomial(n-k-1,n-k-m). - Vladimir Kruchinin, Jan 10 2022
From Peter Bala, Sep 16 2024: (Start)
Riordan array (S(x), x*S(x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the g.f. of the large Schröder numbers A006318.
For integer m and n >= 1, (m + 2)*[x^n] S(x)^(m*n) = m*[x^n] (1/S(-x))^((m+2)*n). For cases of this identity see A103885 (m = 1), A333481 (m = 2) and A370102 (m = 3). (End)

A103884 Square array A(n,k) read by antidiagonals: row n gives coordination sequence for lattice C_n.

Original entry on oeis.org

1, 1, 8, 1, 18, 16, 1, 32, 66, 24, 1, 50, 192, 146, 32, 1, 72, 450, 608, 258, 40, 1, 98, 912, 1970, 1408, 402, 48, 1, 128, 1666, 5336, 5890, 2720, 578, 56, 1, 162, 2816, 12642, 20256, 14002, 4672, 786, 64, 1, 200, 4482, 27008, 59906, 58728, 28610, 7392, 1026, 72
Offset: 2

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Examples

			Array begins:
  1,   8,    16,     24,      32,       40,        48, ... A022144;
  1,  18,    66,    146,     258,      402,       578, ... A010006;
  1,  32,   192,    608,    1408,     2720,      4672, ... A019560;
  1,  50,   450,   1970,    5890,    14002,     28610, ... A019561;
  1,  72,   912,   5336,   20256,    58728,    142000, ... A019562;
  1,  98,  1666,  12642,   59906,   209762,    596610, ... A019563;
  1, 128,  2816,  27008,  157184,   658048,   2187520, ... A019564;
  1, 162,  4482,  53154,  374274,  1854882,   7159170, ... A035746;
  1, 200,  6800,  97880,  822560,  4780008,  21278640, ... A035747;
  1, 242,  9922, 170610, 1690370, 11414898,  58227906, ... A035748;
  1, 288, 14016, 284000, 3281280, 25534368, 148321344, ... A035749;
  ...
Antidiagonals, T(n, k), begins as:
  1;
  1,   8;
  1,  18,   16;
  1,  32,   66,   24;
  1,  50,  192,  146,   32;
  1,  72,  450,  608,  258,   40;
  1,  98,  912, 1970, 1408,  402,  48;
  1, 128, 1666, 5336, 5890, 2720, 578, 56;
		

Crossrefs

Programs

  • Magma
    A103884:= func< n,k | k eq 0 select 1 else 2*(&+[2^j*Binomial(n-k,j+1)*Binomial(2*k-1,j) : j in [0..2*k-1]]) >;
    [A103884(n,k): k in [0..n-2], n in [2..12]]; // G. C. Greubel, May 23 2023
    
  • Mathematica
    nmin = 2; nmax = 11; t[n_, 0]= 1; t[n_, k_]:= 2n*Hypergeometric2F1[1-2k, 1-n, 2, 2]; tnk= Table[ t[n, k], {n, nmin, nmax}, {k, 0, nmax-nmin}]; Flatten[ Table[ tnk[[ n-k+1, k ]], {n, 1, nmax-nmin+1}, {k, 1, n} ] ] (* Jean-François Alcover, Jan 24 2012, after formula *)
  • SageMath
    def A103884(n,k): return 1 if k==0 else 2*sum(2^j*binomial(n-k,j+1)*binomial(2*k-1,j) for j in range(2*k))
    flatten([[A103884(n,k) for k in range(n-1)] for n in range(2,13)]) # G. C. Greubel, May 23 2023

Formula

A(n,k) = Sum_{i=1..2*k} 2^i*C(n, i)*C(2*k-1, i-1), A(n,0) = 1 (array).
G.f. of n-th row: (Sum_{i=0..n} C(2*n, 2*i)*x^i)/(1-x)^n.
T(n, k) = A(n-k, k) (antidiagonals).
T(n, n-2) = A022144(n-2).
T(n, k) = 2*(n-k)*Hypergeometric2F1([1+k-n, 1-2*k], [2], 2), T(n, 0) = 1, for n >= 2, 0 <= k <= n-2. - G. C. Greubel, May 23 2023
From Peter Bala, Jul 09 2023: (Start)
T(n,k) = [x^k] Chebyshev_T(n, (1 + x)/(1 - x)), where Chebyshev_T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
T(n+1,k) = T(n+1,k-1) + 2*T(n,k) + 2*T(n,k-1) + T(n-1,k) - T(n-1,k-1). (End)

Extensions

Definition clarified by N. J. A. Sloane, May 25 2023

A333715 a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.

Original entry on oeis.org

1, 2, 24, 326, 4672, 69002, 1038984, 15856206, 244396544, 3795731282, 59307908024, 931222155030, 14680871849152, 232236016459098, 3684420837693480, 58600075142247326, 934064636705476608, 14917333936933664674, 238641621366613695576, 3823510794994321546214, 61344017874989324388672
Offset: 0

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Author

Peter Bala, Apr 03 2020

Keywords

Comments

a(n) is also equal to [x^n] G(x)^n, where G(x) is the o.g.f. of A027307.
Compare with A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n and A103885(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318.
If we define an operator I acting on power series f(x) = 1 + f_1*x + f_2*x^2 + ... by I(f(x)) = 1/x * Revert( x/f(x) ) then S(x) = I( (1 + x)/(1 - x) ) and G(x) = (I o I)( (1 + x)/(1 - x )).
It can be shown that a(n) 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 a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples of these congruences are given below.
The same congruences may hold more generally for the sequences a(r,s,n) := [x^(r*n)]( (1 + x)/(1 - x) )^(s*n), r a positive integer and s an integer. This is the case a(3,1,n).
For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(6*m,n)*b(n+1) + P(6*m,-n)*b(n-1) = Q(3*m,n^2)*b(n), where the polynomials P(6*m,n) and Q(3*m,n^2) have degree 6*m. Conjecturally, the 6*m zeros of the polynomial Q(3*m,n^2) belong to the interval [-1, 1] and 6*m - 4 of these zeros appear to be approximated by the rational numbers +- k/(4*m), where 1 <= k <= 4*m - 3, k not a multiple of 4.

Examples

			Examples of congruences:
a(11) - a(1) = 931222155030 - 2 = (2^2)*(11^3)*163*1073069 == ( mod 11^3 )
a(3*7) - a(3) = 985413034951400888962602 - 326 = (2^2)*(7^4)*263* 390130947874776863 == 0 ( mod 7^3 )
a(5^2) - a(5) = 66292579025690123511768694002 - 69002 = (2^3)*(5^6)*39461* 13439614612035199009 == 0 ( mod 5^6 )
		

Crossrefs

Programs

  • Maple
    seq(add(binomial(n,k)*binomial(3*n+k-1,n-1), k = 0..n), n = 0..20);
  • Mathematica
    Table[Binomial[3*n-1, n-1] * Hypergeometric2F1[-n, 3*n, 2*n+1, -1], {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2020 *)

Formula

a(n) = Sum_{k = 0..n} C(n,k)*C(3*n+k-1,n-1).
a(n) = (1/3) * Sum_{k = 0..n} C(3*n,n-k)*C(3*n+k-1,k) for n >= 1.
a(n) = (1/3) * [x^n] ( (1 + x)/(1 - x) )^(3*n) for n >= 1.
a(n) = Sum_{k = 1..n} (2^k)*C(n,k)*C(3*n-1,k-1) for n >= 1.
P-recursive:
P(6,n)*a(n+1) + P(6,-n)*a(n-1) = Q(3,n^2)*a(n), where P(6,n) = (2*n-1)*(3*n+1)*(3*n+2)*(3*n+3)*(35*n^2 - 35*n + 6) and the polynomial Q(3,n) = 4*(7805*n^3 - 7132*n^2 + 1559*n - 72).
Congruences: a(p) == 2 ( mod p^3 ) for prime p >= 3.
a(n) ~ (223 + 70*sqrt(10))^n / (2^(3/4) * 5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Apr 04 2020
exp( Sum_{n>=1} a(n)*x^n/n ) = B(x) where B(x) = 1 + x*(B(x)^3 + B(x)^4) is the g.f. of A144097. - Paul D. Hanna, May 31 2023

A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(2*n).

Original entry on oeis.org

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501
Offset: 0

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Author

Peter Bala, Feb 27 2022

Keywords

Comments

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.
The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.
We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.
Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.
More generally, if r is a positive integer and s an integer then the sequence defined by u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

Examples

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)
		

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Maple
    seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);
  • Mathematica
    A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
    a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

Formula

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).
Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).
The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022
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