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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A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819
Offset: 0

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Author

Keywords

Comments

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] 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
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

Examples

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 78 and 163, #19.
  • L. Euler, Exemplum Memorabile Inductionis Fallacis, Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 15, p. 59.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 575.
  • P. Henrici, Applied and Computational Complex Analysis. Wiley, NY, 3 vols., 1974-1986. (Vol. 1, p. 42.)
  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 579.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 74.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 6.3.8.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 22.
  • Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346. See p. 341.

Crossrefs

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

Programs

  • Haskell
    a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013
    
  • Magma
    P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011
    
  • Maple
    A002426 := proc(n) local k;
        sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
    end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    # Alternatively:
    a := n -> simplify(GegenbauerC(n,-n,-1/2)):
    seq(a(n), n=0..29); # Peter Luschny, May 07 2016
  • Mathematica
    Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
    a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
    a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
    Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
    a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
    
  • Maxima
    makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};
    
  • PARI
    /* as lattice paths: same as in A092566 but use */
    steps=[[2, 0], [0, 2], [1, 1]];
    /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013
    
  • Python
    from math import comb
    def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022
  • Sage
    A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
    [simplify(A002426(n)) for n in (0..29)]
    # Peter Luschny, Sep 17 2014
    
  • Sage
    def A():
        a, b, n = 1, 1, 1
        yield a
        while True:
            yield b
            n += 1
            a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
    A002426 = A()
    print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016
    

Formula

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

A008459 Square the entries of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 36, 16, 1, 1, 25, 100, 100, 25, 1, 1, 36, 225, 400, 225, 36, 1, 1, 49, 441, 1225, 1225, 441, 49, 1, 1, 64, 784, 3136, 4900, 3136, 784, 64, 1, 1, 81, 1296, 7056, 15876, 15876, 7056, 1296, 81, 1, 1, 100, 2025, 14400, 44100, 63504, 44100, 14400, 2025, 100, 1
Offset: 0

Views

Author

Keywords

Comments

Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1), having k right turns. - Emeric Deutsch, Nov 23 2003
Product of A007318 and A105868. - Paul Barry, Nov 15 2005
Number of partitions that fit in an n X n box with Durfee square k. - Franklin T. Adams-Watters, Feb 20 2006
From Peter Bala, Oct 23 2008: (Start)
Narayana numbers of type B. Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for associahedra of type B_n. See A001263 for the h-vectors for associahedra of type A_n. The Hilbert transform of this triangular array is A108625 (see A145905 for the definition of this term).
Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i, j <= n + 1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the corresponding array of f-vectors for these type A_n polytopes. See A086645 for the array of h-vectors for type C_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
(End)
The n-th row consists of the coefficients of the polynomial P_n(t) = Integral_{s = 0..2*Pi} (1 + t^2 - 2*t*cos(s))^n/Pi/2 ds. For example, when n = 3, we get P_3(t) = t^6 + 9*t^4 + 9*t^2 + 1; the coefficients are 1, 9, 9, 1. - Theodore Kolokolnikov, Oct 26 2010
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. T(n,k) is the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 231, and 321. T(n,k) is also the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 312, and 321.
T(n,k) is the number of permutations of [n+1] with k descents that avoid the patterns 1342, 3142, 3412, and 3421. (End)
The number of convex polyominoes whose smallest bounding rectangle has size (k+1)*(n+1-k) and which contain the lower left corner of the bounding rectangle (directed convex polyominoes). - Günter Rote, Feb 27 2019
Let P be the poset [n] X [n] ordered by the product order. T(n,k) is the number of antichains in P containing exactly k elements. Cf. A063746. - Geoffrey Critzer, Mar 28 2020

Examples

			Pascal's triangle begins
  1
  1  1
  1  2   1
  1  3   3   1
  1  4   6   4   1
  1  5  10  10   5   1
  1  6  15  20  15   6   1
  1  7  21  35  35  21   7   1
...
so the present triangle begins
  1
  1   1
  1   4    1
  1   9    9     1
  1  16   36    16     1
  1  25  100   100    25    1
  1  36  225   400   225   36   1
  1  49  441  1225  1225  441  49   1
...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.
  • J. Riordan, An introduction to combinatorial analysis, Dover Publications, Mineola, NY, 2002, page 191, Problem 15. MR1949650
  • P. G. Tait, On the Linear Differential Equation of the Second Order, Proceedings of the Royal Society of Edinburgh, 9 (1876), 93-98 (see p. 97) [From Tom Copeland, Sep 09 2010, vol number corrected Sep 10 2010]

Crossrefs

Row sums are in A000984. Columns 0-3 are A000012, A000290, A000537, A001249.
Family of polynomials (see A062145): this sequence (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558, A108625 (Hilbert transform), A145903, A181543, A086645 (logarithmic derivative), A105868 (inverse binomial transform), A093118.

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n,k)^2))); # Muniru A Asiru, Mar 30 2018
    
  • Magma
    /* As triangle */ [[Binomial(n, k)^2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 15 2016
    
  • Maple
    seq(seq(binomial(n, k)^2, k=0..n), n=0..10);
  • Mathematica
    Table[Binomial[n, k]^2, {n, 0, 11}, {k, 0, n}]//Flatten (* Alonso del Arte, Dec 08 2013 *)
  • Maxima
    create_list(binomial(n,k)^2,n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • Maxima
    T(n,k):=if n=k then 1 else if k=0 then 1 else T(n-1,k)*(n+k)/(n-k)+T(n-1,k-1); /* Vladimir Kruchinin, Oct 18 2014 */
    
  • Maxima
    A(x,y):=1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2);
    taylor(x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2),x,0,7,y,0,7); /* Vladimir Kruchinin, Oct 23 2020 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, binomial(n, k)^2)}; /* Michael Somos, May 03 2004 */
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(sum(m=0,n,(2*m)!/m!^2*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(2*m+1)),n,x),k,y)} \\ Paul D. Hanna, Oct 31 2010
    
  • Python
    def A008459(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2))**2 # Chai Wah Wu, Nov 12 2024

Formula

T(n,k) = A007318(n,k)^2. - Sean A. Irvine, Mar 29 2018
E.g.f.: exp((1+y)*x)*BesselI(0, 2*sqrt(y)*x). - Vladeta Jovovic, Nov 17 2003
G.f.: 1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2); g.f. for row n: (1-t)^n P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been modified with the roles of x and y interchanged so that now x corresponds to n and y to k. - Petros Hadjicostas, Oct 22 2017]
G.f. for column k is Sum_{j = 0..k} C(k, j)^2*x^(k+j)/(1 - x)^(2*k+1). - Paul Barry, Nov 15 2005
Column k has g.f. (x^k)*Legendre_P(k, (1+x)/(1-x))/(1 - x)^(k+1) = (x^k)*Sum_{j = 0..k} C(k, j)^2*x^j/(1 - x)^(2*k+1). - Paul Barry, Nov 19 2005
Let E be the operator D*x*D, where D denotes the derivative operator d/dx. Then (1/n!^2) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} binomial(n+k, k)^2*x^k. For example, when n = 3 we have (1/3!)^2*E^3(1/(1 - x)) = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7 = (1/3!)^2 * Sum_{k >= 0} ((k+1)*(k+2)*(k+3))^2*x^k. - Peter Bala, Oct 23 2008
G.f.: A(x, y) = Sum_{n >= 0} (2*n)!/n!^2 * x^(2*n)*y^n/(1 - x - x*y)^(2*n+1). - Paul D. Hanna, Oct 31 2010
From Peter Bala, Jul 24 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Generating function: E(y)*E(x*y) = 1 + (1 + x)*y + (1 + 4*x + x^2)*y^2/2!^2 + (1 + 9*x + 9*x^2 + x^3)*y^3/3!^2 + .... Cf. the unsigned version of A021009 with generating function exp(y)*E(x*y).
The n-th power of this array has the generating function E(y)^n*E(x*y). In particular, the matrix inverse A055133 has the generating function E(x*y)/E(y). (End)
T(n,k) = T(n-1,k)*(n+k)/(n-k) + T(n-1,k-1), T(n,0) = T(n,n) = 1. - Vladimir Kruchinin, Oct 18 2014
Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives Pascal's triangle A007318. - Peter Bala, Dec 21 2014
n-th row polynomial R(n, x) = [z^n] (1 + (1 + x)*z + x*z^2)^n. Note that 1/n*[z^(n-1)] (1 + (1 + x)*z + x*z^2)^n gives the row polynomials of A001263. - Peter Bala, Jun 24 2015
Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2*(1 + t)*x + (1 - t)^2*x^2) denotes the o.g.f. of this array then 1 + x*d/dx log(G(x,t)) = 1 + (1 + t)*x + (1 + 6*t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter Bala, Sep 06 2015
T(n,k) = Sum_{i=0..n} C(n-i,k)*C(n,i)*C(n+i,i)*(-1)^(n-i-k). - Vladimir Kruchinin, Jan 14 2018
G.f. satisfies A(x,y) = x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2). - Vladimir Kruchinin, Oct 23 2020
G.f. satisfies the differential equation y * d^2(A(x,y))/dy^2 - x^2 * d^2(x*A(x,y))/dx^2 + 2*x^2* A(x,y)^3 = 0. - Sergii Voloshyn, Mar 07 2025
T(n,k) = Sum_{i=0..n} C(2*n+1,i)*C(n+k-i,n)^2*(-1)^i. - Natalia L. Skirrow, Apr 14 2025

A109187 Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 6, 0, 1, 6, 0, 12, 0, 1, 0, 30, 0, 20, 0, 1, 20, 0, 90, 0, 30, 0, 1, 0, 140, 0, 210, 0, 42, 0, 1, 70, 0, 560, 0, 420, 0, 56, 0, 1, 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1, 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1, 0, 2772, 0, 11550, 0, 9240, 0, 1980, 0, 110, 0, 1
Offset: 0

Views

Author

Emeric Deutsch, Jun 21 2005

Keywords

Comments

A Grand Motzkin path is a path in the half-plane x >= 0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).
From Peter Bala, Feb 11 2017: (Start)
Consider an infinite 1-dimensional integer lattice with an oriented self-loop at each vertex. Then T(n,k) equals the number of walks of length n from a vertex to itself having k loops. There is a bijection between such walks and Grand Motzkin paths which takes a right step and a left step on the lattice to an up step U and a down step D of a Grand Motzkin path respectively, and takes traversing a loop on the lattice to the horizontal step H. See A282252 for the corresponding triangle of walks on a 2-dimensional lattice with self-loops. (End)

Examples

			T(3,1)=6 because we have hud,hdu,udh,duh,uhd,dhu, where u=(1,1),d=(1,-1), h=(1,0).
Triangle begins:
n\k   [0]  [1]   [2]   [3]   [4]   [5]   [6]  [7]  [8]  [9] [10]
[0]    1;
[1]    0,   1;
[2]    2,   0,    1;
[3]    0,   6,    0,    1;
[4]    6,   0,   12,    0,    1;
[5]    0,  30,    0,   20,    0,    1;
[6]   20,   0,   90,    0,   30,    0,    1;
[7]    0, 140,    0,  210,    0,   42,    0,   1;
[8]   70,   0,  560,    0,  420,    0,   56,   0,   1;
[9]    0, 630,    0, 1680,    0,  756,    0,  72,   0,   1;
[10] 252,   0, 3150,    0, 4200,    0, 1260,   0,  90,   0,   1;
[11] ...
From _Peter Bala_, Feb 11 2017: (Start)
The infinitesimal generator begins
      0
      0    0
      2    0     0
      0    6     0     0
     -6    0    12     0     0
      0  -30     0    20     0   0
     80    0   -90     0    30   0   0
      0  560     0  -210     0  42   0  0
  -2310    0  2240     0  -420   0  56  0  0
  ....
and equals the generalized exponential Riordan array [log(Bessel_I(0,2x)),x], and so has integer entries. (End)
		

Crossrefs

Diagonal of rational function R(x, y, t) = 1/(1 - (x^2 + t*x*y + y^2)) with respect to x,y, i.e., T(n,k) = [(xy)^n*t^k] R(x,y,t). For t=0..7 we have the diagonals: A126869(t=0, column 0), A002426(t=1, row sums), A000984(t=2), A026375(t=3), A081671(t=4), A098409(t=5), A098410(t=6), A104454(t=7).

Programs

  • Maple
    G:=1/sqrt((1-t*z)^2-4*z^2):Gser:=simplify(series(G,z=0,15)): P[0]:=1: for n from 1 to 13 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 13 do seq(coeff(t*P[n],t^k),k=1..n+1) od;
    with(PolynomialTools): CL := p -> CoefficientList(simplify(p), x):
    C := (n,x) -> binomial(2*n,n)*hypergeom([-n,-n],[-n+1/2],1/2-x/4):
    seq(print(CL(C(n,x))), n=0..11); # Peter Luschny, Jan 23 2018
  • Mathematica
    p[0] := 1; p[n_] := GegenbauerC[n, -n , -x/2];
    Flatten[Table[CoefficientList[p[n], x], {n, 0, 11}]] (* Peter Luschny, Jan 23 2018 *)
  • PARI
    T(n,k) = if ((n-k)%2, 0, binomial(n,k)*binomial(n-k, (n-k)/2));
    concat(vector(12, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 06 2018

Formula

G.f.: 1/sqrt((1-tz)^2-4z^2).
Row sums yield the central trinomial coefficients (A002426).
T(2n+1, 0) = 0.
T(2n, 0) = binomial(2n,n) (A000984).
Sum_{k=0..n} k*T(n,k) = A109188(n).
Except for the order, same rows as those of A105868.
Column k has e.g.f. (x^k/k!)*Bessel_I(0,2x). - Paul Barry, Mar 11 2006
T(n,k) = binomial((n+k)/2,k)*binomial(n,(n+k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Sep 18 2007
Coefficient array of the polynomials P(n,x) = x^n*hypergeom([1/2-n/2,-n/2], [1], 4/x^2). - Paul Barry, Oct 04 2008
G.f.: 1/(1-xy-2x^2/(1-xy-x^2/(1-xy-x^2/(1-xy-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
From Paul Barry, Apr 21 2010: (Start)
Exponential Riordan array [Bessel_I(0,2x), x].
Coefficient array of the polynomials P(n,x) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k, k)*x^(n - 2k).
Diagonal sums are the aerated central Delannoy numbers (A001850 with interpolated zeros). (End)
From Peter Bala, Feb 11 2017: (Start)
T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))*(1 + (-1)^(n-k))/2.
T(n,k) = (n/k) * T(n-1,k-1).
T(n,k) = the coefficient of H^k in the expansion of (H + U + 1/U)^n.
n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k) * binomial(2*k,k) * t^(n-2*k) = coefficient of x^n in the expansion of (1 + t*x + x^2)^n.
R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(t - 2)^(n-k).
d/dt(R(n,t)) = n*R(n-1,t).
R(n,t) = (1/Pi) * Integral_{x = 0..Pi} (t + 2*cos(x))^n dx.
Moment representation on a finite interval: R(n,t) = 1/Pi * Integral_{x = t-2 .. t+2} x^n/sqrt((t + 2 - x)*(x - t + 2)) dx.
Recurrence: n*R(n,t) = t*(2*n - 1)*R(n-1,t) - (t^2 - 4)*(n - 1)*R(n-2,t) with R(0,t) = 1 and R(1,t) = t.
R(n,t) = A002426 (t = 1), A000984 (t = 2), A026375 (t = 3), A081671 (t = 4), A098409 (t = 5), A098410 (t = 6) and A104454(t = 7).
The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.
The polynomials R(n,1 + t) are the row polynomials of A171128. (End)
From Peter Luschny, Jan 23 2018: (Start)
These are the coefficients of the polynomials G(n, -n , -x/2) where G(n, a, x) denotes the n-th Gegenbauer polynomial.
These polynomials can also be expressed as C(n, x) = binomial(2*n,n)*hypergeom([-n, -n], [-n+1/2], 1/2-x/4). (End)

A182883 Number of weighted lattice paths of weight n having no (1,0)-steps of weight 1.

Original entry on oeis.org

1, 0, 1, 2, 1, 6, 7, 12, 31, 40, 91, 170, 281, 602, 1051, 1988, 3907, 7044, 13735, 25962, 48643, 94094, 177145, 338184, 647791, 1228812, 2356927, 4500678, 8595913, 16486966, 31521543, 60419872, 115870879, 222045160, 426275647, 818054654
Offset: 0

Views

Author

Emeric Deutsch, Dec 11 2010

Keywords

Comments

L_n is the set of lattice paths of weight n that start at (0,0) end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step with weight 1; an (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.
Diagonal of the rational function 1 / (1 - x^3 - y^3 - (x*y)^2). - Ilya Gutkovskiy, Apr 23 2025

Examples

			a(3)=2. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them, namely ud and du, contain no h steps.
		

References

  • M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
  • E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.

Crossrefs

Programs

  • Maple
    G:=1/sqrt(1-2*z^2+z^4-4*z^3): Gser:=series(G,z=0,40): seq(coeff(Gser,z,n), n=0..35);
    # Alternatively (after Bala):
    seq(add(binomial(n-k,k)*binomial(k,n-2*k), k=ceil(n/3)...floor(n/2)),n=0..35); # Peter Luschny, Feb 07 2017
    # With natural summation bound:
    a := n -> add((-1)^k*binomial(n,k)*hypergeom([-k,k-n,k-n], [1,-n], -1), k=0..n): seq(simplify(a(n)), n=0..35); # Peter Luschny, Feb 13 2018
  • Mathematica
    CoefficientList[Series[1/Sqrt[1-2x^2+x^4-4x^3],{x,0,40}],x] (* Harvey P. Dale, Oct 16 2011 *)

Formula

a(n) = A182882(n,0).
G.f.: 1/sqrt(1-2*z^2+z^4-4*z^3).
It appears that a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k,k)*binomial(k,n-2*k): this gives correct values for a(0) through a(35). If true, then sequence equals antidiagonal sums of triangle A105868. - Peter Bala, Mar 06 2013
D-finite n*a(n) = (2*n - 2)*a(n-2) + (4*n - 6)*a(n-3) - (n - 2)*a(n-4), follows easily by differentiating the o.g.f. Maple's sumrecursion command verifies that Sum_{k = 0..floor(n/2)} binomial(n-k,k)*binomial(k,n-2*k) satisfies the same recurrence with the same initial conditions thus proving the above conjecture. - Peter Bala, Feb 07 2017
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*hypergeom([-k, k-n, k-n], [1, -n], -1). - Peter Luschny, Feb 13 2018
Showing 1-4 of 4 results.