A187075 A Galton triangle: T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1).

1, 2, 3, 4, 18, 15, 8, 84, 180, 105, 16, 360, 1500, 2100, 945, 32, 1488, 10800, 27300, 28350, 10395, 64, 6048, 72240, 294000, 529200, 436590, 135135, 128, 24384, 463680, 2857680, 7938000, 11060280, 7567560, 2027025, 256, 97920, 2904000, 26107200, 105099120, 220041360, 249729480, 145945800, 34459425

Offset

1,2

Comments

This is a companion triangle to A186695.

Let f(x) = (exp(2*x) + 1)^(-1/2); then the n-th derivative of f equals Sum_{k=1..n} (-1)^k*T(n,k)*(f(x))^(2*k+1). - Groux Roland, May 17 2011

Triangle T(n,k), 1 <= k <= n, given by (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 20 2013

Links

G. C. Greubel, Table of n, a(n) for the first 50 rows

Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv:1204.4963v3 [math.CO]

Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11.

E. Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 (2001) 33-51.

Formula

T(n,k) = 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k).

Recurrence relation T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1) with boundary conditions T(1,1) = 1, T(1,k) = 0 for k >= 2.

G.f.: F(x,t) = 1/sqrt((1+x)-x*exp(2*t)) - 1 = Sum_{n >= 1} R(n,x)*t^n/n! = x*t + (2*x+3*x^2)*t^2/2! + (4*x+18*x^2+15*x^3)*t^3/3! + ....

The g.f. F(x,t) satisfies the partial differential equation dF/dt = 2*(x+x^2)*dF/dx + x*F.

The row polynomials R(n,x) satisfy the recursion R(n+1,x) = 2*(x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.

O.g.f. for column k: (2k-1)!!*x^k/Product_{m = 1..k} (1-2*m*x) (compare with A075497). T(n,k) = (2*k-1)!!*A075497(n,k).

The row polynomials R(n,x) = Sum_{k = 1..n} T(n,k)*x^k satisfy R(n,-x-1) = (-1)^n*(1+x)/x*P(n,x) where P(n,x) is the n-th row polynomial of A186695. We also have R(n,x/(1-x)) = (x/(1-x)^n)*Q(n-1,x) where Q(n,x) is the n-th row polynomial of A156919.

T(n,k) = 2^(n-k)*A211608(n,k). - Philippe Deléham, Oct 20 2013

Example

Triangle begins
n\k.|...1.....2......3......4......5......6
===========================================
..1.|...1
..2.|...2.....3
..3.|...4....18.....15
..4.|...8....84....180....105
..5.|..16...360...1500...2100....945
..6.|..32..1488..10800..27300..28350..10395
..
Examples of recurrence relation:
T(4,3) = 6*T(3,3) + 5*T(3,2) = 6*15 + 5*18 = 180;
T(6,4) = 8*T(5,4) + 7*T(5,3) = 8*2100 + 7*1500 = 27300.

Maple

A187075 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 2^(n-1); else 2*k*procname(n-1, k) + (2*k-1)*procname(n-1, k-1) ; end if; end proc:seq(seq(A187075(n, k), k = 1..n), n = 1..10);

Mathematica

Flatten[Table[2^(n - 2*k)*Binomial[2 k, k]*k!*StirlingS2[n, k], {n, 10}, {k, 1, n}]] (* G. C. Greubel, Jun 17 2016 *)

Prog

(Sage) # uses[delehamdelta from A084938]
# Adds a first column (1, 0, 0, 0, ...).
def A187075_triangle(n):
    return delehamdelta([(i+1)*int(is_even(i+1)) for i in (0..n)], [i+1 for i in (0..n)])
A187075_triangle(4)  # Peter Luschny, Oct 20 2013

Crossrefs

Cf. A075497, A156919, A186695, A211402.

Sequence in context: A037430 A370868 A329545 * A154715 A077407 A273002

Adjacent sequences: A187072 A187073 A187074 * A187076 A187077 A187078

Keyword

nonn,easy,tabl

Author

Peter Bala, Mar 27 2011

Status

approved