A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

1, 1, 3, 1, 12, 15, 1, 39, 135, 105, 1, 120, 870, 1680, 945, 1, 363, 4950, 17850, 23625, 10395, 1, 1092, 26565, 159600, 373275, 374220, 135135, 1, 3279, 138285, 1303155, 4795875, 8222445, 6621615, 2027025

Offset

1,3

Comments

The row polynomials R(n,x) of A019538 satisfy the recurrence relation R(n+1,x) = x*d/dx((1+x)*R(n,x)), and have the expansion R(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.

Here we consider a sequence of polynomials P(n,x) (n>=1) defined by means of the similar recursion P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with starting value P(1,x) = x.

The first few polynomials are P(1,x) = x, P(2,x) = x + 3*x^3, P(3,x) = x + 12*x^3 + 15*x^5, and P(4,x) = x + 39*x^3 + 135*x^5 + 105*x^7.

Clearly, the P(n,x) are odd polynomials of the form P(n,x) = Sum_{k = 1..n} T(n,k)*x^(2*k-1).

This triangle lists the coefficients T(n,k). They are related to A039755, the type B Stirling numbers of Suter, by T(n,k) = (2*k-1)!!*A039755(n-1,k-1).

Links

Table of n, a(n) for n=1..36.

Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 8.

Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Tech Report TR 99-05, 1999.

Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).

Formula

T(n+1,k+1) = ((2*k+1)!/(2^k*k!)^2)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.

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

E.g.f.: F(x,t) = (x/(1+x))*(exp(t)/sqrt((1+x) - x*exp(2*t)) - 1) = Sum_{n>=1} R(n,x)*t^n/n! = x*t + (x + 3*x^2)*t^2/2! + (x + 12*x^2 + 15*x^3)*t^3/3! + ....

Compare with the e.g.f. for A019538, which is (x/(1+x))*(exp(t)/((1+x) - x*exp t))-1).

The row polynomials R(n,x) are related to the polynomials P(n,x) of the comments section by P(n,x) = 1/x*R(n,x^2).

The generating function F(x,t) satisfies the partial differential equation d/dt(F) = 2*x*(1+x)*d/dx(F) + (x-1)*F + x.

It follows that the polynomials P(n,x) := Sum_{k = 1..n} T(n,k)*x^(2*k-1) satisfy the recurrence P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with P(1,x) = x. (Cf. the recurrence relation for row polynomials of A185896.)

The recurrence relation for T(n,k) given above now follows.

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)*S(n,x), where S(n,x) is the n-th row polynomial of A187075.

In addition, R(n,1/(x-1)) = (1/(x-1)^n)*Q(n-1,x), where Q(n,x) is the n-th row polynomial of A156919.

Row sums are [1,4,28,280,3616...] = 1/2*A124212(n) for n >= 1.

Main diagonal is [1,3,15,105,...] = A001147(k) for k >= 1.

Put S(n) = sum {k = 1..n} (-1)^k*T(n,k)/(k+1). Then for m>=2, S(2*m-1) = S(2*m) = (4^m-1)*Bernoulli(2*m)/m.

From Peter Bala, Aug 30 2016: (Start)

n-th row polynomial R(n,x) = 1/(1 + x)^(3/2) * Sum_{k >= 0} (1/4)^k*(x/(1 + x))^k*binomial(2*k,k)*(2*k + 1)^n.

R(n,x) = (1/(1 + x))*Sum_{k = 0..n} binomial(2*k,k)*A145901(n,k)*(x/4)^k. (End)

Example

Triangle begins
n\k.|..1.....2.....3......4......5......6
=========================================
..1.|..1
..2.|..1.....3
..3.|..1....12....15
..4.|..1....39...135....105
..5.|..1...120...870...1680....945
..6.|..1...363..4950..17850..23625..10395
..
Examples of recurrence relation
... T(4,3) = 5*(T(3,3)+T(3,2)) = 5*(15+12)= 135;
... T(6,4) = 7*(T(5,4)+T(5,3)) = 7*(1680+870) = 17850.

Maple

A186695 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 1; else (2*k-1)*(procname(n-1, k) + procname(n-1, k-1)) ; end if; end proc: seq(seq(A186695(n, k), k = 1..n), n = 1..10);

Mathematica

T[n_, k_] := (2k-1)! Sum[(-1)^(k-j-1) (2j+1)^(n-1) Binomial[k-1, j], {j, 0, k-1}] / (2^(k-1) (k-1)!)^2;
Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

Crossrefs

Cf. A001147, A019538, A039755, A124212, A145901, A156919, A187075.

Sequence in context: A268298 A291418 A219512 * A210587 A019232 A185697

Adjacent sequences: A186692 A186693 A186694 * A186696 A186697 A186698

Keyword

nonn,easy,tabl

Author

Peter Bala, Mar 26 2011

Status

approved