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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. 5
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 (list; table; graph; refs; listen; history; text; internal format)
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.

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

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..inf} 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 egf 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);

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

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Last modified October 22 22:58 EDT 2018. Contains 316518 sequences. (Running on oeis4.)