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).
LINKS
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
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
KEYWORD
AUTHOR
Peter Bala, Mar 26 2011
STATUS
approved