A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1).

1, -3, 1, 25, -10, 1, -427, 175, -21, 1, 12465, -5124, 630, -36, 1, -555731, 228525, -28182, 1650, -55, 1, 35135945, -14449006, 1782495, -104676, 3575, -78, 1, -2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1, 329655706465, -135565467080, 16724709820, -982532408

Offset: 0

Author

Paul D. Hanna, Feb 28 2005

Keywords

• sign
• tabl

Comments

Column 0 equals signed A009843 (expansion of x/cosh(x)). Row sums form signed A000182 (expansion of tanh(x)).

The matrix logarithm is L(n,k) = -(-1)^(n-k)*A000182(n-k)*A103327(n,k), where A000182 = tangent numbers.

Let E(y) = cosh(sqrt(y)) = 1 + 3*y/3! + 5*y^2/5! + 7*y^3/7! + .... so that 1/E(y) = 1 - 3*y/3! + 25*y^2/5! - 427*y^3/7! + .... Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence (2*n+1)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013

Examples

			Rows begin:
1;
-3, 1;
25, -10, 1;
-427, 175, -21, 1;
12465, -5124, 630, -36, 1;
-555731 ,228525, -28182, 1650, -55, 1;
35135945, -14449006, 1782495, -104676, 3575, -78, 1;
-2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1;
329655706465, -135565467080, 16724709820, -982532408, 33669350, -754936, 11900, -136, 1; ...
From _Peter Bala_, Aug 06 2013: (Start)
The real zeros of the row polynomials R(n,x) seem to converge to the even squares as n increases.
Polynomial |        Real zeros to 6 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 3.999986
R(10,x)    | 4.000000, 15.999978
R(15,x)    | 4.000000, 16.000000, 35.999992, 64.414273, 76.998346
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)
		

Links

• W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Crossrefs

Cf. A000364, A103327, A009843, A000182 (unsigned row sums), A055133, A086645, A086646, A096053, A103364.

Programs

PARI

{T(n,k) = if(n=j, binomial(2*m-1,2*j-1))))^-1)[n+1,k+1])}
for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))

PARI

{T(n,k) = binomial(2*n+1,2*k+1) * polcoeff(1/cosh(x+x*O(x^(2*n))), 2*n-2*k) * (2*n-2*k)!}
for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))

Formula

Column k: Sum_{j=0..n} C(2*n+1, 2*j+1) * T(j, k) = 0 (n>k), or 1 (n=k).

Row n: Sum_{j=0..n} T(n, j) * C(2*j+1, 2*k+1) = 0 (k

Sum_{k=0..n} T(n, k) * 4^k = 1 for n >= 0.

T(n, k) = (-1)^(n-k)*A000364(n-k)*A103327(n, k), where A000364 = Euler numbers.

Sum_{k=0..n} (-1)^(n-k)*T(n, k) = A002084(n). - Philippe Deléham, Aug 27 2005

From Peter Bala, Aug 06 2013: (Start)

Generating function: 1/sqrt(x)*sinh(sqrt(x)*t)/cosh(t) = t + (-3 + x)*t^3/3! + (25 - 10*x + x^2)*t^5/5! + ....

Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} binomial(2*n+1,2*k+1)*R(k,x) with initial value R(0,x) = 1.

It appears that for arbitrary nonzero complex x we have

lim_{n -> inf} R(n,x^2)/R(n,0) = (1/(Pi/2*x))*sin(Pi/2*x).

A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the even squares 4, 16, 36, ... as n increases. Some numerical examples are given below. Cf. A055133, A086646 and A103364.

If p = 2*n + 1 is a prime then all the entries in row n are divisible by p, apart from T(n,n) = 1. Thus the row sum is congruent to 1 modulo p.

Row sums R(n,1) = (-1)^n*A000182(n+1).

R(n,4) = 1; R(n,16) = (1/2)*( 3^(2*n+1) - 1 ) = A096053(n);

R(n,36) = (1/3)*( 5^(2*n+1) - 3^(2*n+1) + 1 );

R(n,64) = (1/4)*( 7^(2*n+1) - 5^(2*n+1) + 3^(2*n+1) - 1 ). (End)

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.