|
| |
|
|
A104033
|
|
Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1).
|
|
5
|
|
|
|
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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Column 0 equals signed A009843 (expansion of x/cosh(x)). Row sums form signed A000182 (expansion of tanh(x)).
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
|
|
|
LINKS
|
|
|
|
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<n), or 1 (k=n).
Sum_{k=0..n} T(n, k) * 4^k = 1 for n >= 0.
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)
|
|
|
EXAMPLE
|
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; ...
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)
|
|
|
PROG
|
(PARI) {T(n, k) = if(n<k || k<0, 0, ((matrix(n+1, n+1, m, j, if(m>=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(""))
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
