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%I #28 Jan 22 2020 05:16:17
%S 1,-3,1,25,-10,1,-427,175,-21,1,12465,-5124,630,-36,1,-555731,228525,
%T -28182,1650,-55,1,35135945,-14449006,1782495,-104676,3575,-78,1,
%U -2990414715,1229758075,-151714563,8912475,-305305,6825,-105,1,329655706465,-135565467080,16724709820,-982532408
%N Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1).
%C Column 0 equals signed A009843 (expansion of x/cosh(x)). Row sums form signed A000182 (expansion of tanh(x)).
%C The matrix logarithm is L(n,k) = -(-1)^(n-k)*A000182(n-k)*A103327(n,k), where A000182 = tangent numbers.
%C 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
%H W. Wang and T. Wang, <a href="https://doi.org/10.1016/j.disc.2007.12.037">Generalized Riordan array</a>, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.
%F Column k: Sum_{j=0..n} C(2*n+1, 2*j+1) * T(j, k) = 0 (n>k), or 1 (n=k).
%F Row n: Sum_{j=0..n} T(n, j) * C(2*j+1, 2*k+1) = 0 (k<n), or 1 (k=n).
%F Sum_{k=0..n} T(n, k) * 4^k = 1 for n >= 0.
%F T(n, k) = (-1)^(n-k)*A000364(n-k)*A103327(n, k), where A000364 = Euler numbers.
%F Sum_{k=0..n} (-1)^(n-k)*T(n, k) = A002084(n). - _Philippe Deléham_, Aug 27 2005
%F From _Peter Bala_, Aug 06 2013: (Start)
%F 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! + ....
%F 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.
%F It appears that for arbitrary nonzero complex x we have
%F lim_{n -> inf} R(n,x^2)/R(n,0) = (1/(Pi/2*x))*sin(Pi/2*x).
%F 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.
%F 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.
%F Row sums R(n,1) = (-1)^n*A000182(n+1).
%F R(n,4) = 1; R(n,16) = (1/2)*( 3^(2*n+1) - 1 ) = A096053(n);
%F R(n,36) = (1/3)*( 5^(2*n+1) - 3^(2*n+1) + 1 );
%F R(n,64) = (1/4)*( 7^(2*n+1) - 5^(2*n+1) + 3^(2*n+1) - 1 ). (End)
%e Rows begin:
%e 1;
%e -3, 1;
%e 25, -10, 1;
%e -427, 175, -21, 1;
%e 12465, -5124, 630, -36, 1;
%e -555731 ,228525, -28182, 1650, -55, 1;
%e 35135945, -14449006, 1782495, -104676, 3575, -78, 1;
%e -2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1;
%e 329655706465, -135565467080, 16724709820, -982532408, 33669350, -754936, 11900, -136, 1; ...
%e From _Peter Bala_, Aug 06 2013: (Start)
%e The real zeros of the row polynomials R(n,x) seem to converge to the even squares as n increases.
%e Polynomial | Real zeros to 6 decimal places
%e = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
%e R(5,x) | 3.999986
%e R(10,x) | 4.000000, 15.999978
%e R(15,x) | 4.000000, 16.000000, 35.999992, 64.414273, 76.998346
%e = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
%e (End)
%o (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])}
%o for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))
%o (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)!}
%o for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))
%Y Cf. A000364, A103327, A009843, A000182 (unsigned row sums), A055133, A086645, A086646, A096053, A103364.
%K sign,tabl
%O 0,2
%A _Paul D. Hanna_, Feb 28 2005
%E Edited by _N. J. A. Sloane_ at the suggestion of _Andrew S. Plewe_, Jun 08 2007
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