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 A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1). 4
 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)). 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 LINKS W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500. 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=0. T(n, k) = (-1)^(n-k)*A000364(n-k)*A103327(n, k), where A000364 = Euler numbers. Sum_{k, 0<=k<=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) . 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; ... 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) PROG (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("")) CROSSREFS Cf. A000364, A103327, A009843, A000182 (unsigned row sums), A055133, A086645, A086646, A096053, A103364. Sequence in context: A309397 A193472 A259208 * A217472 A156654 A098815 Adjacent sequences:  A104030 A104031 A104032 * A104034 A104035 A104036 KEYWORD sign,tabl AUTHOR Paul D. Hanna, Feb 28 2005 EXTENSIONS Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007 STATUS approved

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Last modified August 24 16:15 EDT 2019. Contains 326295 sequences. (Running on oeis4.)