A224069 Matrix inverse of A111636.
1, -1, 1, 3, -4, 1, -25, 36, -12, 1, 543, -800, 288, -32, 1, -29281, 43440, -16000, 1920, -80, 1, 3781503, -5621952, 2085120, -256000, 11520, -192, 1, -1138779265, 1694113344, -629658624, 77844480, -3584000, 64512, -448, 1, 783702329343, -1166109967360, 433693016064, -53730869248, 2491023360, -45875200, 344064, -1024, 1
Offset: 0
Examples
Triangle begins n\k.|......0......1......2......3......4......5 = = = = = = = = = = = = = = = = = = = = = = = = .0..|......1 .1..|.....-1......1 .2..|......3.....-4......1 .3..|....-25.....36....-12......1 .4..|....543...-800....288....-32......1 .5..|.-29281..43440.-16000...1920....-80......1 ... The sequence of zeros of R(10,x) begins 1, 3.280147..., 9.112469..., 23.366923..., 57.084317.... The sequence of zeros of R(20,x) begins 1, 3.280163..., 9.112696..., 23.369274..., 57.105379....
Links
- Vincenzo Librandi, Rows n = 0..50, flattened
- W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500, (2008).
Programs
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Mathematica
max = 8; A111636 = Table[ Binomial[n, k]*2^(k*(n - k)), {n, 0, max}, {k, 0, max}]; t = Inverse[ A111636 ]; Table[ t[[n, k]], {n, 1, max+1}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 10 2013 *)
Formula
Sum_{k = 1..n} k*2^k*T(n,k) = 0 for n >= 1.
Let E(x) = Sum_{n >= 0} x^n/(n!*2^binomial(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + .... Then a generating function for this triangle is E(x*z)/E(z) = 1 + (x - 1)*z + (x^2 - 4*x + 3)*z^2/(2!*2) + (x^3 - 12*x^2 + 36*x - 25)*z^3/(3!*2^3) + ....
This triangle is a generalized Riordan array (1/E(x), x) with respect to the sequence n!*2^C(n,2), as defined by Wang and Wang.
The row polynomials R(n,x) satisfy the recurrence equation R(n,x) = x^n - Sum_{k = 0..n-1} binomial(n,k)*2^(k*(n-k))*R(k,x) with R(0,x) = 1, as well as R'(n,2*x) = n*2^(n-1)*R(n-1,x) (the ' denotes differentiation w.r.t. x). The row polynomials appear to have only positive real zeros of multiplicity 1. Moreover, if r(n,1) < r(n,2) < ... < r(n,n) denotes the zeros of R(n,x) arranged in increasing order then it appears that lim_{n -> oo} r(n,i) exists for each fixed 1 <= i <= n. An example is given below.
Comments