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 A110162 Riordan array ((1-x)/(1+x), x/(1+x)^2). 14

%I

%S 1,-2,1,2,-4,1,-2,9,-6,1,2,-16,20,-8,1,-2,25,-50,35,-10,1,2,-36,105,

%T -112,54,-12,1,-2,49,-196,294,-210,77,-14,1,2,-64,336,-672,660,-352,

%U 104,-16,1,-2,81,-540,1386,-1782,1287,-546,135,-18,1,2,-100,825,-2640,4290,-4004,2275,-800,170,-20,1

%N Riordan array ((1-x)/(1+x), x/(1+x)^2).

%C Inverse of Riordan array A094527. Rows sums are A099837. Diagonal sums are A110164. Product of Riordan array A102587 and inverse binomial transform (1/(1+x), x/(1+x)).

%C Coefficients of polynomials related to Cartan matrices of types C_n and B_n: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2), with p(x,0) = 1; p(x,1) = 2-x; p(x,2) = x^2-4*x-2. - _Roger L. Bagula_, Apr 12 2008

%C From _Wolfdieter Lang_, Nov 16 2012: (Start)

%C The alternating row sums are given in A219233.

%C For n >= 1 the row polynomials in the variable x^2 are R(2*n,x):=2*T(2*n,x/2) with Chebyshev's T-polynomials. See A127672 and also the triangle A127677.

%C (End)

%C From _Peter Bala_, Jun 29 2015: (Start)

%C Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = x/(1 + x)^2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).

%C T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 - 2*x + sqrt(1 - 4*x))/2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

%H G. C. Greubel, <a href="/A110162/b110162.txt">Rows n=0..100 of triangle, flattened</a>

%H P. Peart and W.-J. Woan, <a href="http://dx.doi.org/10.1016/S0166-218X(99)00166-3">A divisibility property for a subgroup of Riordan matrices</a>, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.

%H T. M. Richardson, <a href="http://arxiv.org/abs/1405.6315">The Reciprocal Pascal Matrix</a>, arXiv:1405.6315 [math.CO], 2014.

%F T(n,k) = (-1)^(n-k)*(C(n+k,n-k) + C(n+k-1,n-k-1)), with T(0,0) = 1. - _Paul Barry_, Mar 22 2007

%F From _Wolfdieter Lang_, Nov 16 2012: (Start)

%F O.g.f. row polynomials P(n,x) := Sum(T(n,k)*x^k, k=0..n): (1-z^2)/(1+(x-2)*z+z^2) (from the Riordan property).

%F O.g.f. column No. k: ((1-x)/(1+x))*(x/(1+x)^2)^k, k >= 0.

%F T(0,0) = 1, T(n,k) = (-1)^(n-k)*(2*n/(n+k))*binomial(n+k,n-k), n>=1, and T(n,k) = 0 if n < k. (From the Chebyshev T-polynomial formula due to Waring's formula.)

%F (End)

%F T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham_, Nov 29 2013

%e Triangle T(n,k) begins:

%e m\k 0 1 2 3 4 5 6 7 8 9 10 ...

%e 0: 1

%e 1: -2 1

%e 2: 2 -4 1

%e 3: -2 9 -6 1

%e 4: 2 -16 20 -8 1

%e 5: -2 25 -50 35 -10 1

%e 6: 2 -36 105 -112 54 -12 1

%e 7: -2 49 -196 294 -210 77 -14 1

%e 8: 2 -64 336 -672 660 -352 104 -16 1

%e 9: -2 81 -540 1386 -1782 1287 -546 135 -18 1

%e 10: 2 -100 825 -2640 4290 -4004 2275 -800 170 -20 1

%e ... Reformatted and extended by _Wolfdieter Lang_, Nov 16 2012

%e Row polynomial n=2: P(2,x) = 2 - 4*x + x^2. R(4,x):= 2*T(4,x/2) = 2 - 4*x^2 + x^4. For P and R see a comment above. - _Wolfdieter Lang_, Nov 16 2012.

%t Table[If[n==0 && k==0, 1, (-1)^(n-k)*(Binomial[n+k, n-k] + Binomial[n+k-1, n-k-1])], {n, 0, 15}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Dec 16 2018 *)

%o (MAGMA) /* As triangle */ [[(-1)^(n-k)*(Binomial(n+k,n-k) + Binomial(n+k-1,n-k-1)): k in [0..n]]: n in [0.. 12]]; // _Vincenzo Librandi_, Jun 30 2015

%o (PARI) {T(n,k) = (-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1))};

%o for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Dec 16 2018

%o (Sage) [[(-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1)) for k in range(n+1)] for n in range(12)] # _G. C. Greubel_, Dec 16 2018

%Y Cf. A128411. See A127677 for an almost identical triangle.

%Y Cf. A136674, A053122.

%K easy,sign,tabl

%O 0,2

%A _Paul Barry_, Jul 14 2005

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Last modified September 21 21:15 EDT 2019. Contains 327282 sequences. (Running on oeis4.)