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%I #28 Sep 08 2022 08:45:44
%S 1,-1,1,1,0,1,1,1,1,1,1,2,2,2,1,1,3,4,4,3,1,1,4,7,8,7,4,1,1,5,11,15,
%T 15,11,5,1,1,6,16,26,30,26,16,6,1,1,7,22,42,56,56,42,22,7,1,1,8,29,64,
%U 98,112,98,64,29,8,1,1,9,37,93,162,210,210,162,93,37,9,1,1,10,46,130,255
%N Riordan array ((1-2*x+2*x^2)/(1-x), x/(1-x)).
%C Essentially the same as A087698.
%H Muniru A Asiru, <a href="/A159853/b159853.txt">Table of n, a(n) for n = 0..5151</a>
%H Peter Bala, <a href="/A081577/a081577.pdf">A note on the diagonals of a proper Riordan Array</a>
%F From _Peter Bala_, Mar 20 2018: (Start)
%F T(n,k) = C(n,k) - 2*C(n-1,n-k-1) + 2*C(n-2,n-k-2), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
%F Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + x + x^2/2! + x^3/3!) = 1 + 2*x + 2*x^2/2! + 4*x^3/3! + 8*x^4/4! + 15*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ). (End)
%e Triangle begins:
%e 1;
%e -1, 1;
%e 1, 0, 1;
%e 1, 1, 1, 1;
%e 1, 2, 2, 2, 1;
%e 1, 3, 4, 4, 3, 1;
%e ...
%p C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
%p end proc:
%p for n from 0 to 10 do
%p seq(C(n, n-k) - 2*C(n-1, n-k-1) + 2*C(n-2, n-k-2), k = 0..n);
%p end do; # _Peter Bala_, Mar 20 2018
%t Join[{1, -1}, Rest[T[0, 0]=1; T[n_, k_]:=Binomial[n, n - k] - 2 Binomial[n - 1, n - k - 1] + 2 Binomial[n - 2, n - k - 2]; Table[T[n, k], {n, 1, 15}, {k, 0, n}]//Flatten]] (* _Vincenzo Librandi_, Mar 22 2018 *)
%o (Sage) # uses[riordan_array from A256893]
%o riordan_array((1-2*x+2*x^2)/(1-x), x/(1-x), 8) # _Peter Luschny_, Mar 21 2018
%o (GAP) Flat(List([0..12],n->List([0..n],k->Binomial(n,k)-2*Binomial(n-1,n-k-1)+2*Binomial(n-2,n-k-2)))); # _Muniru A Asiru_, Mar 22 2018
%o (Magma) /* As triangle */ [[Binomial(n, n-k)-2*Binomial(n-1, n-k-1)+2*Binomial(n-2, n-k-2): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Mar 22 2018
%Y Cf. A087698.
%K easy,sign,tabl
%O 0,12
%A _Philippe Deléham_, Apr 24 2009
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