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%I #43 Feb 16 2022 23:21:46
%S 1,3,2,10,15,6,35,84,70,20,126,420,540,315,70,462,1980,3465,3080,1386,
%T 252,1716,9009,20020,24024,16380,6006,924,6435,40040,108108,163800,
%U 150150,83160,25740,3432,24310,175032,556920,1021020,1178100,875160
%N Array read by rows: T(n,k) = binomial(n+k-2,k-1)*binomial(2*n-1,n-k).
%C Alternating sum of elements of n-th row = 1.
%C If a certain event has a probability p of occurring in any given trial, the probability of its occurring at least n times in 2n-1 trials is Sum_{k=1..n} T(n,k)*(-1)^(k-1)*p^(n+k-1). For example, the probability of its occurring at least 4 out of 7 times is 35p^4 - 84p^5 + 70p^6 - 20p^7. - _Matthew Vandermast_, Jun 05 2004
%C With the row polynomial defined as R(n,x) = Sum_{k = 1..n} T(n,k)*x^k, the row polynomial is related to the regularized incomplete Beta function I_x(a,b), through the relation R(n,x) = -(-x)^{-n+1}*I_{-x}(n,n). - _Leo C. Stein_, Jun 06 2019
%F From _Peter Bala_, Apr 10 2012: (Start)
%F O.g.f.: x*t*(1+2*x-sqrt(1-4*t*(x+1)))/(2*(x+t)*sqrt(1-4*t*(x+1))) = x*t + (3*x+2*x^2)*t^2 + (10*x+15*x^2+6*x^3)*t^3 + ....
%F Sum_{k = 1..n} (-1)^(k-1)*T(n,k)*2^(n-k) = 4^(n-1).
%F Row polynomial R(n+1,x) = ((2*n+1)!/n!^2)*x*Integral_{y = 0..1} (y*(1+x*y))^n dy. Row sums A178792. (End)
%e Triangle starts:
%e 1,
%e 3, 2,
%e 10, 15, 6,
%e 35, 84, 70, 20,
%e 126, 420, 540, 315, 70,
%e ...
%t t[n_, k_] := Binomial[n+k-2, k-1]*Binomial[2n-1, n-k]; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Dec 06 2012 *)
%o (PARI) T(x,y)=binomial(x+y-2,y-1)*binomial(2*x-1,x-y)
%o (Magma) [[Binomial(n+k-2,k-1)*Binomial(2*n-1,n-k): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Jun 15 2015
%Y Cf. A001700 (first column), A002740 (second column), A000984 (main diagonal), A033876 (second diagonal), A178792 (row sums).
%K nonn,tabl,nice
%O 1,2
%A _Benoit Cloitre_, Mar 18 2004
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