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Triangle T(n,k) = f(n,k,n-1), n >= 0, 0 <= k <= n, where f is given below.
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%I #10 Jan 17 2014 05:38:48

%S 1,1,1,1,2,1,1,6,6,1,1,12,16,12,1,1,20,35,35,20,1,1,30,66,84,66,30,1,

%T 1,42,112,175,175,112,42,1,1,56,176,328,400,328,176,56,1,1,72,261,567,

%U 819,819,567,261,72,1,1,90,370,920,1540,1820,1540,920,370,90,1,1,110,506,1419,2706,3696,3696,2706,1419,506,110,1,1,132,672

%N Triangle T(n,k) = f(n,k,n-1), n >= 0, 0 <= k <= n, where f is given below.

%H Michel Lassalle, <a href="http://arXiv.org/abs/math.CO/0210208">A new family of positive integers</a>

%F f(n, p, k) = binomial(n, k)*hypergeom([1-k, -p, p-n], [1-n, 1], 1).

%e 1; 1,1; 1,2,1; 1,6,6,1; ...

%p f := proc(n,p,k) convert( binomial(n,k)*hypergeom([1-k,-p,p-n],[1-n,1],1), `StandardFunctions`); end;

%t f[n_, p_, k_] := Binomial[n, k]*HypergeometricPFQ[{1 - k, -p, p-n}, {1-n, 1}, 1]; t[n_, n_] = t[_, 0] = 1; t[n_, k_] := f[n, k, n-1]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 17 2014 *)

%Y Cf. A014410 and A007318 for f(n, k, n), A075779 and A075798 for f(n, k, n-1) and A075780 and A075837 for f(n, k, n-2).

%K nonn,tabl

%O 1,5

%A _N. J. A. Sloane_, Oct 17 2002