login
Triangle based on (0,1/4,1) averaging array.
2

%I #9 Dec 24 2013 01:53:29

%S 1,1,5,1,6,13,1,7,19,29,1,8,26,48,61,1,9,34,74,109,125,1,10,43,108,

%T 183,234,253,1,11,53,151,291,417,487,509,1,12,64,204,442,708,904,996,

%U 1021,1,13,76,268,646,1150,1612,1900,2017,2045,1,14,89,344,914

%N Triangle based on (0,1/4,1) averaging array.

%C See A204201 for a discussion and guide to other averaging arrays.

%F From _Philippe Deléham_, Dec 24 2013: (Start)

%F T(n,n) = A036563(n+1).

%F Sum_{k=1..n} T(n,k) = A014480(n-1).

%F T(n,k) = T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-2*T(n-2,k-2), T(1,1)=1, T(2,1)=1, T(2,2)=5, T(n,k)=0 if k<1 or if k>n. (End)

%e First six rows:

%e 1

%e 1...5

%e 1...6...13

%e 1...7...19...29

%e 1...8...26...48...61

%e 1...9...34...74...109...125

%t a = 0; r = 1/4; b = 1; t[1, 1] = r;

%t t[n_, 1] := (a + t[n - 1, 1])/2;

%t t[n_, n_] := (b + t[n - 1, n - 1])/2;

%t t[n_, k_] := (t[n - 1, k - 1] + t[n - 1, k])/2;

%t u[n_] := Table[t[n, k], {k, 1, n}]

%t Table[u[n], {n, 1, 5}] (* averaging array *)

%t u = Table[(1/2) (1/r) 2^n*u[n], {n, 1, 12}];

%t TableForm[u] (* A204203 triangle *)

%t Flatten[u] (* A204203 sequence *)

%Y Cf. A204201.

%K nonn,tabl

%O 1,3

%A _Clark Kimberling_, Jan 12 2012