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Triangle based on (0,1/3,1) averaging array.
7

%I #12 Dec 24 2013 01:51:13

%S 1,1,4,1,5,10,1,6,15,22,1,7,21,37,46,1,8,28,58,83,94,1,9,36,86,141,

%T 177,190,1,10,45,122,227,318,367,382,1,11,55,167,349,545,685,749,766,

%U 1,12,66,222,516,894,1230,1434,1515,1534,1,13,78,288,738,1410

%N Triangle based on (0,1/3,1) averaging array.

%C For a<r<b, let t(1,1)=r, and for n>1, let

%C t(n,1)=[a+t(n-1,1)]/2,

%C t(n,n)=[b+t(n-1,n-1)]/2,

%C t(n,k)=[t(n-1,k-1)+t(n-1,k)]/2 for 2<=k<=n-1.

%C We call (t(n,k)) the (a,r,b) averaging array. If a and b

%C are integers and r is a rational number, then multiplying

%C row n of (t(n,k)) by the LCM of its denominators yields a

%C triangle of integers; A204201 arises in this manner from

%C (a,r,b)=(0,1/3,1).

%C ...

%C Guide to related arrays:

%C (a,r,b).........triangle

%C (0,1/2,1).......A054143

%C (0,1/3,1).......A204201

%C (0,2/3,1).......A204202

%C (0,1/4,1).......A204203

%C (0,3/4,1).......A204204

%C (0,1/5,1).......A204205

%C (1,3/2,2).......A204206

%C (1,2,3).........A204207

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

%F T(n,n) = A033484(n-1).

%F Sum{k=1..n} T(n,k) = A053220(n).

%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)=4, T(n,k)=0 if k<1 or if k>n. (End)

%e The (0,1/3,1) averaging array has these first four rows:

%e 1/3

%e 1/6....2/3

%e 1/12...5/12...5/6

%e 1/24...1/4....5/8...11/12.

%e Multiplying those rows by 3,6,12,24, respectively:

%e 1

%e 1...4

%e 1...5...10

%e 1...6...15...22

%e The first nine rows:

%e 1

%e 1...4

%e 1...5...10

%e 1...6...15...22

%e 1...7...21...37...46

%e 1...8...28...58...83...94

%e 1...9...36...86...141..177..190

%e 1...10..45...122..227..318..367..382

%e 1...11..55...167..349..545..685..749..766

%t a = 0; r = 1/3; b = 1;

%t 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] (* A204102 triangle *)

%t Flatten[u] (* A204201 sequence *)

%Y Cf. A204202.

%K nonn,tabl

%O 1,3

%A _Clark Kimberling_, Jan 12 2012