A116414 Riordan array (1/((1-x)(1-3x)),x/((1-x)(1-3x))).
1, 4, 1, 13, 8, 1, 40, 42, 12, 1, 121, 184, 87, 16, 1, 364, 731, 496, 148, 20, 1, 1093, 2736, 2454, 1040, 225, 24, 1, 3280, 9844, 11064, 6170, 1880, 318, 28, 1, 9841, 34448, 46738, 32624, 13015, 3080, 427, 32, 1, 29524, 118101, 188208, 158724, 79044, 24381, 4704
Offset: 0
Examples
Triangle begins
1;
4, 1;
13, 8, 1;
40, 42, 12, 1;
121, 184, 87, 16, 1;
364, 731, 496, 148, 20, 1;
Triangle T(n,k), 0 <= k <= n, given by (0, 4, -3/4, 3/4, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins:
1;
0, 1;
0, 4, 1;
0, 13, 8, 1;
0, 40, 42, 12, 1;
0, 121, 184, 87, 16, 1;
0, 364, 731, 496, 148, 20, 1;
... - _Philippe Deléham_, Jan 18 2012
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..11324 (rows 0 <= n <= 150)
- Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.
Programs
-
Mathematica
With[{n = 10}, DeleteCases[#, 0] & /@ Rest@ CoefficientList[Series[(1 - 4 x + 3 x^2)/(1 - 4 x + 3 x^2 - x y), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)
Formula
Riordan array (1/(1-4x+3x^2), x/(1-4x+3x^2)); number triangle T(n,k) = Sum_{j=0..n} binomial(n-j,k)*binomial(k+j,j)*3^j.
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k), T(0,0) = T(1,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, T(2,1) = 4, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Oct 31 2013
G.f.: (1-4*x+3*x^2)/(1-4*x+3*x^2-x*y). - Philippe Deléham, Oct 31 2013
From Peter Bala, Oct 07 2019: (Start)
O.g.f.: 1/(1 - 4*x + 3*x^2 - x*y) = 1 + (4 + y)*x + (13 + 8*y + y^2)*x^2 + ....
Recurrence for row polynomials: R(n,y) = (4 + y)*R(n-1,y) - 3*R(n-2,y) with R(0,y) = 1 and R(1,y) = 4 + y.
The row reverse polynomial y^n*R(n,1/y) is equal to the numerator polynomial of the finite continued fraction 1 + y/(1 + 3*y/(1 + ... + y/(1 + 3*y/(1)))) (with 2*n partial numerators). Cf. A110441. (End)
Comments