A106513 A Pell-Pascal matrix.

1, 2, 1, 5, 3, 1, 12, 8, 4, 1, 29, 20, 12, 5, 1, 70, 49, 32, 17, 6, 1, 169, 119, 81, 49, 23, 7, 1, 408, 288, 200, 130, 72, 30, 8, 1, 985, 696, 488, 330, 202, 102, 38, 9, 1, 2378, 1681, 1184, 818, 532, 304, 140, 47, 10, 1, 5741, 4059, 2865, 2002, 1350, 836, 444, 187, 57, 11, 1

Offset: 0

Author

Paul Barry, May 05 2005

Keywords

• easy
• nonn
• tabl

Comments

This triangle gives the iterated partial sums of the Pell sequence A000129(n+1), n>=0. - Wolfdieter Lang, Oct 05 2014

Examples

			The triangle T(n,k) begins:
n\k    0    1    2    3    4   5   6   7  8  9 10 ...
0:     1
1:     2    1
2:     5    3    1
3:    12    8    4    1
4:    29   20   12    5    1
5:    70   49   32   17    6   1
6:   169  119   81   49   23   7   1
7:   408  288  200  130   72  30   8   1
8:   985  696  488  330  202 102  38   9  1
9:  2378 1681 1184  818  532 304 140  47 10  1
10: 5741 4059 2865 2002 1350 836 444 187 57 11  1
... Reformatted and extended. - _Wolfdieter Lang_, Oct 05 2014
-----------------------------------------------------
Recurrence from the Z-sequence (see the formula above) for T(0,n) in terms of the entries of row n-1. For example, 29 = T(4,0) = 2*12 + 1*8 + (-1)*4 + 1*1 = 29. - _Wolfdieter Lang_, Oct 05 2014
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A000129, A001333, A106514 (row sums), A106515 (antidiagonal sums), A248156.

Programs

Magma

[ (&+[Binomial(n+1, 2*j+k+1)*2^j: j in [0..Floor((n+1)/2)]]) : k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021

Mathematica

T[n_, k_]= Sum[Binomial[n+1, 2*j+k+1]*2^j, {j, 0, Floor[(n+1)/2]}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 05 2021 *)

Sage

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return lucas_number1(n+1, 2, -1)
    else: return T(n-1,k-1) + T(n-1,k)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Formula

Riordan array (1/(1-2*x-x^2), x/(1-x)).

Number triangle T(n,0) = A000129(n+1), T(n,k) = T(n-1,k-1) + T(n-1,k).

T(n,k) = Sum_{j=0..floor((n+1)/2)} binomial(n+1, 2*j+k+1)*2^j.

Sum_{k=0..n} T(n, k) = A106514(n).

Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n).

T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-3,k) - T(n-3,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 14 2014

From Wolfdieter Lang, Oct 05 2014: (Start)

O.g.f. for row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k: (1 - z)/((1 - 2*z - z^2)*(1 - (1+x)*z)).

O.g.f. column m: (1/(1 - 2*z - z^2))*(z/(1 - z))^m, m >= 0. (Riordan property).

The alternating row sums are shown in A001333.

A-sequence: [1, 1] (see the three term recurrence given above). Z-sequence has o.g.f. (2 + 3*x)/(1 + x), [2, 1, repeat(-1,1)] (unsigned A054977). See the W. Lang link under A006232 for Riordan A- and Z-sequences.

The inverse Riordan triangle is shown in A248156. (End)