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

0,2

Comments

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

Links

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

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))*(x/(1 - x)))^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)

Example

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

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 *)

Prog

(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
(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

Crossrefs

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

Sequence in context: A120095 A327631 A130197 * A054446 A164981 A366858

Adjacent sequences: A106510 A106511 A106512 * A106514 A106515 A106516

Keyword

easy,nonn,tabl

Author

Paul Barry, May 05 2005

Status

approved