A106516 - OEIS

A106516

A Pascal-like triangle based on 3^n.

1, 3, 1, 9, 4, 1, 27, 13, 5, 1, 81, 40, 18, 6, 1, 243, 121, 58, 24, 7, 1, 729, 364, 179, 82, 31, 8, 1, 2187, 1093, 543, 261, 113, 39, 9, 1, 6561, 3280, 1636, 804, 374, 152, 48, 10, 1, 19683, 9841, 4916, 2440, 1178, 526, 200, 58, 11, 1, 59049, 29524, 14757, 7356, 3618, 1704, 726, 258, 69, 12, 1

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OFFSET

0,2

COMMENTS

Row sums are A027649. Antidiagonal sums are A106517.

From Wolfdieter Lang, Jan 09 2015: (Start)

Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.

The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)

LINKS

Table of n, a(n) for n=0..65.

FORMULA

Riordan array (1/(1-3x), x/(1-x)); Number triangle T(n, 0)=A000244(n), T(n, k)=T(n-1, k-1)+T(n-1, k); T(n, k)=sum{j=0..n, binomial(n, k+j)2^j}.

From Peter Bala, Jul 16 2013: (Start)

T(n,k) = binomial(n,k) + 2*sum {i = 1..n} 3^(i-1)*binomial(n-i,k).

O.g.f.: (1 - t)/( (1 - 3*t)*(1 - (1 + x)*t) ) = 1 + (3 + x)*t + (9 + 4*x + x^2)*t^2 + ....

The n-th row polynomial R(n,x) = 1/(x - 2)*( x*(x + 1)^n - 2*3^n ). (End)

Closed-form formula for arbitrary left and right borders of Pascal-like triangle see A228196. - Boris Putievskiy, Aug 19 2013

T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 3*T(n-2,k-1), T(0,0)=1, T(1,0)=3, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 26 2013

From Peter Bala, Dec 23 2014: (Start)

exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(27 + 13*x + 5*x^2/2! + x^3/3!) = 27 + 40*x + 58*x^2/2! + 82*x^3/3! + 113*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).

Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143495 (but with a different offset). See the Example section. Cf. A055248. (End)

EXAMPLE

The triangle T(n,k) begins:

n\k 0 1 2 3 4 5 6 7 8 9 10 ...

0: 1

1: 3 1

2: 9 4 1

3: 27 13 5 1

4: 81 40 18 6 1

5: 243 121 58 24 7 1

6: 729 364 179 82 31 8 1

7: 2187 1093 543 261 113 39 9 1

8: 6561 3280 1636 804 374 152 48 10 1

9: 19683 9841 4916 2440 1178 526 200 58 11 1

10: 59049 29524 14757 7356 3618 1704 726 258 69 12 1

... reformatted and extended. - Wolfdieter Lang, Jan 06 2015

----------------------------------------------------------

With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins

/ 1 \/1 \/1 \ /1 \

| 3 1 ||0 1 ||0 1 | | 3 1 |

| 9 4 1 ||0 3 1 ||0 0 1 |... = | 9 7 1 |

|27 13 5 1 ||0 9 4 1 ||0 0 3 1 | |27 37 12 1 |

|... ||0 27 13 5 1 ||0 0 9 4 1| |... |

|... ||... ||... | |... |

= A143495. - Peter Bala, Dec 23 2014

MATHEMATICA

a106516[n_] := Block[{a, k},

a[x_] := Flatten@ Last@ Reap[For[k = -1, k < x, Sow[Binomial[x, k] +

2 Sum[3^(i - 1)*Binomial[x - i, k], {i, 1, x}]], k++]]; Flatten@Array[a, n, 0]]; a106516[11] (* Michael De Vlieger, Dec 23 2014 *)

CROSSREFS

Columns 1, 2, 3, 4, 5: A003462, A000340, A052150, A097786, A097787.

Cf. A055248, A143495.

Sequence in context: A370150 A162852 A054448 * A140071 A285280 A285266

Adjacent sequences: A106513 A106514 A106515 * A106517 A106518 A106519

KEYWORD

easy,nonn,tabl

AUTHOR

Paul Barry, May 05 2005

STATUS

approved