A073370 Convolution triangle of A001045(n+1) (generalized (1,2)-Fibonacci), n>=0.

1, 1, 1, 3, 2, 1, 5, 7, 3, 1, 11, 16, 12, 4, 1, 21, 41, 34, 18, 5, 1, 43, 94, 99, 60, 25, 6, 1, 85, 219, 261, 195, 95, 33, 7, 1, 171, 492, 678, 576, 340, 140, 42, 8, 1, 341, 1101, 1692, 1644, 1106, 546, 196, 52, 9, 1

Offset: 0

Author

Wolfdieter Lang, Aug 02 2002

Keywords

• nonn
• easy
• tabl

Comments

The g.f. for the row polynomials P(n,x) = Sum_{m=0..n} T(n,m)*x^m is 1/(1-(1+x+2*z)*z). See Shapiro et al. reference and comment under A053121 for such convolution triangles.

Riordan array (1/(1-x-2*x^2), x/(1-x-2*x^2)). - Paul Barry, Mar 15 2005

Subtriangle (obtained by dropping the first column) of the triangle given by (0, 1, 2, -2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 19 2013

The number of ternary words of length n having k letters equal 2 and 0,1 avoid runs of odd lengths. - Milan Janjic, Jan 14 2017

Examples

			Triangle begins as:
    1;
    1,   1;
    3,   2,   1;
    5,   7,   3,   1;
   11,  16,  12,   4,   1;
   21,  41,  34,  18,   5,   1;
   43,  94,  99,  60,  25,   6,   1;
   85, 219, 261, 195,  95,  33,   7,   1;
  171, 492, 678, 576, 340, 140,  42,   8,   1;
The triangle (0, 1, 2, -2, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins:
  1;
  0,  1;
  0,  1,  1;
  0,  3,  2,  1;
  0,  5,  7,  3,  1;
  0, 11, 16, 12,  4,  1;
  0, 21, 41, 34, 18,  5,  1; - _Philippe Deléham_, Feb 19 2013
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened
• W. Lang, First 10 rows.
• Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Crossrefs

Columns: A001045 (k=0), A073371 (k=1), A073372 (k=2), A073373 (k=3), A073374 (k=4), A073375 (k=5), A073376 (k=6), A073377 (k=7), A073378 (k=8), A073379 (k=9).

Cf. A002605 (row sums), A006130 (diagonal sums), A073399, A073400.

Cf. A000045, A077957.

Programs

Magma

A073370:= func< n,k | (&+[Binomial(n-j,k)*Binomial(n-k-j,j)*2^j: j in [0..Floor((n-k)/2)]]) >;
[A073370(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 01 2022

Maple

# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> (2^n - (-1)^n) / 3); # Peter Luschny, Oct 07 2022

Mathematica

T[n_, k_]:= T[n, k]= Sum[Binomial[n-j,k]*Binomial[n-k-j,j]*2^j, {j,0,Floor[(n- k)/2]}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 01 2022 *)

SageMath

def A073370(n,k): return binomial(n,k)*sum( 2^j * binomial(2*j,j) * binomial(n-k,2*j)/binomial(n,j) for j in range(1+(n-k)//2))
flatten([[A073370(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Oct 01 2022

Formula

T(n, m) = Sum_{k=0..floor((n-m)/2)} binomial(n-k, m)*binomial(n-m-k, k)*2^k, if n > m, else 0.

Sum_{k=0..n} T(n, k) = A002605(n+1).

T(n, m) = (1*(n-m+1)*T(n, m-1) + 2*2*(n+m)*T(n-1, m-1))/((1^2 + 4*2)*m), n >= m >= 1, T(n, 0) = A001045(n+1), n >= 0, else 0.

T(n, m) = (p(m-1, n-m)*1*(n-m+1)*T(n-m+1) + q(m-1, n-m)*2*(n-m+2)*T(n-m))/(m!*9^m), n >= m >= 1, with T(n) = T(n, m=0) = A001045(n+1), else 0; p(k, n) = Sum_{j=0..k} (A(k, j)*n^(k-j) and q(k, n) = Sum_{j=0..k} B(k, j)*n^(k-j), with the number triangles A(k, m) = A073399(k, m) and B(k, m) = A073400(k, m).

G.f.: 1/(1-(1+2*x)*x)^(m+1) = 1/((1+x)*(1-2*x))^(m+1), m >= 0, for column m (without leading zeros).

T(n, 0) = A001045(n), T(1, 1) = 1, T(n, k) = 0 if k>n, T(n, k) = T(n-1, k-1) + 2*T(n-2, k) + T(n-1, k) otherwise. - Paul Barry, Mar 15 2005

G.f.: (1+x)*(1-2*x)/(1-x-2*x^2-x*y) for the triangle including the 1, 0, 0, 0, 0, ... column. - R. J. Mathar, Aug 11 2015

From Peter Bala, Oct 07 2019: (Start)

Recurrence for row polynomials: R(n,x) = (1 + x)*R(n-1,x) + 2*R(n-2,x) with R(0,x) = 1 and R(1,x) = 1 + x.

The row reverse polynomial x^n*R(n,1/x) is equal to the numerator polynomial of the finite continued fraction 1 + x/(1 - 2*x/(1 + ... + x/(1 - 2*x/(1)))) (with 2*n partial numerators). Cf. A110441. (End)

From G. C. Greubel, Oct 01 2022: (Start)

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

T(n, k) = binomial(n, k)*Hypergeometric2F1([(k-n)/2, (k-n+1)/2], [-2*n], -8).

Sum_{k=0..n} (-1)^k * T(n, k) = A077957(n).

Sum_{k=0..floor(n/2)} T(n-k, k) = A006130(n).

Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = A000045(n+1). (End)