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
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
Programs
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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
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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
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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)
Comments