A060920 Bisection of Fibonacci triangle A037027: even-indexed members of column sequences of A037027 (not counting leading zeros).

1, 2, 1, 5, 5, 1, 13, 20, 9, 1, 34, 71, 51, 14, 1, 89, 235, 233, 105, 20, 1, 233, 744, 942, 594, 190, 27, 1, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 1597, 6865, 12473, 12402, 7285, 2534, 490, 44, 1, 4181, 20284, 42447, 49963, 36122, 16407, 4578, 726, 54, 1

Offset: 0

Author

Wolfdieter Lang, Apr 20 2001

Keywords

• nonn
• easy
• tabl

Comments

Companion triangle (odd-indexed members) A060921.

Examples

			Triangle begins as:
      1;
      2,     1;
      5,     5,      1;
     13,    20,      9,      1;
     34,    71,     51,     14,      1;
     89,   235,    233,    105,     20,     1;
    233,   744,    942,    594,    190,    27,     1;
    610,  2285,   3522,   2860,   1295,   315,    35,    1;
   1597,  6865,  12473,  12402,   7285,  2534,   490,   44,    1;
   4181, 20284,  42447,  49963,  36122, 16407,  4578,  726,   54,  1;
  10946, 59155, 140109, 190570, 163730, 91959, 33705, 7776, 1035, 65, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened
• Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, Nov. 2005, pp. 359-370.

Crossrefs

Column sequences: A001519 (k=0), A054444 (k=1), A061178 (k=2), A061179 (k=3), A061180 (k=4), A061181 (k=5).

Cf. A000045, A001519, A007583, A037027, A060921, A061176.

Programs

Magma

A060920:= func< n,k | (&+[Binomial(2*n-k-j, j)*Binomial(2*n-k-2*j, k): j in [0..2*n-k]]) >;
[A060920(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 06 2021

Mathematica

A060920[n_, k_]:= Sum[Binomial[2*n-k-j, j]*Binomial[2*n-k-2*j, k], {j,0,2*n-k}];
Table[A060920[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 06 2021 *)

Sage

def A060920(n,k): return sum(binomial(2*n-k-j, j)*binomial(2*n-k-2*j, k) for j in (0..2*n-k))
flatten([[A060920(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 06 2021

Formula

T(n, k) = A037027(2*n-k, k).

T(n, k) = ((2*(n-k) + 1)*A060921(n-1, k-1) + 4*n*T(n-1, k-1))/(5*k), n >= k >= 1.

T(n, 0) = F(n)^2 + F(n+1)^2 = A001519(n), with the Fibonacci numbers F(n) = A000045(n).

Sum_{k=0..n} T(n, k) = (2^(2*n + 1) + 1)/3 = A007583(n).

G.f. for column m >= 0: x^m*pFe(m+1, x)/(1-3*x+x^2)^(m+1), where pFe(n, x) := Sum_{m=0..n} A061176(n, m)*x^m (row polynomials of signed triangle A061176).

G.f.: (1-x*(1+y))/(1 - (3+2*y)*x + (1+y)^2*x^2). - Vladeta Jovovic, Oct 11 2003