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A067418
Triangle A067330 with rows read backwards.
13
1, 2, 1, 5, 3, 2, 10, 7, 5, 3, 20, 15, 12, 8, 5, 38, 30, 25, 19, 13, 8, 71, 58, 50, 40, 31, 21, 13, 130, 109, 96, 80, 65, 50, 34, 21, 235, 201, 180, 154, 130, 105, 81, 55, 34, 420, 365, 331, 289, 250, 210, 170, 131, 89, 55, 744, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1308, 1164, 1075, 965
OFFSET
0,2
COMMENTS
The column m (without leading 0's) gives the convolution of Fibonacci numbers F(n+1) := A000045(n+1), n>=0, with those with m-shifted index: a(n+m,m)=sum(F(k+1)*F(m+n+1-k),k=0..n), n>=0, m=0,1,...
The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(z)*(A(z)-x*A(x*z))/(1-x), with A(x) := 1/(1-x-x^2) (g.f. for Fibonacci F(n+1), n>=0).
The columns give A001629(n+2), A023610, A067331-4, A067430-1, A067977-8 for m= 0..9, respectively. Row sums give A067988.
LINKS
FORMULA
a(n, m)=A067330(n, n-m), n>=m>=0, else 0.
a(n, m)= (((3*(n-m)+5)*F(n-m+1)+(n-m+1)*F(n-m))*F(m+1)+((n-m)*F(n-m+1)+2*(n-m+1)*F(n-m))*F(m))/5.
G.f. for column m=0, 1, ...: (x^m)*(F(m+1)+F(m)*x)/(1-x-x^2)^2, with F(m) := A000045(m) (Fibonacci).
a(n, m) = ((-1)^m*F(n-2*m+1)-m*L(n+2)+n*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(n-m+1)*L(n+2)-A067990(n, m))/5, n>=m>=0. - Ehren Metcalfe, Apr 11 2016
EXAMPLE
{1}; {2,1}; {5,3,2}; {10,7,5,3}; ...; p(2,n)=5+3*x+2*x^2.
MATHEMATICA
Reverse /@ Table[Sum[Fibonacci[k + 1] Fibonacci[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)
CROSSREFS
Sequence in context: A171176 A332529 A213849 * A287548 A067323 A106534
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Feb 15 2002
STATUS
approved