login
Triangle read by rows of incomplete convolutions of Fibonacci numbers F(n+1) = A000045(n+1), n>=0.
13

%I #18 Apr 11 2016 23:12:14

%S 1,1,2,2,3,5,3,5,7,10,5,8,12,15,20,8,13,19,25,30,38,13,21,31,40,50,58,

%T 71,21,34,50,65,80,96,109,130,34,55,81,105,130,154,180,201,235,55,89,

%U 131,170,210,250,289,331,365,420,89,144,212,275,340,404,469,532,600,655,744,144,233,343,445

%N Triangle read by rows of incomplete convolutions of Fibonacci numbers F(n+1) = A000045(n+1), n>=0.

%C The diagonals d>=0 (d=0: main diagonal) give convolutions of Fibonacci numbers F(n+1), n>=0, with those with d-shifted index: a(d+n,d)=sum(F(k+1)*F(d+n+1-k),k=0..n), n>=0.

%C The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(x*z)*(A(z)-x*A(x*z))/(1-x), with A(x) := 1/(1-x-x^2) (g.f. Fibonacci F(n+1), n>=0).

%C The diagonals give A001629(n+2), A023610, A067331-4, A067430-1, A067977-8 for d= n-m= 0..9, respectively.

%C A row with n terms = the dot product of vectors with n terms: (1,1,2,3,...)dot(...3,2,1,1) with carryovers; such that (3, 5, 7, 10) = (1*3=3), (1*2+3=5), (2*1+5=7), (3*1+7=10).

%H Michael De Vlieger, <a href="/A067330/b067330.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n, m)= sum(F(k+1)*F(n-k+1), k=0..m), n>=m>=0, else 0.

%F a(n, m)= (((3*m+5)*F(m+1)+(m+1)*F(m))*F(n-m+1)+(m*F(m+1)+2*(m+1)*F(m))*F(n-m))/5.

%F G.f. for diagonals d=n-m>=0: (x^d)*(F(d+1)+F(d)*x)/(1-x-x^2)^2, with F(n) := A000045(n) (Fibonacci).

%F a(n, m) = ((-1)^m*F(n-2*m-1)+m*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(m+1)*L(n+2)-A067979(n, m))/5, n>=m>=0. - _Ehren Metcalfe_, Apr 11 2016

%e {1}; {1,2}; {2,3,5}; {3,5,7,10}; ...; p(2,n)= 2+3*x+5*x^2.

%t 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 *)

%Y Cf. A067418 (triangle with rows read backwards).

%K nonn,easy,tabl

%O 0,3

%A _Wolfdieter Lang_, Feb 15 2002