F[1]==F[2]==1
F[n+2]==F[n+1]+F[n]
F[n]==((1+Sqrt[5])^n-(1-Sqrt[5])^n)/(2^n Sqrt[5])

(*diagonals in Pascal's triangle*)
Column@Table[ Binomial[n - i, i], {n, 0, 10}, {i, 0, Floor[n/2]}]
{1}
{1,1}
{1,2}
{1,3,1}
{1,4,3}
{1,5,6,1}
{1,6,10,4}
{1,7,15,10,1}
{1,8,21,20,5}
{1,9,28,35,15,1}
F[n]==Sum[ Binomial[n - i - 1, i], {i, 0, Floor[(n-1)/2]}]