4,6,9.10.14.15,……的通项公式

a(n)=a(n-1)+a(n-2)-1
a(1)=4
a(2)=6
The auxiliary equation of a(n)=a(n-1)+a(n-2)
is x^2-x-1=0
x=(1+sqrt(5))/2 or (1-sqrt(5))/2
So a(n)=c*((1+sqrt(5))/2)^n + d*((1-sqrt(5))/2)^n where c and d are constant
Solve c*((1+sqrt(5))/2) + d*((1-sqrt(5))/2)=4
c*((1+sqrt(5))/2)^2 + d*((1-sqrt(5))/2)^2=6
c=(3-sqrt(5))/sqrt(5) d=(4*(2 sqrt(5)-5))/(5*(sqrt(5)-1))
通项公式=c*((1+sqrt(5))/2)^n+d*((1-sqrt(5))/2)^(n)-2n+3(For n>2)