∴an+ 1-an=an-an- 1

∴an-an- 1=an- 1-an-2=…=a2-a 1=6

∴an=an- 1+6

(n≥2)

When ∴n≥2,

an = an- 1+6 = an-2+6×2 =…= a2+6×(n-2)= 6n-4

When n= 1, a 1=6× 1-4=2.

find

To sum up, an=6n-4.

(2) When c = 1, an+ 1+an- 1=an (n≥2)

∴an+3=an+2-an+ 1=-an，

a3=a2-a 1=6，a4=a3-a2=-2

∴an+6=an+3+3=-an+3=an

∴ Sequence {an} is a sequence with a period of 6.

∵20 14=335×6+

∴a20 14=a4=-2.

(3) Assuming that there is a constant c, an+3=an holds.

∫an+3 = an

an+2+an=can+ 1

∴an- 1+an=can+ 1

①

An+ 1+an- 1 = Yes.

②

If ① is subtracted from ②, (an+ 1-an)( 1+c)=0.

∴an+ 1-an=0, or 1+c=0.

When n∈N*, an+ 1-an=0, the sequence {an} is a constant sequence, which does not meet the requirements.

∴ 1+c=0

∴c=- 1

When c=- 1, there are:

An+ 1+an- 1=-an, that is, for n∈N and n≥2, an+ 1 =-An- 1 exists.

∴an+3=-an+2-an+ 1,an+2=-an+ 1-an.

∴an+3=-an+2-an+ 1,=an+ 1+an-an+ 1=an(n≥ 1).

So there is a constant c=- 1, which makes a +3 = constant.