**Final Proof of the Identity and of the Second Order Recurrence**

Let be any solution of the second order recurrence which has been obtained for the left-hand side:

`> `
**rec[left];**

Thus:

`> `
**h(n+2)=collect(solve(",h(n+2)),h,normal);**

`> `
**subs(n=n+1,");**

and

`> `
**subs(n=n+1,");**

Then, also solves the recurrence for the right-hand side:

`> `
**collect(subs(","",""",rec[right]),h,normal);**

At this point, we have proved that both sides of the equation satisfy the same recurrence of order 4. To prove the announced equality, we simply need to check 4 initial conditions, since the leading coefficient of the recurrence of order 4,

`> `
**coeff(rec[right],h(n+3));**

never vanishes for non-negative . Now the proof of the identity

`> `

simplify follows from

`> `
**eval(subs(n=0,Sum=add,eq));**

`> `
**eval(subs(n=1,Sum=add,eq));**

`> `
**eval(subs(n=2,Sum=add,eq));**

`> `
**eval(subs(n=3,Sum=add,eq));**

Therefore, the Apéry numbers also satisfy the announced second order recurrence.