Final Proof of the Identity and of the Second Order Recurrence

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

> rec[left];

[Maple Math]

Thus:

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

[Maple Math]

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

[Maple Math]

and

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

[Maple Math]

Then, [Maple Math] also solves the recurrence for the right-hand side:

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

[Maple Math]

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));

[Maple Math]

never vanishes for non-negative [Maple Math] . Now the proof of the identity

> [Maple Math]

simplify follows from

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

[Maple Math]

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

[Maple Math]

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

[Maple Math]

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

[Maple Math]

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