FullSimplify gives wrong answer when operating on HypergeometricPFQ

Question

When I try to simplify this hypergeometric function

FullSimplify[HypergeometricPFQ[{1/2 - n/2}, {-(1/2) - n/2}, -a^2]]

Mathematica 10.0.1.0 returns

(Exp[-a^2] (1 + n + 2 a^2))/(1 + n)

but if I compare the two expressions for different integer values of n,

Table[(Exp[-a^2] (1 + 2 a^2 + n))/(1 + n) ==
      HypergeometricPFQ[{1/2 - n/2}, {-(1/2) - n/2}, -a^2], {n, 0, 6}]

they are different for odd n

{True, 1/2 (2 + 2 a^2) E^-a^2 == 1, True, 1/4 (4 + 2 a^2) E^-a^2 == 1 - a^2/2,
 True, 1/6 (6 + 2 a^2) E^-a^2 == 1 - (2 a^2)/3 + a^4/6, True}

What's going on?

(This behavior persists if I use FunctionExpand[] in place of FullSimplify[], since the latter calls the former.)

Answer 1

To recap, the result returned by FunctionExpand[Hypergeometric1F1[1/2 - n/2, -(1/2) - n/2, -a^2]] is a generic solution, which is intended to be correct in most cases. Unfortunately, the case of odd n is not part of those cases:

Table[Hypergeometric1F1[1/2 - n/2, -(1/2) - n/2, -a^2] == 
      Exp[-a^2] (1 + 2 a^2 + n)/(1 + n) // FullSimplify // TrueQ, {n, 10}]
   {False, True, False, True, False, True, False, True, False, True}

To handle the odd n case, we first do a preliminary transformation:

Hypergeometric1F1[1/2 - n/2, -(1/2) - n/2, -a^2] /. n -> 2 k + 1 // Simplify
   Hypergeometric1F1[-k, -1 - k, -a^2]

which shows the source of the failure: the case of odd n results in a Kummer function with a nonpositive numerator parameter, which is precisely the degenerate polynomial case. To explicitly deal with this, we go back to the defining series for ${}_1 F_1$:

$${}_1 F_1\left({{a}\atop{b}}\mid z\right)=\sum_{j=0}^\infty \frac{(a)_j}{(b)_j}\frac{z^j}{j!}$$

For the special case under consideration, we have

$${}_1 F_1\left({{-k}\atop{-1-k}}\mid -a^2\right)=\sum_{j=0}^\infty \frac{(-k)_j}{(-1-k)_j}\frac{\left(-a^2\right)^j}{j!}$$

Since $(-k)_j=0$ if $j>k$, the series terminates:

$${}_1 F_1\left({{-k}\atop{-1-k}}\mid -a^2\right)=\sum_{j=0}^k \frac{(-k)_j}{(-1-k)_j}\frac{\left(-a^2\right)^j}{j!}$$

Using

FullSimplify[Pochhammer[-k, j]/Pochhammer[-1 - k, j]]
   1 - j/(1 + k)

we thus have

Sum[(1 - j/(1 + k)) (-a^2)^j/j!, {j, 0, k}] // FullSimplify
   ((-a)^k a^(4 + k) E^a^2 + (1 + a^2 + k) Gamma[2 + k, -a^2])/
   (E^a^2 (1 + k)*Gamma[2 + k])

We can now undo the earlier substitution:

s[n_] = % /. k -> (n - 1)/2 // FullSimplify
   (2 ((-a)^((-1 + n)/2) a^((7 + n)/2) E^a^2 + ((1 + 2 a^2 + n)
    Gamma[(3 + n)/2, -a^2])/2))/(E^a^2 (1 + n) Gamma[(3 + n)/2])

The rather complicated closed form gives no indication that it is in fact a polynomial. Nevertheless,

Table[Hypergeometric1F1[1/2 - n/2, -(1/2) - n/2, -a^2] == s[n] // 
      FullSimplify, {n, 1, 11, 2}]
   {True, True, True, True, True, True}

s[7] // FullSimplify
   (24 - 18*a^2 + 6*a^4 - a^6)/24