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ocfnash
The poor Maple developer who tried to fix the "bug" is active on MathOverflow and commented briefly on his experience of this in a comment below this answer: https://mathoverflow.net/a/11607
twtw
Unfortunately I can't think through it now, but I would bet there is a fun connection to be made with 3*5=15. Something about roots more frequently getting repeated, or zeros in the s-plane going to a higher order maybe? I'd be delighted if someone can chime in with the connection, if there is one.
SidiousL
You can compute this integral as follows. First, you remember that sinc(x) = sin(x)/x is the Fourier transform of 1/2 * indicator function of [-1, 1]. In general you have:

  sinc(x/n) = n/2 \int_{-1/n}^{1/n} d t e^{i t x}.
Next, notice that

  \int_0^\infty sinc(x) \dots sinc(x/n) = 1/2 \int_{-\infty}^\infty sinc(x) \dots sinc(x/n).
Next, replace the sinc functions by their Fourier representation, change the order of integration and use the formula

  \int_{-\infty}^\infty d x e^{i a x} = 2 pi \delta(a),
where \delta(a) is the Dirac delta. You are left with an integral of a delta function over a product of intervals. What remains to be done is just some tedious computation.

\int_{-1}^1 d t_0 \dots \int_{-1/(2 k + 1)}^{1/(2 k + 1)} d t_k 1/2 \dots (2 k + 1)/2 2 \pi \delta(t_1 + \dots + t_k).

Next, the question is whether t_1 + \dots + t_k = 0 has a solution for t's in the integration domain. This holds until k = 7. To see this, compute

1/3 + 1/5 = 8/15 < 1

1/3 + 1/5 + 1/7 = 71/105 < 1

1/3 + 1/5 + 1/7 + 1/9 = 248/315 < 1

1/3 + 1/5 + 1/7 + 1/9 + 1/11 = 3043/3465 <1

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 = 43024/45045 < 1

but

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 = 46027/45045 > 1.