show/hide this revision's text 3 correct typo

Using Robbins' [1] form of Stirling's formula,

$$\sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n+1))< n!< \sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n))$$

we get

$$\left\lceil\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-11))\right\rceil$$ $$\le (n-1)!\le$$ $$\left\lfloor\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-12))\right\rfloor$$

which is accurate enough to distinguish prime from composite for $n\le8$. For larger numbers, the error bound is too large.


This can be extended further using a modification of Wilson's theorem: for n > 9, $$\lfloor n/2\rfloor!\equiv0\pmod n$$ if and only if n is composite. This allows testing 10 through 15, plus (with some cleverness) 17.

With tighter explicit bounds and high-precision evaluation, it might be possible to test as high as 100 with related methods: direct evaluation up to 25 and the 'divide by 4' variant of the above for n > 25.

This is not so much 'using a cannon to swat a fly' (using methods more powerful than needed) as it is 'using the space station to swat a fly': the methods must be extremely powerful and accurate to do very little.


[1] H. Robbins, "A Remark of on Stirling's Formula." The American Mathematical Monthly 62 (1955), pp. 26-29.

show/hide this revision's text 2 extended to {5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17}

Using Robbins' [1] form of Stirling's formula,

$$\sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n+1))< n!< \sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n))$$

we get

$$\left\lceil\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-11))\right\rceil$$ $$\le (n-1)!\le$$ $$\left\lfloor\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-12))\right\rfloor$$

which is accurate enough to distinguish prime from composite for $n\le8$. For larger numbers, the error bound is too large.


This can be extended further using a modification of Wilson's theorem: for n > 9, $$\lfloor n/2\rfloor!\equiv0\pmod n$$ if and only if n is composite. This allows testing 10 through 15, plus (with some cleverness) 17.

With tighter explicit bounds and high-precision evaluation, it might be possible to test as high as 100 with related methods: direct evaluation up to 25 and the 'divide by 4' variant of the above for n > 25.

This is not so much 'using a cannon to swat a fly' (using methods more powerful than needed) as it is 'using the space station to swat a fly': the methods must be extremely powerful and accurate to do very little.


[1] H. Robbins, "A Remark of Stirling's Formula." The American Mathematical Monthly 62 (1955), pp. 26-29.

show/hide this revision's text 1

Using Robbins' [1] form of Stirling's formula,

$$\sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n+1))< n!< \sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n))$$

we get

$$\left\lceil\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-11))\right\rceil$$ $$\le (n-1)!\le$$ $$\left\lfloor\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-12))\right\rfloor$$

which is accurate enough to distinguish prime from composite for $n\le8$. For larger numbers, the error bound is too large.

[1] H. Robbins, "A Remark of Stirling's Formula." The American Mathematical Monthly 62 (1955), pp. 26-29.