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Let S(n) =sum_ Sum_{i=0,..n-1} (i!)^2. Note that 6 divides S(n) for n>1. For prime p=20879, p divides S(p-1). Hence p divides S(n) for all n >= p-1 and all prime values of S(n)/6 are for n < p-1. These n yield provable primes for n <= 93. No other n < 4000.
No other n < 8000. [From __T. D. Noe_, Jul 31 2008]
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No other n < 8000. [From _T. D. Noe (noe(AT)sspectra.com), _, Jul 31 2008]
_T. D. Noe (noe(AT)sspectra.com), _, Dec 18 2004
No other n < 8000. [From T. D. Noe (noe(AT)sspectra.com), Jul 31 2008]
fini,nonn,new
f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]
fini,nonn,new
Numbers n such that ((0!)^2+(1!)^2+(2!)^2+...+(n!)^2)/6 is prime.
3, 4, 5, 6, 7, 19, 40, 56, 93
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Let S(n)=sum_{i=0,..n-1} (i!)^2. Note that 6 divides S(n) for n>1. For prime p=20879, p divides S(p-1). Hence p divides S(n) for all n >= p-1 and all prime values of S(n)/6 are for n < p-1. These n yield provable primes for n <= 93. No other n < 4000.
f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]
fini,nonn
T. D. Noe (noe(AT)sspectra.com), Dec 18 2004
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