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A101747
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Numbers n such that ((0!)^2+(1!)^2+(2!)^2+...+(n!)^2)/6 is prime.
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1
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OFFSET
| 1,1
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COMMENTS
| 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.
No other n < 8000. [From T. D. Noe (noe(AT)sspectra.com), Jul 31 2008]
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MATHEMATICA
| f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]
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CROSSREFS
| Cf. A061062 (S(n)), A100288 (primes of the form S(n)-1), A100289 (n such that S(n)-1 is prime), A101746 (primes of the for S(n)/6).
Sequence in context: A072599 A095138 A026475 * A134338 A084919 A153100
Adjacent sequences: A101744 A101745 A101746 * A101748 A101749 A101750
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KEYWORD
| fini,nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), Dec 18 2004
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