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A101746
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Primes of the form ((0!)^2+(1!)^2+(2!)^2+...+(n!)^2)/6.
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2
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OFFSET
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1,1
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COMMENTS
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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.
The next term (a(7)) has 96 digits. The largest term (a(9)) has 288 digits. - Harvey P. Dale, Aug 31 2021
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LINKS
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MATHEMATICA
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f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]
Select[Accumulate[(Range[0, 25]!)^2]/6, PrimeQ] (* Harvey P. Dale, Aug 31 2021 *)
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CROSSREFS
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KEYWORD
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fini,nonn
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AUTHOR
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STATUS
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approved
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