

A101746


Primes of the form ((0!)^2+(1!)^2+(2!)^2+...+(n!)^2)/6.


1




OFFSET

1,1


COMMENTS

Let S(n)=sum_{i=0,..n1} (i!)^2. Note that 6 divides S(n) for n>1. For prime p=20879, p divides S(p1). Hence p divides S(n) for all n >= p1 and all prime values of S(n)/6 are for n < p1.


LINKS

Table of n, a(n) for n=1..6.


MATHEMATICA

f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]


CROSSREFS

Cf. A061062 (S(n)), A100288 (primes of the form S(n)1), A100289 (n such that S(n)1 is prime), A101747 (n such that S(n)/6 is prime).
Sequence in context: A188946 A234292 A177752 * A318398 A318815 A195246
Adjacent sequences: A101743 A101744 A101745 * A101747 A101748 A101749


KEYWORD

fini,nonn


AUTHOR

T. D. Noe, Dec 18 2004


STATUS

approved



