A101747 Numbers k such that ((0!)^2+(1!)^2+(2!)^2+...+(k!)^2)/6 is prime.

3, 4, 5, 6, 7, 19, 40, 56, 93

Offset

1,1

Comments

Let S(k) = Sum_{i=0,..k} (i!)^2. Note that 6 divides S(k) for k>1. For prime p=20879, p divides S(p-1). Hence p divides S(k) for all k >= p-1 and all prime values of S(k)/6 are for k < p-1. These k yield provable primes for k <= 93. No other k < 4000.

No other k < 8000. [T. D. Noe, Jul 31 2008]

Links

Table of n, a(n) for n=1..9.

Maple

S:= proc(k) local i; add((i!)^2, i=0..k) end proc:
select(k -> isprime(S(k)/6), [$2..100]); # Robert Israel, Jan 25 2026

Mathematica

f2=1; s=2; Do[f2=f2*n*n; s=s+f2; If[PrimeQ[s/6], Print[{n, s/6}]], {n, 2, 100}]
Flatten[Position[Accumulate[Table[(n!)^2/6, {n, 0, 100}]], _?(PrimeQ[#]&)]]-1 (* Harvey P. Dale, Jan 09 2026 *)

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 form S(n)/6).

Sequence in context: A072599 A095138 A026475 * A383190 A355703 A134338

Adjacent sequences: A101744 A101745 A101746 * A101748 A101749 A101750

Keyword

fini,nonn

Author

T. D. Noe, Dec 18 2004

Status

approved