A129978 Numbers k such that A120265(k) = numerator(Sum_{j=1..k} 1/j!) is a prime.

2, 3, 4, 5, 6, 7, 12, 16, 19, 21, 22, 25, 41, 114, 181, 236, 2003, 6138

Offset: 1

Author

Alexander Adamchuk, Jun 13 2007

Keywords

• more
• nonn

Comments

Corresponding primes are A120265(a(n)) = {3, 5, 41, 103, 1237, 433, 164611949, 35951249665217, 52255141388393, 43894318766250120011, 386270005143001056097, 53952693026046706215979, 1249584099900912571604389306768231303904375213027, ...}.

a(17) > 1000; A120265(1000) ~ 2.9*10^2564 = (e-1)*A061355(1000). - M. F. Hasler, Jun 18 2007

Crossrefs

Cf. A120265, A061354, A061355.

Programs

Mathematica

Do[ f=Numerator[ Sum[ 1/k!, {k,1,n} ] ]; If[ PrimeQ[f], Print[{n,f}] ], {n,1,236} ]
Flatten[Position[Numerator[Accumulate[1/Range[2150]!]],?PrimeQ]] (* _Harvey P. Dale, May 03 2013 *)

PARI

my(t=0); for( n=1,1000, if( ispseudoprime( numerator( t+=1/n!)), print1( n", " ))) \\ M. F. Hasler, Jun 18 2007

Extensions

Edited by M. F. Hasler, Jun 18 2007

a(17) from Alexander Adamchuk, May 02 2010

a(18) from Michael S. Branicky, Sep 24 2024