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Numbers n such that the denominator of n!/!n (= A000142(n)/A000166(n)) is prime.
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%I #18 Nov 03 2017 03:48:11

%S 4,5,6,7,10,11,15,16,34,44,63,66,168,427,575,928,1094,1218,1363,1713

%N Numbers n such that the denominator of n!/!n (= A000142(n)/A000166(n)) is prime.

%C That is, numbers n such that A053557(n) is prime. - _Michel Marcus_, Aug 28 2013

%t a[n_] := Sum[(-1)^k/k!, {k, 0, n}]; Select[Range[100], PrimeQ[Numerator[a[#]]] &] (* _G. C. Greubel_, Oct 31 2017 *)

%o (PARI) isok(n) = isprime(numerator(sum(k=0, n, (-1)^k/k!))); \\ _Michel Marcus_, Aug 28 2013

%Y Cf. A000142, A000166, A053557.

%K nonn,more

%O 1,1

%A _Ed Pegg Jr_, Jul 11 2008

%E a(16)-a(20) from _G. C. Greubel_, Nov 01 2017