A242916 Numbers n such that (k!+n)/(k+n) is prime for some k.

1, 6, 11, 16, 34, 110, 113, 351, 586, 708, 839, 1286, 5026, 6518, 9404, 10070, 10154, 10909, 19382, 19951, 20148, 23547

Offset

1,2

Comments

For all a(n) > 1, there is only one k that makes (k!+n)/(k+n) a prime (up to k <= 5000). For a(2) through a(11), they are given by {4, 6, 4, 12, 5, 6, 6, 12, 6, 44} respectively.

From Robert Israel, Aug 09 2016: (Start)

If n > 1 and (k!+n)/(k+n) is prime, we must have k < n.

Proof: if k >= n, then n | k! + n, so (after eliminating some small cases) for (k!+n)/(k+n) to be prime we need n | k+n and thus n | k. But if k = mn,

mn-1 >= m+1 so m+1 | (mn-1)!, and (k!+n)/(k+n)=(m(mn-1)!+1)/(m+1) is not an integer. (End)

Links

Table of n, a(n) for n=1..22.

Example

(k!+6)/(k+6) is prime for some k (let k = 4). Thus 6 is a member of this sequence.

Maple

f:= proc(n) ormap(proc(k) local v; v:= (k!+n)/(k+n); v::integer and isprime(v) end proc, [$2..n-1]) end proc:
f(1):= true:
select(f, [$1..1500]); # Robert Israel, Aug 09 2016

Mathematica

fQ[n_] := Block[{k = 1}, While[k < n && !PrimeQ[(k! + n)/(k + n)], k++]; k < n]; k = 2; lst = {1}; While[k < 25001, If[fQ@k, Print@k; AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Aug 14 2016 *)

Prog

(PARI) a(n)=for(k=1, 5000, s=(k!+n)/(k+n); if(floor(s)==s, if(ispseudoprime(s), return(k))))
n=1; while(n<1000, if(a(n), print1(n, ", ")); n+=1)

Crossrefs

Cf. A242565.

Sequence in context: A191158 A208719 A208775 * A256429 A024730 A024952

Adjacent sequences: A242913 A242914 A242915 * A242917 A242918 A242919

Keyword

nonn,hard,more

Author

Derek Orr, May 26 2014

Extensions

a(12)-a(15) from Robert Israel, Aug 09 2016

a(16)-a(21) from Robert G. Wilson v, Aug 11 2016

a(22) from Robert G. Wilson v, Aug 13 2016

Status

approved