A245695 -id:A245695 - cp's OEIS frontend

A245696 Least number k >= 0 such that (n!-k)/n is prime.

0, 4, 5, 42, 7, 8, 279, 130, 121, 156, 13, 322, 15, 752, 901, 1062, 779, 2020, 651, 682, 1679, 2136, 1825, 3874, 999, 1204, 2929, 930, 31, 1952, 33, 34, 6755, 4068, 4699, 3686, 39, 2920, 3403, 5502, 3397, 4796, 4905, 2438, 4183, 3792, 5047, 2950, 4947, 9308, 3551, 3186, 6985, 3416, 26277, 16066, 6431, 8220, 8479, 4402, 4473, 6464, 23335, 8382, 21239, 12988, 17319, 7210, 6887, 54072, 11899, 27602

Offset: 3

Derek Orr, Jul 29 2014

nonn

a(n) < n! for all n > 2.

a(n) = n times (least m >= 0 such that (n-1)!-m is prime) = n*A033933(n-1). - Jens Kruse Andersen, Jul 30 2014 (This shows that a(n) always exists.)

			(6!-42)/6 = 113 is prime. Since 42 is the smallest number to produce a prime, a(6) = 42.

Jens Kruse Andersen, Table of n, a(n) for n = 3..2001

Cf. A033933, A245695, A245697.

lnk[n_]:=Module[{k=0},While[!PrimeQ[(n!-k)/n],k++];k]; Array[lnk,80,3] (* Harvey P. Dale, Jan 30 2023 *)

a(n)=for(k=0,10^6,s=(n!-k)/n;if(floor(s)==s,if(ispseudoprime(s),return(k))))
n=3;while(n<100,print1(a(n),", ");n++)

A245697 Least number k such that (n!+k)/n and (n!-k)/n are both prime.

Original entry on oeis.org

0, 4, 25, 42, 133, 152, 279, 170, 121, 204, 1079, 938, 5295, 3632, 2771, 1062, 1159, 2260, 7413, 682, 33281, 13704, 9725, 4966, 9099, 24724, 2929, 54690, 20429, 22688, 5379, 46274, 15365, 11052, 40441, 65854, 97149, 42520, 44731, 83958, 61877, 4796, 123885, 27922, 122999, 12912, 5047

Offset: 3

Derek Orr, Jul 29 2014

nonn

a(n) < n! for all n > 2.
It is believed that a(n) exists for all n > 2.
a(n) = n times (least m such that (n-1)!+m and (n-1)!-m are both prime) = n*A075409(n-1). - Jens Kruse Andersen, Jul 30 2014 [Goldbach's conjecture would then imply that a(n) always exists.]

			(4!+4)/4 = 7 is prime and (4!-4)/4 = 5 is prime. Thus a(4) = 4.

Jens Kruse Andersen, Table of n, a(n) for n = 3..101

Cf. A075409, A245695, A245696.

a(n)=for(k=0,10^7,s1=(n!-k)/n;s2=(n!+k)/n;if(floor(s1)==s1&&floor(s2)==s2,if(ispseudoprime(s1)&&ispseudoprime(s2),return(k))))
n=3;while(n<100,print1(a(n),", ");n++)

cp's OEIS Frontend

A245696 Least number k >= 0 such that (n!-k)/n is prime.

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