A245696 -id:A245696 - cp's OEIS frontend

A245695 Least number k >= 0 such that (n!+k)/n is prime.

1, 2, 0, 4, 25, 42, 49, 88, 207, 170, 121, 12, 377, 938, 285, 688, 391, 558, 703, 1780, 609, 682, 713, 2328, 3275, 1066, 1593, 28, 1943, 6690, 3317, 4064, 2607, 1258, 3395, 2196, 4847, 38, 1677, 3880, 2173, 42, 4171, 3124, 2115, 10994, 4747, 11184, 2597, 4150, 3111, 14092, 2809, 3834, 12265, 3976, 8493, 6206, 16697, 17580, 16531, 47678, 8253, 17344, 4355, 12738, 18961, 4964, 5727, 9170, 9869, 61704, 7373

Offset: 1

Derek Orr, Jul 29 2014

nonn

a(n) = M*n for some integer M >= 0.

a(n) = n times least m >= 0 such that (n-1)!+m is prime. - Jens Kruse Andersen, Jul 30 2014

			(4!+0)/4 = 6 is not prime.
(4!+1)/4 = 25/4 is not prime.
(4!+2)/4 = 26/4 is not prime.
(4!+3)/4 = 27/4 is not prime.
(4!+4)/4 = 7 is prime. Thus a(4) = 4.

Jens Kruse Andersen, Table of n, a(n) for n = 1..2001

Cf. A033932, A245696, A245697.

a(n)=for(k=0,10^6,s=(n!+k)/n;if(floor(s)==s,if(ispseudoprime(s),return(k))))
n=1;while(n<100,print1(a(n),", ");n++)

a(n) = n*A033932(n-1), except a(3) = 0 where A033932 demands positive values. - Jens Kruse Andersen, Jul 30 2014

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

A245695 Least number k >= 0 such that (n!+k)/n is prime.

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