OFFSET
1,1
COMMENTS
Primes p such that nextprime(p)-p >= 4.
Primes p such that p+2 divides (p-1)!.
Odd primes n such that n!*B(n+1) is an integer, where B(k) are the Bernoulli numbers. - Benoit Cloitre, Feb 06 2002
Sequence appears also to give all n > 1 such that there is no prime p satisfying the inequality n < p < n+tau(n)^2 where tau(n)=A000005(n). - Benoit Cloitre, Apr 13 2002
Conjecture: start from any initial value f(1) >= 2 and define f(n) to be the largest prime factor of f(1) +f(2) + ... +f(n-1); then f(n) = n/2 + O(log(n)) and there are infinitely many primes p such that f(2p)=p. Conjecture: current sequence gives primes satisfying f(2p)=p when f(1)=3. - Benoit Cloitre, Jun 04 2003
Numbers k such that 2((k-1)! + 1) + k is divisible by k(k+2). For 7 and 13, the respective quotients are also in the sequence. Are there any other such k? - Ivan N. Ianakiev, Aug 03 2019. The next values of k with respective quotients in this sequence are 103, 1531, 1637. - Amiram Eldar, Jun 08 2020
Numbers k such that 4((k-1)! + 1) == k^2 (mod k(k+2)). - Thomas Ordowski, May 09 2020
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
K. Soundararajan, Small gaps between prime numbers: the work of Goldston-Pintz-Yildirim, Bull. Amer. Math. Soc., 44 (2007), 1-18.
EXAMPLE
13 is here because it is prime and 15 is composite. Also 15 divides 12!.
MAPLE
d:=4; M:=1000; t0:=[]; for n from 1 to M do p:=ithprime(n); if nextprime(p) - p >= d then t0:=[op(t0), p]; fi; od: t0;
MATHEMATICA
Select[Prime[Range[100]], NextPrime[#] -#>=4 &] (* G. C. Greubel, Aug 22 2019 *)
PROG
(PARI) isok(p) = isprime(p) && (p % 2) && !isprime(p+2); \\ Michel Marcus, Feb 25 2014
(Magma) [k:k in PrimesInInterval(3, 400)| not IsPrime(k+2)]; // Marius A. Burtea, Aug 03 2019
(Sage) [nth_prime(n) for n in (1..100) if (nth_prime(n+1) - nth_prime(n)) >= 4] # G. C. Greubel, Aug 22 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Benoit Cloitre, Jun 04 2003
Edited by Don Reble, Dec 20 2006
STATUS
approved