A090975 Least integer k such that n!+1-k is prime.

0, 0, 0, 0, 2, 8, 2, 2, 32, 14, 12, 0, 2, 24, 2, 48, 54, 60, 42, 102, 32, 32, 74, 90, 74, 150, 38, 0, 102, 32, 2, 62, 2, 2, 194, 114, 128, 0, 2, 74, 84, 0, 80, 110, 110, 54, 90, 80, 104, 60, 98, 180, 68, 60, 128, 62, 462, 278, 110, 138, 140, 72, 72, 102, 360, 128, 318, 192

Offset: 0

Author

Frederick Magata (frederick.magata(AT)uni-muenster.de), Feb 28 2004

Keywords

• nonn

Comments

The (n-1) consecutive numbers n!+2,...,n!+n (for n>=2) are not prime. This fact implies that there are arbitrarily large gaps in the distribution of the prime numbers. n!+1 itself may be a prime number as in the case of n=3, 11, 27 (see A002981 for all such n). Now a(n) measures, when the first prime number previous to n!+2 appears. Thus a(n)=8 means that n!+1-3 is prime and so on. Obviously, the values of a(n) are always even numbers. Conjectures: |a(n)-1| is either 1 or a prime number. Is the growth of b(n) := sum(a(k),k=0..n) quadratic, that is b(n)=O(n^2)?

Examples

			a(3)=0 because 3!+1-0=7 is prime.
a(4)=2 because 4!+1-2=23 is prime and 24 and 25 are not.
		

Crossrefs

Cf. A090786, A002981, A005095.

Programs

Maple

a := proc(n) option remember;n!+1-prevprime(n!+2); end;