A081545 Smallest prime which is 1 more than the product of n distinct composite numbers.

2, 5, 37, 193, 2161, 23041, 241921, 2903041, 55987201, 958003201, 17915904001, 250822656001, 5518098432001, 142216445952001, 2897001676800001, 90386452316160001, 1807729046323200001, 52563198423859200001

Offset: 0

Author

Amarnath Murthy, Apr 01 2003

Keywords

• nonn
• nice

Comments

Let K(m) be the smallest possible K satisfying the Theorem. Conjecture: K(m) ~ m, i.e., a(k) ~ A002808(1)*...*A002808(k), only very few of the last factors will be insignificantly larger.

Let K(b,r) be the smallest possible K satisfying the Corollary, i.e., the index from which on all a(k)-1 are multiples of b^r. With the preceding conjecture, there are asymptotically at least (k+PrimePi(A002808(k)))/b multiples of b among the factors of a(k)-1, so this is an (asymptotic) lower bound on r.

Experimentally, a(k) = 1 + Product_{i>=1} prime(i)^e(i), with e(1)~k*3/2, e(2)~k*2/3, e(3)~k/4, e(4)~k/5, ... (Here ~ is not asymptotic equivalence.) Is there a simple formula? - M. F. Hasler, Jun 16 2007

Examples

			Writing c(n) for the n-th composite number A002808(n):
a(0) = (Product_{i=1..0} c(i))+1 = 1+1 = 2 (empty product).
a(1) = c(1)+1 = 4+1 = 5.
a(2) = c(1)*c(4)+1 = 4*9+1 = 37, since c(1)*c(k)+1 is not prime for k < 4.
a(3) = c(1)*c(2)*c(3)+1 = 4*6*8 + 1 = 193.
a(4) = c(1)*c(2)*c(4)*c(5)+1 = 2161, nothing better since c(6)*c(3) > c(5)*c(4).
a(5) = c(1)*c(2)*c(3)*c(5)*c(6)+1 = 23041, none better since c(7)*c(4) > c(5)*c(6).
a(6) = c(1)*c(2)*c(3)*c(4)*c(5)*c(7)+1 = 241921, best since c(1)*...*c(6)+1 is not prime.
a(7) = p(7)+1 = 2903041 with p(n) = Product_{i=1..n} c(i). - _M. F. Hasler_, Jun 16 2007
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 0..300 (terms 0..99 from M. F. Hasler)

Crossrefs

Cf. A002808 (composite numbers), A073918, A081546, A131100.

Programs

PARI

A081545(n, b=0 /*best*/, p=1 /*product*/, f=[]/*factors*/)={ if( #f= f[n+1] ) || !b = A081545( n-1, b, p*f[n], f), while( isprime( f[n]++ ),) /* next composite */ ); b } /* then vector(30,n,A081545(n-1)) gives the first 30 terms */ \\ M. F. Hasler, Jun 16 2007

Formula

Theorem: For any m > 0 there is a K > 0 such that for all k > K, a(k)-1 is divisible by the first m composite numbers.

Corollary: For any b > 1, r > 0 there is a K > 0 such that for all k > K, a(k) == 1 (mod b^r). Taking b=10 shows that all a(k) > a(8) end in 0..01 with an increasing number of zeros. - M. F. Hasler, Jun 16 2007

Extensions

More terms from Michel ten Voorde Jun 13 2003

Terms beyond a(8) by M. F. Hasler Jun 16 2007