A239374 Smallest product of consecutive distinct prime factors of t = prime(n)^2 - 1 in ascending order that provides more than 1/3 factored parts for Brillhart-Lehmer-Selfridge primality test for prime(n).

2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 6

Offset: 2

Author

Lei Zhou, Mar 17 2014

Keywords

• nonn
• easy

Comments

The first greater than 2 element of this sequence is a(99).

Examples

			n = 2: prime(2) = 3, 3^2 - 1 = 8 = 2^3, 2^3 > 3, 100% factorization.  So a(2) = 2.
n = 45: prime(45) = 197, 197^2 - 1 = 38808 = 2^3*3^2*7^2*11, 2^3 = 8,  log_197(8) = 0.3936 > 1/3, 39.36% factorization.  So a(45) = 2.
n = 99: prime(99) = 523, 523^2 - 1 = 273528 = 2^3*3^2*29*131, 2^3 = 8, log_523(8) = 0.3322 < 1/3, log_523(2^3*3^2) = 0.6832 > 1/3, 68.32% factorization. So a(99) = 6.
		

Links

• Lei Zhou, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A000040, A177854.

Programs

Mathematica

Table[p = Prime[n]; ck = p^(1/3); sp = p^2 - 1; dp = sp; prod = 1; fp = Union[Transpose[FactorInteger[p + 1]][[1]], Transpose[FactorInteger[p - 1]][[1]]]; i = 0; While[i++; m = fp[[i]]; prod = prod*m; While[Divisible[sp, m], sp = sp/m]; (dp/sp) < ck]; prod, {n, 2, 100}]