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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).
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%I #11 Mar 21 2014 12:28:25

%S 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,

%T 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,

%U 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

%N 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).

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

%H Lei Zhou, <a href="/A239374/b239374.txt">Table of n, a(n) for n = 2..10000</a>

%e n = 2: prime(2) = 3, 3^2 - 1 = 8 = 2^3, 2^3 > 3, 100% factorization. So a(2) = 2.

%e 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.

%e 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.

%t 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}]

%Y Cf. A000040, A177854.

%K nonn,easy

%O 2,1

%A _Lei Zhou_, Mar 17 2014