|
%I #26 Feb 24 2013 13:08:08
%S 2,2,2,2,3,2,3,2,2,3,3,2,3,3,2,3,3,2,3,3,5,3,3,2,3,3,2,3,3,2,3,2,5,3,
%T 3,2,3,3,2,3,3,2,3,3,5,3,3,2,3,3,5,3,3,2,5,3,2,3,3,2,3,3,5,2,5,2,3,3,
%U 2,3,3,2,3,3,5,3,3,2,5,3,5,3,3,2,3,3,7
%N a(n) is the smallest possible largest prime factor of the difference of two prime numbers q > p such that p + q = 2n.
%C Test up to n=1000000 shows that,
%C when n is odd, a(n)<=prime(m+3) such that m is the number of prime factors of n that are smaller or equal to prime(m+3);
%C when n is even, a(n)<=prime(m+4) such that m is the number of prime factors of n that are smaller or equal to prime(m+4).
%C This is hypothesized true for all n >= 4.
%C The first appearance of prime(k) in this sequence is A217016.
%H Lei Zhou, <a href="/A217403/b217403.txt">Table of n, a(n) for n = 4..10000</a>
%e For n<4, there is not a pair of different prime numbers such that p+q=2n.
%e When n=4, we have p=3 and q=5 such that 3+5=2n=8 (the only case). The largest prime factor of q-p=2 is 2, so a(4)=2;
%e ...
%e When n=90, we have prime number pairs (7,173) (13,167), (17,163), (23,157), (29,151), (31,149), (41,139), (43,137), (53,127), (67,113), (71,109), (73,107), (79,101), (83,97), totalled 14 pairs such that p1+p2=2n=180. The difference of each pairs are 173-7=166=2*83, 163-17=154=2*7*11, 163-17=146=2*73, 157-23=134=2*67, 151-29=122=2*61, 149-31=118=2*59, 139-41=98=2*7^2, 137-43=94=2*47, 127-53=74=2*37, 113-67=46=2*23, 109-71=38=2*19, 107-73=34=2*17, 101-79=22=2*11, and 97-83=14=2*7 respectively. Among these prime factorizations, the largest prime factors for each pairs are, 83, 11, 73, 67, 61, 59, 7, 47, 37, 23, 19, 17, 11, and 7 respectively. In the fourteen prime numbers, the smallest one is 7. So a(90)=7.
%e Additionally, 90=2*3^2*5, prime(3)=5, a(90)=7<prime(3+4)=17, consistent with the hypothesis.
%t Table[a = i; Do[If[PrimeQ[i - j] && PrimeQ[i + j], f = Last[FactorInteger[j*2]][[1]]; If[a > f, a = f]], {j, 1, i - 1}]; a, {i, 4, 100}]
%Y Cf. A002375, A217016.
%K nonn,easy
%O 4,1
%A _Lei Zhou_, Feb 18 2013
|
