rev(2*(10^k-4)) = 3*(10^k-3). If 10^k-3 is prime, then phi(3*(10^k-3)) = 2*(10^k-4), so 2*(10^k-4) is a term. 10^1-3=7 is prime, so 2*(10^1-4)=12 is a term, a(2). 10^2-3=97 is prime, so 2*(10^2-4)=192 is a term, a(4). 10^3-3=997 is prime, so 2*(10^3-4)=1992 is a term, a(5). 10^17-3 is prime, so 2*(10^17-4)=199999999999999992 is a term. 10^140-3 is prime, so 2*(10^140-4) is a term. 10^990-3 is prime, so 2*(10^990-4) is a term. Conjecture: sequence is infinite. -
Ray Chandler, Jul 20 2003
Let f(m,n,r,t)=((9).(m).78.(0)(n).21.(9)(m))(r).(9)(t).7 where m, n, r & t are nonnegative integers; dot between numbers means concatenation and "(m)(n)" means number of m's is n. If r*t=0 & p=f(m,n,r,t) is prime then reversal(3*p) = 1.((9)(m).56.(0)(n).43.(9)(m))(r).(9)(t).2 is in the sequence. For example p1=f(0,0,0,0)=7 so reversal(3*p1) = 12 is in the sequence, p2=f(0,0,2,0)=(7821)(2).7=782178217 so reversal(3*p2) = 1.(5643)(2).2 = 1564356432 is in the sequence & p3=f(0,0,674,0) so reversal(3*p3) = 1.(5643)(674).2 is in the sequence. Primes of the form f(m,n,r,t) are a generalized form of primes of the form 10^j-3 that were already related to this sequence by Ray Chandler. For all n,
A085331(n) = reversal(
A072395(n)). -
Farideh Firoozbakht, Jan 08 2005
