
COMMENTS

If n is a term greater than 2 in this sequence and m = 3*(78*10^n + 217) then phi(m) = reversal(m) (m is in the sequence A069215) because phi(m) = 2*(78*10^n + 216) = 156*10^n + 432 = reversal(234*10^n + 651) = reversal(m).
For example since 8>2 & 8 is in this sequence, for m = 3* (78*10^8 + 217) = 23400000651 phi(m) = reversal(m), so 23400000651 is a term of A069215.
Let f(n,m,r,t) = ((9)(n).78.(0)(m).21.(9)(n))(r).(9)(t).7 where dot between numbers means concatenation and "(m)(n)" means number of m's is n.
In fact I proved that for nonnegative integers n, m, r & t such that r*t = 0 if p = f(n,m,r,t) is prime then phi(3*p) = reversal (3*p). (3*p is in the sequence A069215, some special cases:
Case I, p = f(0,0,0,n1) = (9)(n1).7 = 10^n  3 (see A089675). Case II, p = f(0,n3,0,0) = 78.(0)(n3).217 = 78*10^n + 217. Case III, p = f(0,0,n,0) = (7821)(n).7. In this case I found only three such prime p1 = (78217)(0).7 = 7, p2 = (7821)(2).7 = 782178217 & p3 = (7821)(674).7, p3 is a prime with length 2697.
Next term is greater than 8280.
