|
%I #13 Mar 22 2023 08:42:30
%S 1,2,8,10,13,21,22,36,57,80,149,484,505,642,806,974,1674
%N Numbers n such that 78*10^n + 217 is prime.
%C 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).
%C 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.
%C 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.
%C 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:
%C Case I, p = f(0,0,0,n-1) = (9)(n-1).7 = 10^n - 3 (see A089675). Case II, p = f(0,n-3,0,0) = 78.(0)(n-3).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.
%C Next term is greater than 8280.
%C Next term is greater than 24000. - _Michael S. Branicky_, Mar 22 2023
%e 8 is in the sequence because 78.(8-3)(0).217 = 7800000217 is prime.
%t Do[If[PrimeQ[78*10^n + 217], Print[n]], {n, 8280}]
%o (PARI) is(n)=ispseudoprime(78*10^n+217) \\ _Charles R Greathouse IV_, May 22 2017
%Y Cf. A089675, A069215, A085331, A101700.
%K more,nonn
%O 1,2
%A _Farideh Firoozbakht_, Jan 04 2005
|
