

A202286


Smallest prime which is equal to a multiple of its reversal +/ a prime smaller than itself in exactly n ways.


4




OFFSET

1,1


COMMENTS

The sequence is finite, more specifically there cannot be more than 19 terms. Proof: To have p = k*R(p) /+ q, with q < p, we must have 0 < k = (p +/ q) / R(p) < 2p / (p/10) = 20, since the prime p cannot end in 0 and therefore R(p) > p/10. (R(p) and p have the same number of digits.) Thus, for a given prime p, there cannot be more than 19 solutions (k=1,...,19) to p  k*R(p) = +/ q, and therefore no a(n) beyond n=19.  M. F. Hasler, Mar 13 2012, improved following remarks from Hans Havermann, Mar 14 2012
A refined analysis shows that the maximal number is less. On one hand, if R(p) is odd, then only even k can yield a prime. Therefore the maximal number of solutions can only be obtained for p starting with an even digit <= 8, and therefore p/R(p) < 9, thus k < 2p/R(p) < 18, k <= 17. Moreover, R(p) is not a multiple of 3 (since p isn't), therefore 1/3 of the kvalues lead to q=0 (mod 3) and are excluded, which leaves at most 11 possibilities. Other kvalues lead to q=0 (mod 5), unless p starts with '5'. In any case there cannot be more than 9 solutions.  M. F. Hasler, Mar 14 2012


LINKS

Table of n, a(n) for n=1..8.
Claudio Meller, Números y algo mas...: 833  Igual a un múltiplo del "inverso" mas/ menos un primo


EXAMPLE

a(4)=61 because 61 = 16 x 2 + 29, 61 = 16 x 3 + 13, 61 = 16 x 4  3, and 61 = 16 x 5  19.


PROG

(PARI) A202286(n)={ forprime(p=1, default(primelimit), my(r=A004086(p)); 2*p > n*r & sum(k=1, (2*p1)\r, isprime(abs(pk*r)))==n & return(p))} \\  M. F. Hasler, Mar 14 2012


CROSSREFS

Cf. A099180, A209063, A182239.
Sequence in context: A089442 A243704 A060327 * A285805 A141180 A176371
Adjacent sequences: A202283 A202284 A202285 * A202287 A202288 A202289


KEYWORD

nonn,base,fini,more


AUTHOR

Claudio Meller, Dec 15 2011, a(3) from Olivier Gérard


EXTENSIONS

a(7)a(8) from Hans Havermann, Mar 12 2012


STATUS

approved



