|
| |
|
|
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 k-values lead to q == 0 (mod 3) and are excluded, which leaves at most 11 possibilities. Other k-values 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
|
|
|
|
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*p-1)\r, isprime(abs(p-k*r)))==n & return(p))} \\ M. F. Hasler, Mar 14 2012
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,base,fini,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
