A099180 -id:A099180 - cp's OEIS frontend

A209063 Let p = A099180(n) be the n-th prime which is greater than its reversal p'; a(n) = number of ways to write p as a multiple of p' +- q where q is a prime < p.

1, 2, 0, 1, 4, 3, 0, 2, 1, 1, 1, 1, 1, 1, 0, 0, 3, 2, 4, 0, 0, 2, 0, 2, 1, 3, 1, 1, 1, 1, 1, 4, 1, 4, 2, 0, 1, 2, 1, 1, 1, 2, 1, 0, 1, 1, 1, 4, 4, 1, 0, 1, 1, 2, 0, 3, 2, 0, 0, 3, 1, 3, 0, 0, 1, 3, 0, 0, 2, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1

Offset: 1

N. J. A. Sloane, Mar 13 2012

Sequence is bounded - see A202286.

			For n=6, p = 71, p' = 17 and there are three solutions:
71 = 2*17 + 3771 = 4*17 + 371 = 6*17 - 31, so a(6) = 3.

Cf. A099180, A209914. A202286 gives smallest m such that a(m) = n.

A209063(n)=A209914(A099180(n)) \\ M. F. Hasler, Mar 15 2012

Equals A209914 o A099180. - M. F. Hasler, Mar 15 2012

A202286 Smallest prime which is equal to a multiple of its reversal +- a prime smaller than itself in exactly n ways.

Original entry on oeis.org

31, 41, 71, 61, 6421, 8501, 8116001, 845534401

Offset: 1

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

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
a(9) > 8*10^15, if it exists. - Giovanni Resta, Oct 29 2018

			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.

Claudio Meller, Números y algo mas...: 833 - Igual a un múltiplo del "inverso" mas/ menos un primo

Cf. A099180, A209063, A182239.

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

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

cp's OEIS Frontend

A209063 Let p = A099180(n) be the n-th prime which is greater than its reversal p'; a(n) = number of ways to write p as a multiple of p' +- q where q is a prime < p.

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