

A235923


Minimal k > 1 such that the basek representation of the nth Ramanujan prime (A104272), read in decimal, is also a Ramanujan prime.


0



3, 10, 4, 7, 10, 2, 10, 6, 5, 6, 5, 10, 7, 10, 8, 10, 2, 10, 10, 2, 10, 7, 4, 10, 10, 8, 3, 4, 10, 3, 3, 3, 10, 10, 4, 10, 3, 10, 10, 6, 7, 10
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OFFSET

1,1


COMMENTS

Conjecture 1. Every number 2,...,10 occurs infinitely many times.
Conjecture 2. There exists limit of average (a(1)+...+a(n))/n.
However note that, already for the eleventh Ramanujan prime 101 the binary representation, read in decimal, is rather large prime: 1100101. We cannot decide, if it is a Ramanujan prime, using bfile in A104272. Therefore, the calculation of the sequence here ends off. It is a general problem: how to decide, if a large prime is a Ramanujan one?


LINKS

Table of n, a(n) for n=1..42.


EXAMPLE

The third Ramanujan prime is 17. If k=2, we have 10001, if k=3, we have 122, if k=4, we have 101. In this list, read in decimal, 101 is the first prime. Since 101 is a Ramanujan prime, then a(3)=4.


PROG

(PARI) ramanujan_prime_list(n) = {my(L=vector(n), s=0, k=1); for(k=1, prime(3*n)1, if(isprime(k), s++); if(k%2==0 && isprime(k/2), s); if(s<n, L[s+1] = k+1)); L; } \\ A104272
a(n, vp, pmax) = {my(p=vp[n], d, np); for (b=2, 10, d = digits(p, b); np = fromdigits(d, 10); if (np > pmax, return (0)); if (vecsearch(vp, np), return (b)); ); }
lista(nn) = {my(vp = ramanujan_prime_list(nn), pmax = vecmax(vp)); for (n=1, nn, my(result = a(n, vp, pmax)); if (result, print1(result, ", "), break); ); } \\ use nn=10^7 to get 42 terms \\ Michel Marcus, Dec 17 2018


CROSSREFS

Cf. A104272, A235354.
Sequence in context: A320583 A131814 A003620 * A192028 A111229 A100984
Adjacent sequences: A235920 A235921 A235922 * A235924 A235925 A235926


KEYWORD

nonn,base,more


AUTHOR

Vladimir Shevelev, Jan 17 2014


EXTENSIONS

a(11)a(42) from Michel Marcus, Dec 17 2018


STATUS

approved



