

A232127


Maximal number of digits that can be appended to prime(n) preserving primality at each step.


4



7, 7, 7, 7, 1, 6, 3, 8, 6, 6, 3, 6, 1, 5, 3, 0, 6, 5, 5, 4, 6, 1, 1, 0, 2, 4, 9, 0, 4, 0, 5, 1, 1, 5, 3, 1, 2, 1, 0, 2, 0, 4, 2, 3, 7, 5, 2, 3, 4, 3, 5, 4, 5, 0, 4, 3, 4, 5, 3, 1, 1, 5, 1, 2, 2, 0, 6, 3, 0, 4, 5, 2, 4, 5, 1, 2, 0, 0, 3, 10, 0, 3, 0, 2, 4, 0, 3, 0, 0, 6
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OFFSET

1,1


COMMENTS

Consider chains (p^(0),p^(1),p^(2),...p^(L)) of primes such that p^(k1) = floor(p^(k)/10), or otherwise said, p^(k+1) is obtained from p^(k) by appending a digit. Then a(n) is one less than the number of primes in the longest possible such chain with p^(0)=prime(n).


LINKS

Table of n, a(n) for n=1..90.
Archimedes' Lab, What's Special About This Number?, section about 43.


FORMULA

a(n)=A232128(A000040(n)).
a(n) > 0 if and only if there is a prime p between 10*prime(n)+1 and 10*prime(n)+9, in which case a(n) >= 1+a(primepi(p))
a(n) = max { L in N  exists (p[0],...,p[L]) in P^(L+1) (P = the primes A000040), such that p[0] = prime(n) and for k=1,...,L : p[k1] = floor(p[k]/10) }


EXAMPLE

a(14)=5 because for prime(14)=43, one can add at most 5 digits to the right preserving primality at each step: 439 is prime, 4391 is prime, 43913 is prime, 439133 is prime, 4391339 is prime. There is no longer chain possible starting with 43.


PROG

(PARI) {howfar(p)=my(m); forstep(d=1, 9, 2, d==5&&next; isprime(p*10+d)next; m=max(1+howfar(10*p+d), m)); m}


CROSSREFS

Cf. A232125.
Sequence in context: A216159 A276615 A103983 * A195413 A083947 A269349
Adjacent sequences: A232124 A232125 A232126 * A232128 A232129 A232130


KEYWORD

nonn,base


AUTHOR

M. F. Hasler and Michel Marcus, Nov 19 2013


STATUS

approved



