A232129 Largest prime that can be obtained from n by successively appending digits to the right with the constraint that each of the numbers obtained that way must be prime; a(n)=0 if there is no such prime at all.

1979339339, 29399999, 37337999, 4391339, 59393339, 6733997, 73939133, 839, 9719, 103997939939, 113, 12791333, 13999133, 149399, 15797, 1637, 17333, 1811993, 1979339339, 0, 21139, 2273993, 23399339, 24179399, 2579939, 2699393, 27191939, 2837, 29399999, 3079, 31379, 0, 331999799, 3491333, 35393999

Offset: 1

Author

M. F. Hasler, Nov 19 2013

Keywords

• nonn
• base

Comments

See A232128 for the number of steps required to reach a(n), equal to the length of a(n) minus the length of n. See A232126 for a variant "working backwards", where truncation is considered.

Examples

			Starting with 8, one can get the primes 83 and 89 which is larger, but 83 allows one further extension to 839 while 89 does not (no prime in the range 890..899). No further extension is possible, since there are no primes in the range 8390,...,8399. Therefore a(8)=839 and A232128(8)=2.
a(20)=a(42)=0 since no prime can be obtained by appending one digit to 20 or 42.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Archimedes' Lab, What's Special About This Number?, section about 43.

Crossrefs

Cf. A232128, A232127, A232126, A232125.

Programs

PARI

{A232129(n)=local(t(p)=my(m,r=[0,p]);forstep(d=1,9,2,isprime(p*10+d)&&(m=t(10*p+d)+[1,0])[1]>=r[1]&&r=m);r);n<(n=t(n))[2]&&return(n[2])}

Python

from sympy import isprime, nextprime
def a(n):
    while True:
        extends, reach, maxp = -1, {n}, 0
        while len(reach) > 0:
            candidates = (int(str(e)+d) for d in "1379" for e in reach)
            reach1 = set(filter(isprime, candidates))
            extends, reach, maxp = extends+1, reach1, max({maxp}|reach1)
        return maxp
print([a(n) for n in range(1, 36)]) # Michael S. Branicky, Sep 07 2021