A232128 Maximal number of digits that can be appended to n such that each step yields a prime.
9, 7, 7, 6, 7, 6, 7, 2, 3, 10, 1, 6, 6, 4, 3, 2, 3, 5, 8, 0, 3, 5, 6, 6, 5, 5, 6, 2, 6, 2, 3, 0, 7, 5, 6, 5, 6, 3, 1, 11, 1, 4, 5, 4, 1, 7, 3, 4, 6, 4, 0, 5, 0, 6, 4, 4, 6, 2, 6, 7, 5, 0, 4, 2, 3, 3, 5, 2, 5, 4, 4, 1, 6, 2, 4, 4, 1, 7, 1, 4, 4, 10, 1, 0, 5, 1, 6, 5, 0, 1, 4
Offset: 1
Examples
a(1)=9 because "1" can be extended with at most 9 digits to the right such that each extension is prime; the least one of the possible 1+9 digit primes is 1979339333, the largest one is given in A232129.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
f:= proc(n) local V,k; V:= select(isprime, [seq(10*n+k, k=[1,3,7,9])]); if V = [] then 0 else 1 + max(map(procname,V)) fi end proc: map(f, [$1..100]); # Robert Israel, Oct 16 2020
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PARI
a(n)=my(m,r=[0,n]);forstep(d=1,9,2,d==5&&next;isprime(n*10+d)||next;m=[1,0]+howfar(10*n+d);m[1]>r[1]&&r=m);r \\ Note: this returns the list [a(n), minimal longest prime]
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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 extends print([a(n) for n in range(1, 92)]) # Michael S. Branicky, Sep 07 2021
Formula
If a(n) > 0, then there is some prime p in the range 10n+1,...,10n+9 such that a(p)=a(n)-1. If a(n)=0, then there is no prime in that range.
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