

A101115


Beginning with the nth prime, the number of successive times a new prime can be formed by prepending the smallest nonzero digit.


5



0, 5, 0, 9, 5, 4, 8, 4, 5, 9, 4, 6, 2, 7, 6, 8, 9, 7, 6, 3, 14, 5, 5, 2, 4, 10, 1, 5, 7, 3, 4, 3, 5, 5, 0, 6, 5, 8, 5, 13, 4, 5, 4, 5, 3, 8, 4, 4, 5, 8, 3, 6, 1, 4, 4, 2, 5, 2, 2, 3, 4, 9, 8, 7, 4, 7, 3, 3, 5, 5, 7, 8, 4, 3, 3, 2, 1, 7, 0, 4, 3, 5, 3, 7, 9, 6, 6, 5, 6, 8
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OFFSET

1,2


COMMENTS

It is possible the procedure described would generate some lefttruncatable primes (A024785). Although zero digits cannot be added, it is possible the starting prime may contain zeros. Therefore the possible number of digit additions is not limited by the length of the largest known lefttruncatable prime. Further, because the smallest digit that satisfies the requirement is used each time, it is possible that choosing a larger digit would allow more single digits to be added. Therefore although some of the set of lefttruncatable primes may be generated by this practice, not all of them will.
In principle it is possible that some a(n) is undefined because the process could go on indefinitely, but this is very unlikely. The largest a(n) for n <= 300000 is a(49120) = 18.  Robert Israel, Jun 29 2015


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000
I. O. Angell, and H. J. Godwin, On Truncatable Primes, Math. Comput. 31, 265267, 1977.
Index entries for sequences related to truncatable primes


EXAMPLE

a(2) is 5 because the second prime is 3, to which single nonzero digits can be prepended 5 times yielding a new prime each time (giving preference to the smallest digit that satisfies the requirement): 13, 113, 2113, 12113, 612113 (see A053583). There is no nonzero digit which can be prepended to 612113 to yield a new prime.
a(21) = 14 because the 21st prime (73) can be prepended with single nonzero digits 14 times yielding a new prime each time: 73, 173, 6173, 66173, ..., 4818372912366173.


MAPLE

f:= proc(n) local p, nd, d, count, x, success;
p:= ithprime(n); nd:= ilog10(p);
for count from 0 do
nd:= nd+1;
success:= false;
for d from 1 to 9 do
x:= 10^nd * d + p;
if isprime(x) then
success:= true;
break
fi
od;
if not success then return(count) fi;
p:= x;
od
end proc:
map(f, [$1..100]); # Robert Israel, Jun 29 2015


CROSSREFS

Cf. A053583, A024785, A000040, A101116, A101117, A101118.
Sequence in context: A271175 A196769 A019925 * A200633 A196820 A176325
Adjacent sequences: A101112 A101113 A101114 * A101116 A101117 A101118


KEYWORD

base,nonn


AUTHOR

Chuck Seggelin (seqfan(AT)plastereddragon.com), Dec 02 2004


STATUS

approved



