|
|
A241499
|
|
a(n)= k is the number of consecutive primes of the form 10^d - prime(n), 10^(d+1)- prime(n),...,10^(d+k-1)- prime(n) where d is the number of decimal digits of prime(n).
|
|
0
|
|
|
0, 3, 1, 1, 1, 0, 2, 0, 0, 2, 0, 0, 1, 0, 2, 2, 3, 0, 0, 4, 0, 0, 1, 2, 1, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The growth of a(n) is very slow. The smallest prime p such that the number of consecutive primes is equal to n for n = 0, 1, 2,... is given by the sequence b(n) = 2, 5, 17, 3, 71, 535673,... (hard).
|
|
LINKS
|
|
|
EXAMPLE
|
a(2) = 3 because prime(2)= 3 => 10^1-3 = 7, 10^2-3 = 97 and 10^3-3 = 997 with three consecutive primes, but 10^4-3 = 9997 = 13*769 is composite.
|
|
MAPLE
|
with(numtheory):for n from 1 to 100 do:p:=ithprime(n):c:=0:ii:=0:l:=length(p):for k from l to 100 while(ii=0) do:q:=10^k - p:if type(q, prime)=true then c:=c+1:else ii:=1:fi:od: printf(`%d, `, c):od:
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|