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%I #11 Apr 05 2015 13:05:51
%S 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,
%T 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,
%U 1,0,1,1,0,0,0,1,0,0,1,0,0,0,1,0,0,1,0
%N 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).
%C 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).
%e 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.
%p 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:
%Y Cf. A000040.
%K nonn,base
%O 1,2
%A _Michel Lagneau_, Apr 24 2014
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