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Number of successive primes beginning with n at the first instance.
1

%I #9 Aug 09 2015 00:43:18

%S 4,1,1,3,1,2,1,2,1,4,1,1,1,1,2,2,1,1,1,14,1,3,1,1,2,2,2,2,1,1,1,11,2,

%T 2,2,1,1,2,1,2,1,1,1,2,1,3,1,1,2,2,11,2,1,2,1,2,2,1,1,2,1,13,1,3,2,1,

%U 1,1,1,2,1,1,1,1,2,2,1,1,1,1,1,4,1,8,3,1,1,3,1,1,2,1,1,2,1,1,1,1,2,1

%N Number of successive primes beginning with n at the first instance.

%C a(p) = 1. If a composite n has k digits and there is no prime beginning with n of k+1 digits then a(n) shoots up and a(n-1) < a(n) > a(n+1). a(n) must exceed n at some values of n. as a(20) = 14, etc. at what point a(n) exceeds n? and we get a subsidiary sequence.

%e a(1) = 4 as there are four primes 11,13,17,19, that begin with 1.

%e a(2) = 1 as 2 follows by 3 hence 23 is not considered for a count.

%e a(20) = 14, 14 primes from 2003 to 2099.

%p A111311 := proc(n) local p, nstrt,a ; if isprime(n) then 1; else nstrt := 10 ; a := 0 ; while a = 0 do for p from nstrt*n to nstrt*n+nstrt -1 do if isprime(p) then a := a+1 ; fi ; od ; if a <> 0 then RETURN(a) ; fi ; nstrt := nstrt*10 ; od: fi ; end: seq(A111311(n),n=1..100) ; # _R. J. Mathar_, Feb 08 2008

%K base,nonn

%O 1,1

%A _Amarnath Murthy_, Nov 11 2005

%E Corrected and extended by _R. J. Mathar_, Feb 08 2008