login
Let b(i) = A134204(i) and c(n) = A133242(n); a(n) is the number of primes p <= c(n) such that p is not in {b(0), b(1), ..., b(c(n)-1)}.
2

%I #28 Jan 03 2024 07:41:17

%S 1,1,2,1,1,2,2,3,2,1,2,3,6,7,6,7,7,7,6,5,7,12,11,10,10,9,10,12,11,12,

%T 11,10,9,9,8,8,8,9,8,8,8,7,10,16,16,16,19,18,17,16,15,15,16,16,17,16,

%U 15,16,16,19,19,20,20,19,18,17,16,17,20,19,20,19,18,18,19,23,24,23,25,24,25,27,26,27,27,26,25,25

%N Let b(i) = A134204(i) and c(n) = A133242(n); a(n) is the number of primes p <= c(n) such that p is not in {b(0), b(1), ..., b(c(n)-1)}.

%C Computed by _David Applegate_, Oct 2007.

%C Arises from studying the question of whether A134204 is an infinite sequence.

%H David Applegate, <a href="/A232992/b232992.txt">Table of n, a(n) for n = 1..10000</a>

%H David Applegate, <a href="/A134204/a134204.cc.txt">C++ Program</a>

%H David Applegate, <a href="/A232992/a232992.notes.txt">Notes on programs and output</a>

%H David Applegate, <a href="/A232992/a232992.out2.txt">The first 106394 lines of output</a>. The first 3 columns give the first 106394 terms of A133242, A133243 and A232992 (the present sequence), and establish that at least 800 million terms of A134204 exist.

%e Terms b(0) through b(12) of A134202 are (ignore the periods, which are just for alignment):

%e i:... 0, 1, 2, 3,. 4,. 5,. 6,. 7,. 8,. 9, 10, 11, 12

%e b(i): 2, 3, 5, 7, 13, 17, 19, 23, 41, 31, 29, 37, 11

%e c(1) = 12 is the first i for which b(i)<i.

%e Then a(1) is the number of primes p <= 12 that are not in the set {b(0), ..., b(11)} = {2, 3, 5, 7, 13, 17, 19, 23, 41, 31, 29, 37}.

%e Only p = 11 is missing, so a(1)=1.

%Y Cf. A134204, A133242.

%K nonn

%O 1,3

%A _N. J. A. Sloane_, Dec 13 2013