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%I #8 Dec 05 2013 19:55:32
%S 1,2,4,6,3,8,9,10,5,12,14,15,16,18,7,20,21,22,24,25,26,27,11,28,30,32,
%T 33,34,35,36,38,39,40,42,13,44,45,46,48,49,50,51,52,54,55,56,57,58,17,
%U 60,62,63,64,65,66,68,69,70,72,74,75,76
%N a(1) = 1; then k-th prime prime(k) followed by prime(k) consecutive composite numbers not occurring earlier.
%C Primes grow faster than composites. Question: For what (the smallest) value of m is a(m) prime and bigger than the previous term which is obviously composite? Answer: There is no such m. Proof: Assume a(m) is the n-th prime and a(m-1) < a(m). Checking manually gives n>10. Then a(m) < 2*n*log(n). The number of composite numbers appearing before a(m) is apparently the sum of the first n-1 primes, which is bigger than (n-1)^2. This means that a(m-1) is definitely bigger than (n-1)^2. Therefore we have a(m) < 2*n*log(n) as well as a(m-1) > (n-1)^2. Therefore a(m-1) > a(m).
%t a = {1}; For[n = 1, n < 9, n++, AppendTo[a, Prime[n]]; For[j = 1, j < Prime[n] + 1, j++,i = 4; While[PrimeQ[i] || Length[Intersection[a, {i}]] == 1, i++ ]; AppendTo[a, i]]]; a
%o (PARI) { nonprim = listcreate(50000) ; for(n=2,50000, if( !isprime(n),listput(nonprim,n)) ; ) ; print("1,") ; k=2 ; indxn = 1 ; for (n = 2, 80, pr=prime(k-1); print1(pr,",") ; for(i=1,pr, print1(nonprim[indxn],",") ; indxn++ ; ); print("") ; k++ ; ) } - _R. J. Mathar_, Mar 14 2006
%Y Cf. A073900.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Aug 18 2002
%E More terms from _R. J. Mathar_, Mar 14 2006
%E Edited by _Stefan Steinerberger_, Aug 13 2007
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