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A073899
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a(1) = 1; then k-th prime prime(k) followed by prime(k) consecutive composite numbers not occurring earlier.
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1
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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, 33, 34, 35, 36, 38, 39, 40, 42, 13, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 17, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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MATHEMATICA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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