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A214018
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Least k >= 1, such that prime(n) + k has the form 2^m * q, m >= 0, where q >= 2 is prime.
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2
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1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 3, 3, 4, 2, 1, 2, 3, 3, 1, 3, 2, 3, 2, 1, 1, 1, 5, 3, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 3, 2, 3, 3, 5, 5, 2, 1, 1, 2, 1, 3, 4, 2, 1, 3, 1, 7, 2, 3, 3, 3, 1, 3, 3, 1, 5, 1, 3, 3, 2, 1, 2, 3, 4, 3, 3
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OFFSET
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1,7
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COMMENTS
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By the definition, a(n)<=p_(n+1)-p_n. It is well known that, for large n, p_(n+1)-p_n on average is approximately log(n). What is the average behavior of a(n)? By the Broughan-Qizhi inequality, A192869(n)>>n*(log(n))^2. Besides, they conjecture that A192869(n)=O(n*(log(n))^2). But in the case of this sequence, we have, on average, log(n)/2 possible odd values of k< p_(n+1)-p_n.
Therefore, we conjecture that, on average, a(n) is approximately c*log(n) with c in (0,1). Calculations up to 10^6 (Peter J. C. Moses) show that, most likely, c < 0.53 (cf. comment in A213892).
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LINKS
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EXAMPLE
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a(1)=1, since 2+1=3=2^0*3; a(2)=1, since 3+1=2^1*2.
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MATHEMATICA
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Table[NestWhile[#+1&, 1, Not[Apply[Or, PrimeQ[(Prime[n]+#)/(2^Range[0, Floor[Log[Prime[n]]/Log[2]]])]]]&], {n, 100}] (* Peter J. C. Moses, Jul 09 2012 *)
Table[p = Prime[n]; k = 1; While[q = (p + k)/2^IntegerExponent[p + k, 2]; ! (q == 1 || PrimeQ[q]), k++]; k, {n, 100}] (* T. D. Noe, Jul 10 2012 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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