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A108234
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Minimum m such that n*2^m+k is prime, for k < 2^m. In other words, assuming you've read n out of a binary stream, a(n) is the minimum number of additional bits (appended to the least significant end of n) you must read before it is possible to obtain a prime.
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4
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1, 0, 0, 2, 0, 1, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 2, 2, 1, 2, 2, 0, 1, 0, 2, 1, 2, 1, 1, 0, 3, 1, 2, 0, 3, 0, 1, 2, 3, 0, 1, 2, 1, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 0, 2, 1, 2, 1, 4, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 2, 1, 0, 3, 1, 2, 0, 2, 3, 1, 2, 2, 0, 1, 2, 3, 2, 2, 1, 1, 0, 1, 1, 2
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OFFSET
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1,4
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COMMENTS
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Somewhat related to the Riesel problem, A040081, the minimum m such that n*2^m-1 is prime.
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LINKS
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EXAMPLE
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a(12) = 3 because 12 = 1100 in binary and 97 = 1100001 is the first prime that starts with 1100, needing 3 extra bits.
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PROG
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(Octave/MATLAB) for n=1:100; m=0; k=0; while(~isprime(n*2^m+k))k=k+1; if k==2^m k=0; m=m+1; end; end; x(n)=m; end; x
(PARI) A108234(n) = { my(m=0, k=0); while(!isprime((n*2^m)+k), k=k+1; if(2^m==k, k=0; m=m+1)); m; }; \\ Antti Karttunen, Dec 16 2017, after Octave/MATLAB code
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CROSSREFS
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KEYWORD
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easy,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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