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A247134
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Least k such that 2n+1 = 2^k - p^m for some prime p and integer m > 0, or 0 if no such k exists.
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1
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2, 3, 3, 4, 4, 4, 4, 5, 6, 5, 5, 5, 5, 5, 5, 7, 6, 6, 6, 6, 6, 29, 6, 6, 7, 6, 6, 6, 6, 6, 6, 8, 8, 7, 7, 10, 9, 7, 8, 7, 7, 8, 7, 7, 8, 7, 8, 10, 7, 7, 7, 7, 7, 8, 7, 7, 10, 7, 7, 7, 7, 7, 7, 47, 8, 8, 9, 8, 10, 9, 10, 8, 9, 8, 8, 9, 8, 8, 15, 8, 10, 9, 9, 8
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OFFSET
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0,1
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COMMENTS
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Since even numbers cannot be of this form unless they are powers of 2, only odd 2n+1 are considered.
Except for n=3, 11, 18, 19, 27, 39, 50, 51, ..., the value of k corresponds to m=1, i.e., 2^k - (2n+1) is prime.
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LINKS
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FORMULA
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a(n) = min{ k | exists m>1: 2n+1 = A000079(k) - A000961(m) } > log_2(n).
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PROG
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(PARI) a(n)=for(k=log(n=n*2+1)\log(2)+1, 9e9, isprimepower(2^k-n)&&return(k))
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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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