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A051445
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Smallest k such that phi(k) = 2n, or 0 if there is no such k.
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4
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3, 5, 7, 15, 11, 13, 0, 17, 19, 25, 23, 35, 0, 29, 31, 51, 0, 37, 0, 41, 43, 69, 47, 65, 0, 53, 81, 87, 59, 61, 0, 85, 67, 0, 71, 73, 0, 0, 79, 123, 83, 129, 0, 89, 0, 141, 0, 97, 0, 101, 103, 159, 107, 109, 121, 113, 0, 177, 0, 143, 0, 0, 127, 255, 131, 161, 0, 137
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| The zero values are easy to prove because of the bounds on the phi function.
phi(15)=2*4, so a(4)=15.
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LINKS
| T. D. Noe, Table of n, a(n) for n = 1..10000
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EXAMPLE
| sigma(4)=7, 4 is the smallest, so a(7)=4.
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PROG
| (PARI) a(n)=n+=n; for(k=n+1, solve(x=n, if(n<20, 99, 5*n*log(log(n))), x/(exp(Euler)*log(log(x))+3/log(log(x)))-n), if(eulerphi(k)==n, return(k))); 0 \\ Charles R Greathouse IV, Dec 19 2011
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CROSSREFS
| Cf. A002181. For records see A132012, A132115.
Sequence in context: A114980 A024372 A061390 * A002181 A073692 A132012
Adjacent sequences: A051442 A051443 A051444 * A051446 A051447 A051448
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KEYWORD
| nonn
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AUTHOR
| Jud McCranie (JudMcCranie(AT)ugaalum.uga.edu)
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