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Search: id:A056777
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| A056777 |
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Composite n such that both Phi(n+12)=Phi(n)+12 and Sigma(n+12)=Sigma(n)+12. |
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| 65, 209, 11009, 38009, 680609, 2205209, 3515609, 4347209, 10595009, 12006209, 31979009, 89019209, 169130009, 244766009, 247590209, 258084209, 325622009, 357777209, 377330609, 441630209, 496175609, 640343009, 1006475609
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OFFSET
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1,1
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COMMENT
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It is easy to show that if p, p+2, p+6, and p+8 are all prime (a prime quadruple as defined in A007530, which lists the values of p) with x=p(p+8), x+12=(p+2)(p+6), then x is in the sequence. I conjecture that all members of the sequence are of this form - Jud McCranie
Numbers so far are all 65 mod 72. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Jul 07 2003
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EXAMPLE
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n = 209 = 11.19, n+12 = 221 = 13.17, Phi(n+12) = 192 = 180+12 = Phi(n)+12, also Sigma(221) = 252 = Sigma(209)+12 = 240+12.
phi(65)+12 = 60 = phi(65+12), sigma(65)+12 = 96 = sigma(65+12), 65 is composite.
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CROSSREFS
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Cf. A000010, A001838, A015917, A054902, A046133.
Adjacent sequences: A056774 A056775 A056776 this_sequence A056778 A056779 A056780
Sequence in context: A044778 A054902 A051968 this_sequence A048512 A038637 A115342
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Aug 17 2000
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EXTENSIONS
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More terms from Jud McCranie (j.mccranie(AT)comcast.net), Oct 11 2000
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