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A132365
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Least number k such that the Lucas number A000032(k) contains n.
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0
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1, 0, 2, 3, 13, 9, 4, 6, 7, 24, 5, 10, 15, 26, 20, 25, 49, 6, 11, 16, 13, 12, 10, 21, 45, 40, 20, 36, 7, 31, 50, 12, 35, 19, 17, 15, 41, 36, 22, 23, 39, 39, 14, 21, 41, 60, 8, 32, 19, 56, 20, 13, 45, 37, 51, 44, 17, 56, 42, 22, 25, 62, 35, 15, 71, 47, 25, 24, 43, 32, 17, 45, 49, 38
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Values such that a(n)=n (fixed points) are 1, 62. I don't know if there are any other fixed points. The first time a(n)=a(n+1) occurs because L(39)=141422324 which includes both 41 and 42 (and later on in the sequence, because it contains 141 and 142). [From Sean A. Irvine (sairvin(AT)xtra.co.nz), Nov 30 2009]
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FORMULA
| a(n) = Min{k such that A000032(k) contains the decimal digit substring which represents the integer n}.
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CROSSREFS
| Cf. A000032, A000045, A000350, A050816, A038546, A052000, A023172.
Sequence in context: A057776 A110362 A074478 * A129671 A117684 A056445
Adjacent sequences: A132362 A132363 A132364 * A132366 A132367 A132368
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KEYWORD
| base,easy,nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 08 2007
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EXTENSIONS
| Removed incorrect comment Sean A. Irvine (sairvin(AT)xtra.co.nz), Nov 30 2009
Corrected and extended by Sean A. Irvine (sairvin(AT)xtra.co.nz), Nov 30 2009
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