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Search: id:A101202
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| 142857, 285714, 428571, 571428, 714285, 857142, 999999, 1142856, 1285713, 1428570, 1571427, 1714284, 1857141, 1999998, 2142855, 2285712, 2428569, 2571426, 2714283, 2857140, 2999997, 3142854, 3285711, 3428568, 3571425
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OFFSET
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1,1
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COMMENT
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This sequence is interesting because the first six terms are cyclic shifts of one another (in decimal).
The digits of each term can be partitioned into two parts with the right part always consisting of 3 digits and the left part containing all the remaining digits; adding then those two numbers will yield a sum which is multiple of 999: (142+857)/999 = 1 .. (857+142)/999 = 1 (999+999)/999 = 2 (1142+856)/999 = 2 .. (1857+141)/999 = 2 (2142+855)/999 = 3 .. (2857+140)/999 = 3 (2999+997)/999 = 4 .. As a result of above operations the sequence A054896 starting with the term 8 is produced. - Alexander R. Povolotsky (pevnev(AT)juno.com), Nov 02 2007
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LINKS
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Tanya Khovanova, Recursive Sequences
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FORMULA
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G.f. = 142857/(x^2 - 2*x + 1) - Alexander R. Povolotsky (pevnev(AT)juno.com), Apr 26 2008
{-a(n) + 142857 n + 142857, a(0) = 142857}. - Robert Israel, May 14 2008
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CROSSREFS
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Cf. A054896.
Sequence in context: A086999 A023089 A166320 this_sequence A144504 A146754 A004042
Adjacent sequences: A101199 A101200 A101201 this_sequence A101203 A101204 A101205
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KEYWORD
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easy,nonn
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AUTHOR
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Paul Boddington (psb(AT)maths.warwick.ac.uk), Dec 12 2004
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