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A053697
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a(n+1)=a(n)+a^(n), where the addition is in base 11 and where a^(n) is obtained from a(n) by replacing each digit by its multiplicative inverse modulo 11. Zero digits, if any, are deleted.
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0
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1, 2, 8, 14, 27, 94, 137, 284, 947, 1384, 2847, 9484, 13847, 28484, 94847, 138484, 284847, 948484, 1384847, 2848484, 9484847, 13848484, 28484847, 94848484, 138484847, 284848484, 948484847, 1384848484, 2848484847, 9484848484
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Conjecture. For any positive integer a(1), the sequence generated according to the above rule eventually cycles through the forms a(k)=[1][4^a][3][(84)^b],..., a(k+6)=[1][4^a][3][(84)^(b+1)], or through a(k)=[1][5^a][4][(84)^b],..., a(k+6)=[1][5^a][4][(84)^(b+1)], for nonnegative integers a and b. The sequence listed above, with a(1)=1, is an example of the first type.
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CROSSREFS
| Sequence in context: A125902 A173974 A056677 * A092346 A196227 A166045
Adjacent sequences: A053694 A053695 A053696 * A053698 A053699 A053700
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KEYWORD
| nonn,base,easy
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Feb 14 2000
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