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A086695
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Digits of the inverse of these numbers gives the Fibonacci numbers. More precisely the digits of 1/(10^(2*n)-10^n-1) give the Fibonacci numbers up to 10^n.
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0
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OFFSET
| 0,1
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COMMENTS
| More generally, if x_1,x_2, x_n=x_(n-1)-x_(n-2) is any Lucas sequence, then the digits of the numbers (x_1*10^n-(x_1-x_2))/(10^(2*n)-10^n-1) gives the x_n up to 10^n
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FORMULA
| a(n) = 10^(2*n)-10^n-1
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CROSSREFS
| Sequence in context: A020472 A093948 A116254 * A056568 A174758 A181681
Adjacent sequences: A086692 A086693 A086694 * A086696 A086697 A086698
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KEYWORD
| base,easy,full,fini,nonn
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AUTHOR
| Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003
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