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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 with 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
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OFFSET
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1,2
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COMMENTS
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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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LINKS
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FORMULA
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a(n) = a(n-2) + 10*a(n-3) - 10*a(n-5) for n>7.
G.f.: x*(1+2*x+7*x^2+2*x^3-x^4+10*x^5-10*x^6+10*x^8) / ((1-x)*(1+x)*(1-10*x^3)). (End)
For k >= 1:
a(6*k-4) = 2*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
a(6*k-3) = 94*10^(2*k-3) + 84*(10^(2*k-3)-10)/99 + 7.
a(6*k-2) = 13*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
a(6*k-1) = 2*10^(2*k-1) + 84*(10^(2*k-1)-10)/99 + 7.
a(6*k) = 94*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
a(6*k+1) = 13*10^(2*k-1) + 84*(10^(2*k-1)-10)/99 + 7.
Colin Barker's conjectures follow from these. (End)
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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