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A060460
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Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives position (or phase) of the maximal value seen in these n digits.
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2
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3, 12, 53, 254, 1255, 6256, 31257, 156258, 781259, 3906260, 19531261, 97656262, 488281263, 2441406264, 12207031265, 61035156266, 305175781267, 1525878906268, 7629394531269, 38146972656270, 190734863281271
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OFFSET
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1,1
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COMMENTS
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The last n digits of 2^a(n) are predictable if maximal values of periods are known.
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LINKS
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FORMULA
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a(1) = 3, a(n) = 5*a(n-1)-(3+4*(n-2)).
a(n) = a(n) = 2*5^(n-1) + n.
G.f.: (-3 + 9 x - 2 x^2)/((-1 + x)^2 (-1 + 5 x)) - Harvey P. Dale, Aug 01 2021
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EXAMPLE
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a(2) = 5*3-(3+4*0) = 15-3 = 12, etc...
For n=2, the last 2 digits of powers of 2 have the period {2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,4,8,16,32} displayed in A000855. The maximum is 96 and it occurs at 2^12=4096. So a(2)=12.
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MATHEMATICA
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nxt[{n_, a_, b_}]:={n+1, b, 5b-(3+4(n-1))}; NestList[nxt, {2, 3, 12}, 20][[All, 2]] (* or *) Table[2*5^(n-1)+n, {n, 30}] (* or *) LinearRecurrence[{7, -11, 5}, {3, 12, 53}, 30] (* Harvey P. Dale, Aug 01 2021 *)
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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