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A136308
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a(n) = (10^2^n - 1)/9.
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4
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OFFSET
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0,2
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COMMENTS
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More generally, reading in base B >= 2: a(n) = (B^2^n - 1)/(B-1).
Recurrence: a(n) = a(n-1)*(B^K + 1) and a(0)=1 where K = floor(log_B a(n-1)) + 1.
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LINKS
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FORMULA
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a(n) = a(n-1)*(10^K + 1) and a(0)=1 where K=floor(log_10 a(n-1)) + 1 = 2^n + 1.
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MATHEMATICA
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PROG
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(Sage) [(10^2^n -1)/9 for n in (0..10)] # G. C. Greubel, Apr 19 2021
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CROSSREFS
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KEYWORD
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easy,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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