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A091831
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Pierce expansion of 1/sqrt(2).
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2
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1, 3, 8, 33, 35, 39201, 39203, 60245508192801, 60245508192803, 218662352649181293830957829984632156775201, 218662352649181293830957829984632156775203
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OFFSET
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0,2
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COMMENTS
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If u(0)=exp(1/m) m integer>1 and u(n+1)=u(n)/frac(u(n)) then floor(u(n))=m*n.
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LINKS
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FORMULA
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Let u(0)=sqrt(2) and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)).
1/sqrt(2)= 1/a(1) - 1/a(1)/a(2) + 1/a(1)/a(2)/a(3) - 1/a(1)/a(2)/a(3)/a(4)...
limit n -> infinity a(n)^(1/n) = e.
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MATHEMATICA
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PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[2^(-1/2), 7!], 17] (* G. C. Greubel, Nov 13 2016 *)
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PROG
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(PARI) r=sqrt(2); for(n=1, 10, r=r/(r-floor(r)); print1(floor(r), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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