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A167155
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Exponential primorial constant Sum_{k>=0} 1/A140319(k).
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2
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1, 6, 1, 1, 1, 1, 1, 6, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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1,2
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COMMENTS
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This is a Liouville number and therefore transcendental.
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LINKS
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EXAMPLE
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1 + 1/2^1 + 1/3^2 + 1/5^9 + 1/7^(5^9)+ ... = 1.6111116231111111111111111111111111111111...
Since 1/9 = 0.11111... and 1/5^9 = 512*10^(-9), the initial 10 digits are 1.611111623.
Since 1/A140319(4) = 1/7^1953125 = 7.7731519...*10^(-1650583), these digits are followed by a string of 1650573 "1"s, then followed by digits 8884263011....
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MATHEMATICA
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Clear[ep, s]; ep[0] = 1; ep[n_] := Prime[n]^ep[n-1]; s[n_] := s[n] = RealDigits[Sum[1/ep[k], {k, 0, n}], 10, 105] // First; s[n=1]; While[s[n] != s[n-1], n++]; s[n] (* Jean-François Alcover, Feb 13 2013 *)
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PROG
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(PARI) 1+1/2+1/3^2+1/5^9+1/7^5^9. /* The final dot is part of the code! */
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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