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Search: id:A100441
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| A100441 |
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Consider the sequence of fractions f(n) defined by: f(1) = 2/1; f(n+1) is chosen so that f(n+1) + Sum_{i=1..n} f(i) = f(n+1) * Product_{i=1..n} f(i); sequence gives denominator of f(n). |
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1
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| 1, 1, 3, 13, 217, 57073, 3811958497, 16605534578235736513, 309708098978072051970763989442580255617, 106322990835084829467725909226560893968664147958670035553130958199430801942273
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Let E(0) = x + 1, let E(n+1) = 1 - E(n) + E(n)^2. Let e(n) = discrim(E(n),x) and let f(n) = e(n+1)/e(n)^2. Then f(1,2,3,...) = -3,13,217,57073,381195849,... which looks like this sequence (I do not have a proof yet). - Daniel R. L. Brown (dbrown(AT)certicom.com), Nov 18 2005
This sequence gives the next number in a sequence where the sum and the product of the terms of the sequence are equal.
It happens that the sum or product of the terms of this sequence match A001146 for the numerator of the sum or product and A076628 for the denominator of the sum or product of the sequence.
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FORMULA
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Let F(n) = Product_{i=1..n} f(i) = p/q (say). Then f(n+1) = p/(p-q).
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EXAMPLE
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2, 2, 4/3, 16/13, 256/217, 65536/57073, 4294967296/3811958497, 18446744073709551616/16605534578235736513, ... = A001146/A100441 (essentially).
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MAPLE
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f:=proc(n) option remember; local i, k, k1, k2; if n = 1 then return(2); fi; k:=mul(f(i), i=1..n-1); k1:=numer(k); k2:=denom(k); k1/(k1-k2); end;
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CROSSREFS
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Sequence in context: A112093 A085010 A165903 this_sequence A036680 A111431 A015701
Adjacent sequences: A100438 A100439 A100440 this_sequence A100442 A100443 A100444
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KEYWORD
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nonn,frac,nice
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AUTHOR
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Gilbert Boily (sgbl(AT)escape.ca), Nov 21 2004, Sep 03 2007
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