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Search: id:A090633
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| A090633 |
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Start with the sequence [1, 1/2, 1/3, ..., 1/n]; form new sequence of n-1 terms by taking averages of successive terms; repeat until reach a single number F(n); a(n) = numerator of F(n). |
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2
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| 1, 3, 7, 15, 31, 21, 127, 255, 511, 1023, 2047, 1365, 8191, 16383, 32767, 65535, 131071, 29127, 524287, 209715, 299593, 4194303, 8388607, 5592405, 33554431, 67108863, 134217727, 268435455, 536870911, 357913941, 2147483647, 4294967295, 8589934591, 17179869183
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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The resistance across a single resistor of an n-dimensional hypercube made of 1 ohm resistors. - Peter J. C. Moses (mows(AT)mopar.freeserve.co.uk), May 27 2004
Also, numerators of BINOMIAL transform of sequence [1, 1/2, 1/3, 1/4, ...]. - Gary Adamson, Apr 26 2005
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REFERENCES
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Putnam Competition, 2003, Problem B2.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..200
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FORMULA
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a(n)=(2-2^(1-n))/n gf:=2*(Log(1-x/2)-Log(1-x)) egf:=2*(e^x-e^(x/2))/x - Peter J. C. Moses (mows(AT)mopar.freeserve.co.uk), May 27 2004
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EXAMPLE
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n=3: [1, 1/2, 1/3] -> [3/4, 5/6] -> [7/12], so F(3) = 7/12. Sequence of F(n)'s begins 1, 3/4, 7/12, 15/32, 31/80, 21/64, 127/448, 255/1024, ...
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MAPLE
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f := proc(L) local t1, i; t1 := []; for i from 1 to nops(L)-1 do t1 := [op(t1), (L[i]+L[i+1])/2]; od: t1; end; f2 := n->[seq(1/i, i=1..n)];
F := proc(n) local L, i; L := f2(n); for i from 1 to n-1 do L := f(L); od: op(L); end;
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MATHEMATICA
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a[n_]:=(2-2^(1-n))/n; a[1]:=1; a[n_]:=a[n-1]+(2^(1-n)*(1+n)-2)/((n-1)*n); a[1]:=1; a[n_]:=a[n-1]*(2^n-1)*(n-1)/(n*(2^n-2));
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CROSSREFS
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Cf. A090634.
Sequence in context: A120411 A069112 A064084 this_sequence A098583 A043729 A137170
Adjacent sequences: A090630 A090631 A090632 this_sequence A090634 A090635 A090636
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KEYWORD
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nonn,frac
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AUTHOR
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njas, Dec 13 2003
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