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%I #31 Jan 04 2024 06:37:57
%S 1,4,12,32,80,64,448,1024,2304,5120,11264,8192,53248,114688,245760,
%T 524288,1114112,262144,4980736,2097152,3145728,46137344,96468992,
%U 67108864,419430400,872415232,1811939328,3758096384,7784628224,5368709120,33285996544,68719476736
%N 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) = denominator of F(n).
%C a(n) is the denominator of the resistance of the n-dimensional cube between two adjacent nodes, when the resistance of each edge is 1. See Nedermeyer and Smorodinsky. - _Michel Marcus_, Sep 13 2019
%H Alois P. Heinz, <a href="/A090634/b090634.txt">Table of n, a(n) for n = 1..3312</a> (first 200 terms from T. D. Noe)
%H F. Nedermeyer and Y. Smorodinsky, <a href="http://static.nsta.org/pdfs/QuantumV7N1.pdf">Resistance in the multidimensional cube</a>, Quantum, Sept/October 1996, pp. 12-15 (beware file is 75Mb).
%H Putnam Competition, <a href="https://kskedlaya.org/putnam-archive/2003.pdf">Problem B2</a>, <a href="https://kskedlaya.org/putnam-archive/2003s.pdf">Solutions</a>, 2003.
%F a(n) = A131135(n)/2. - _Paul Barry_, Jun 17 2007
%F a(n) = denominator(2*(1-1/2^n)/n) (conjectured). - _Michel Marcus_, Sep 12 2019
%e 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, ...
%p a:= n-> denom(coeff(series(2*log((x/2-1)/(x-1)), x, n+1), x, n)):
%p seq(a(n), n=1..35); # _Alois P. Heinz_, Aug 02 2018
%t f[s_list] := Table[(s[[k]] + s[[k+1]])/2, {k, 1, Length[s]-1}];
%t a[n_] := Nest[f, 1/Range[n], n-1] // First // Denominator;
%t Array[a, 40] (* _Jean-François Alcover_, Aug 02 2018 *)
%o (Haskell)
%o import Data.Ratio (denominator, (%))
%o a090634 n = denominator z where
%o [z] = (until ((== 1) . length) avg) $ map (1 %) [1..n]
%o avg xs = zipWith (\x x' -> (x + x') / 2) (tail xs) xs
%o -- _Reinhard Zumkeller_, Dec 08 2011
%Y Cf. A090633 (numerators).
%K nonn,frac
%O 1,2
%A _N. J. A. Sloane_, Dec 13 2003
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