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Largest denominator used in the Egyptian fraction representation of n/(n + 1) by the greedy algorithm.
5

%I #22 Oct 27 2023 20:06:21

%S 2,6,4,20,3,42,24,18,15,231,12,156,231,10,240,32640,9,342,180,126,99,

%T 34362,8,600,312,216,168,24360,120,633759288,96,88,16728,9240,72,6808,

%U 5016,6552,60,28536,7,1806,924,630,483,779730,336,294,138075,238,79716

%N Largest denominator used in the Egyptian fraction representation of n/(n + 1) by the greedy algorithm.

%C a(n) = A247765(n,A100678(n)). - _Reinhard Zumkeller_, Sep 25 2014

%H Alois P. Heinz, <a href="/A100695/b100695.txt">Table of n, a(n) for n = 1..5000</a> (first 100 terms from Reinhard Zumkeller)

%H <a href="https://oeis.org/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>

%e a(16) = 32640 because 16/17 = 1/2 + 1/3 + 1/10 + 1/128 + 1/32640.

%t a[n_] := Module[{m = n/(n+1)}, While[Numerator[m]>1, m = m-1/Ceiling[1/m]]; 1/m]; Array[a, 100] (* _Jean-François Alcover_, Mar 12 2019, after _M. F. Hasler_ *)

%o (PARI) a(n)={n/=(n+1);while(numerator(n)>1,n-=1/ceil(1/n));1/n} \\ _M. F. Hasler_, Sep 24 2014

%o (Haskell)

%o a100695 = last . a247765_row -- _Reinhard Zumkeller_, Sep 25 2014

%Y Cf. A100678.

%Y Cf. A247765.

%K nonn

%O 1,1

%A _John W. Layman_, Dec 08 2004