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%I #4 Oct 18 2019 16:36:15
%S 6,20,57960,3145940416080,5765760,
%T 288680192354725622464710969631440008928,
%U 20484953806009937929429725901717124022833778640,59553628273094395440,102119994931499628863688098762720537989600
%N Denominator of the last term of the Egyptian fraction sum (using the greedy algorithm) which satisfies 1 = 1/n + 1/(n+1) + 1/(n+2) ... 1/a(n).
%C The next term in the series, a(11), is 7*10^192.
%H Amiram Eldar, <a href="/A069257/b069257.txt">Table of n, a(n) for n = 2..14</a>
%e Since 1 = 1/3 + 1/4 + 1/5 + 1/6 + 1/20, a(3) = 20.
%t a[n_] := Module[{s = 1/n, k = n}, While[s < 1, k = Max[k + 1, Ceiling[1/(1 - s)]]; s += 1/k]; k]; Array[a, 9, 2] (* _Amiram Eldar_, Oct 18 2019 *)
%K easy,frac,nonn
%O 2,1
%A Christopher Lund (clund(AT)san.rr.com), Apr 14 2002
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