%I #7 Oct 18 2019 16:37:04
%S 3,5,8,11,12,16,18,18,21,25,29,27,28,36,35,35,41,42,42,47,51,52,49,52,
%T 56,59,61,67,68,69,72,72,76,80,78,79,89,86,82,92,94,92,101,100,99,102,
%U 101,109,108,114,115,114,114,120,125
%N The number of terms in the greedy Egyptian fraction expansion of 1 = 1/n + 1/(n+1) + 1/(n+2) + ... + 1/A069257(n).
%e a(3) = 5 since the expansion 1 = 1/3 + 1/4 + 1/5 + 1/6 + 1/20 has 5 terms.
%t a[n_] := Module[{c = 1, s = 1/n, k = n}, While[s < 1, k = Max[k + 1, Ceiling[1/(1 - s)]]; s += 1/k; c++]; c]; Array[a, 20, 2]
%Y Cf. A069257.
%K nonn,more
%O 2,1
%A _Amiram Eldar_, Oct 18 2019