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a(n) is the largest denominator when the greedy algorithm for Egyptian fractions is applied to 1/n + 1/(n+1).
0

%I #30 Apr 15 2023 23:32:12

%S 2,3,12,180,30,1428,56,2520,90,2310,132,100292556,182,9240,240,119952,

%T 306,614444040,380,23100,462,42190274940,552,77390453400,650,201474,

%U 756,23370247110,870,200880,992,14523137084239067683872,1122,2206260,1260,104845560637757648698080

%N a(n) is the largest denominator when the greedy algorithm for Egyptian fractions is applied to 1/n + 1/(n+1).

%F a(n) = A050210(n*(n+1), 2*n+1). - _Michel Marcus_, Apr 14 2023

%e For n=16, 1/16 + 1/17 = 33/272 which written in Egyptian fractions is 1/9 + 1/98 + 1/119952 and the largest denominator is 119952.

%t egyptFraction[f_] := Ceiling[1/Most[NestWhileList[# - 1/Ceiling[1/#] &, f, # != 0 &]]]; a[n_] := egyptFraction[1/n + 1/(n + 1)][[-1]]; Array[a, 40] (* _Amiram Eldar_, Apr 14 2023 *)

%Y Cf. A050210.

%K nonn

%O 1,1

%A _Sebastian F. Orellana_, Apr 14 2023

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Last modified September 20 18:47 EDT 2024. Contains 376075 sequences. (Running on oeis4.)