A154429 - OEIS

%I #13 Oct 04 2023 15:30:58

%S 2,2,2,5,3,4,13,2,2,7,5,51,4,4,5,2,3,5,5,7,5,6,2,5,11,4,3,5,5,2,2,7,4,

%T 5,29,2,2,2,5,8,4,11,2,2,6,4,11,5,3,11,2,5,5,5,7,4,37,2,3,3,4,7,5,5,2,

%U 2,17,5,5,54,2,2,2,5,7,4,11,2,2,6,5,3,4,5,10,2,7,5,5,7,5,12,2,3,10,4,7,5,5,2

%N a(n) is the least k such that the greedy algorithm (for Egyptian fractions) on 4k/(24n+1) terminates in at most three steps.

%D J. Steuding, Diophantine Analysis, Chapman & Hall/CRC, 2005, pp. 39-40, 50.

%H Seiichi Manyama, <a href="/A154429/b154429.txt">Table of n, a(n) for n = 1..10000</a>

%H D. Eppstein, <a href="http://www.ics.uci.edu/~eppstein/numth/egypt/">Egyptian fractions</a>

%e For n=3, the Greedy Algorithm gives 8/73=1/10+1/105+1/15330

%t GreedyPart[q_Integer] := 0;

%t GreedyPart[Rational[1, y_]] := 0;

%t GreedyPart[q_Rational] := q - If[q < 0 || q > 1, Floor[q], Rational[1, 1 + Quotient[1, q]]];

%t SubtractShifted[l_] := Drop[l, -2] - Take[l, {2, -2}];

%t EgyptGreedy[q_] := SubtractShifted[FixedPointList[GreedyPart, q]];

%t terms := 200;

%t For[i = 25, i <= 24*terms + 1, i = i + 24,k = 2;While[Length[EgyptGreedy[4k/i]]> 3, k++ ];Print[k]]

%K nonn

%O 1,1

%A Matthew McMullen (mmcmullen(AT)otterbein.edu), Jan 09 2009

%E More terms from _Seiichi Manyama_, Sep 21 2022