A154429 a(n) is the least k such that the greedy algorithm (for Egyptian fractions) on 4k/(24n+1) terminates in at most three steps.

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, 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, 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

Offset: 1

Author

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

Keywords

• nonn

Examples

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

References

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

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• D. Eppstein, Egyptian fractions

Programs

Mathematica

GreedyPart[q_Integer] := 0;
GreedyPart[Rational[1, y_]] := 0;
GreedyPart[q_Rational] := q - If[q < 0 || q > 1, Floor[q], Rational[1, 1 + Quotient[1, q]]];
SubtractShifted[l_] := Drop[l, -2] - Take[l, {2, -2}];
EgyptGreedy[q_] := SubtractShifted[FixedPointList[GreedyPart, q]];
terms := 200;
For[i = 25, i <= 24*terms + 1, i = i + 24,k = 2;While[Length[EgyptGreedy[4k/i]]> 3, k++ ];Print[k]]

Extensions

More terms from Seiichi Manyama, Sep 21 2022