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A261527 Irregular triangular array giving minimum number of reciprocal steps in the boomerang fractions process needed to return to 1 if a returning path exists, otherwise 0. 0
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 1, 2, 20, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 2, 24, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,8
COMMENTS
The boomerang fractions process is defined as follows. Fix a rational number q, 0<q<1. Starting with 1, on the first step add q and on subsequent steps either add q or take the reciprocal.
Let q(n) be the n-th rational number in the interval (0,1) in the canonical ordering, that is, q(n)=A038566(n+1)/A038567(n+1). Then a(n) is obtained by applying the sequence definition with q=q(n).
The value of a(n) is 1 if and only if q(n) is the difference of two unit fractions.
If q(m) = k q(n) for some positive integer k, then a(n) <= a(m).
The first rational number in the canonical ordering for which it is not known whether a(n) is nonzero is q(40)=9/11. If a(40) is nonzero, then a(40) >= 55.
LINKS
EXAMPLE
a(1) = 1 since q(1) = 1/2 and there is the returning path 1 --> 1+2*(1/2) = 2 --> 1/2 --> 1/2+1/2 = 1, which uses the reciprocal operation once.
a(8) = 2 since q(8) = 3/5, which cannot be written as the difference of two unit fractions (ruling out a(8) = 1) and because there is the returning path 1 --> 1+15*(3/5) = 10 --> 1/10 --> 1/10+4*(3/5) = 5/2 --> 2/5 --> 2/5+3/5 = 1, which uses the reciprocal operation twice.
Triangle starts:
1;
1, 1;
1, 1;
1, 1, 2, 1;
1, 1;
1, 1, 1, 2, 4, 1;
1, 1, 2, 1;
1, 1, 1, 2, 20, 1;
1, 1, 4, 1;
1, 1, 2, 2, 1, 2, 24, 2,
...
MATHEMATICA
(* In the following code, Alpha is the operation "add q" and Beta is the operation "take the reciprocal and add q". The set L(j) is defined to be the set of positive rational numbers r such that there is a path from r to 1 that uses Beta exactly j times. The program computes L(1), L(2), and so on, until an L(j) is found that contains 1, in which case it returns j, or until maxIterations is exceeded, in which case it returns 0. The function iterateUntilOne can generate the result for all q up to 6/11 rather quickly, but for q = 7/11, which corresponds to a(38) = 24, it requires considerable time; it is not capable of ruling out the existence of a returning path that uses Beta more than maxIterations times. *)
applyBetaInverse[q_, x_] := 1/(x - q)
applyAlphaPowerInverse[q_, x_] :=
Table[x - q j, {j, 0, Ceiling[x/q] - 1}]
iterateUntilOne[q_, maxIterations_] :=
Module[{list, listOld, oneFound, it, betaInverseResult},
listOld = Flatten[applyAlphaPowerInverse[q, #] & /@ {1}];
oneFound = False;
For[it = 1, ! oneFound && it <= maxIterations, it++,
betaInverseResult =
applyBetaInverse[q, #] & /@ Select[listOld, # > q &];
list = Flatten[applyAlphaPowerInverse[q, #] & /@ betaInverseResult];
oneFound = MemberQ[list, 1];
Print["L(", it, ") : length ", Length[list],
If[oneFound, ", contains 1", ", does not contain 1"]];
listOld = list
];
If[oneFound,
it - 1,
0
]
]
iterateUntilOne[#, 20] & /@Flatten[Join[
Table[Select[Range[1, d], CoprimeQ[d, #] &]/d, {d, 2, 10}],
Range[1, 6]/11]]
CROSSREFS
Cf. A256174.
Sequence in context: A324350 A175466 A214403 * A292436 A184097 A345932
KEYWORD
nonn,tabf,more
AUTHOR
William P. Orrick, Aug 21 2015
STATUS
approved

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Last modified July 20 18:09 EDT 2024. Contains 374459 sequences. (Running on oeis4.)