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%I #31 Mar 18 2022 21:49:58
%S 4,11,14,29,36,76,82,140,199,234,364,393,478,521,536,756,1030,1364,
%T 1764,2236,2786,3420,3571,3775,4144,4287,4964,5886,6916,8060,8886,
%U 9324,9349,10714,12236,13896,15700,16238,17654,18557,19764,22036,24476,27090,29884
%N Indices of metallic means that are powers of other metallic means.
%C Metallic mean k is mm(k) = (k + sqrt(k^2 + 4))/2.
%C The 4th metallic mean (mm), sometimes called the "copper" mean, is mm(4) = (4 + sqrt(16 + 4))/2 = 4.236... This value is also = 1.618...^3, where 1.618... is the 1st mm, the "golden" one, or (1 + sqrt(1 + 4))/2. This can be shown algebraically.
%C Any odd power of an mm will give another mm. The odd powers of the 1st mm are:
%C phi^1 = 1.618..., the 1st mm,
%C phi^3 = 4.236..., the 4th mm,
%C phi^5 = 11.090..., the 11th mm,
%C phi^7 = 29.034..., the 29th mm,
%C etc.
%C The indices of these mm's are 1, 4, 11, 29, 76, ... (A002878).
%C In parallel, the powers of the 2nd mm are:
%C slv^1 = 2.414..., the 2nd mm,
%C slv^3 = 14.071..., the 14th mm,
%C slv^5 = 82.012..., the 82nd mm,
%C etc.
%C The indices of these mm's are 2, 14, 82, 478, 2786, ... (A077444).
%C The indices of the mm's for the 3rd mm are A259131. The 4th mm's are A267797.
%C Every mm produces such a sequence.
%C The union of all such sequences (excluding their first terms) is this sequence.
%e 76 is a term since mm(76) = mm(1)^9 is a power of an earlier mean (the golden ratio in this case, and 76 is a Lucas number).
%e From _Peter Luschny_, Mar 16 2022: (Start)
%e A representation of the values claimed in the definition is given for the first 8 terms by:
%e ( 4 + 2*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^3.
%e (11 + 5*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^5.
%e (14 + 5*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^3.
%e (29 + 13*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^7.
%e (36 + 10*sqrt(13))/ 2 = ((3 + sqrt(13))/ 2)^3.
%e (76 + 34*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^9.
%e (82 + 29*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^5.
%e (140 + 26*sqrt(29))/ 2 = ((5 + sqrt(29))/ 2)^3.
%e (End)
%t getMetallicMean[n_] := (n + Power[Power[n, 2] + 4, 1 / 2]) / 2;
%t getMetallicCompositesUpTo[maxCandidateIndex_] := Module[
%t {sequence, metallicMeanIndex, metallicMean, oddPower, candidateIndex},
%t sequence = {};
%t metallicMeanIndex = 1;
%t While[
%t True,
%t (* skip metallic means already shown to be a power of another *)
%t If[MemberQ[sequence, metallicMeanIndex], metallicMeanIndex++];
%t metallicMean = getMetallicMean[metallicMeanIndex];
%t oddPower = 3;
%t While[
%t True,
%t candidateIndex = Floor[Power[metallicMean, oddPower]];
%t If[
%t candidateIndex <= maxCandidateIndex,
%t AppendTo[sequence, candidateIndex];
%t oddPower += 2,
%t Break[]
%t ]
%t ];
%t If[
%t oddPower == 3,
%t (* no chance of finding further results below the max, if even the first candidate at this index exceeded it *)
%t Break[],
%t metallicMeanIndex++
%t ];
%t ];
%t Sort[sequence]
%t ];
%t getMetallicCompositesUpTo[50000]
%Y Union of A002878, A077444, A259131, A267797, etc., minus each sequence's first entry.
%K nonn
%O 1,1
%A _Douglas Blumeyer_, Mar 14 2022
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