A352403 Indices of metallic means that are powers of other metallic means.
4, 11, 14, 29, 36, 76, 82, 140, 199, 234, 364, 393, 478, 521, 536, 756, 1030, 1364, 1764, 2236, 2786, 3420, 3571, 3775, 4144, 4287, 4964, 5886, 6916, 8060, 8886, 9324, 9349, 10714, 12236, 13896, 15700, 16238, 17654, 18557, 19764, 22036, 24476, 27090, 29884
Offset: 1
Keywords
Examples
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). From _Peter Luschny_, Mar 16 2022: (Start) A representation of the values claimed in the definition is given for the first 8 terms by: ( 4 + 2*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^3. (11 + 5*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^5. (14 + 5*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^3. (29 + 13*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^7. (36 + 10*sqrt(13))/ 2 = ((3 + sqrt(13))/ 2)^3. (76 + 34*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^9. (82 + 29*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^5. (140 + 26*sqrt(29))/ 2 = ((5 + sqrt(29))/ 2)^3. (End)
Programs
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Mathematica
getMetallicMean[n_] := (n + Power[Power[n, 2] + 4, 1 / 2]) / 2; getMetallicCompositesUpTo[maxCandidateIndex_] := Module[ {sequence, metallicMeanIndex, metallicMean, oddPower, candidateIndex}, sequence = {}; metallicMeanIndex = 1; While[ True, (* skip metallic means already shown to be a power of another *) If[MemberQ[sequence, metallicMeanIndex], metallicMeanIndex++]; metallicMean = getMetallicMean[metallicMeanIndex]; oddPower = 3; While[ True, candidateIndex = Floor[Power[metallicMean, oddPower]]; If[ candidateIndex <= maxCandidateIndex, AppendTo[sequence, candidateIndex]; oddPower += 2, Break[] ] ]; If[ oddPower == 3, (* no chance of finding further results below the max, if even the first candidate at this index exceeded it *) Break[], metallicMeanIndex++ ]; ]; Sort[sequence] ]; getMetallicCompositesUpTo[50000]
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