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A349688
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Numbers k such that the sequence of elements of the continued fraction of the abundancy index of k is palindromic.
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2
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1, 6, 24, 28, 30, 42, 54, 66, 70, 78, 84, 90, 96, 102, 114, 120, 138, 140, 174, 186, 220, 222, 246, 258, 264, 270, 282, 308, 318, 330, 342, 348, 354, 364, 366, 402, 426, 438, 474, 476, 496, 498, 532, 534, 582, 606, 618, 642, 644, 654, 660, 672, 678, 744, 760, 762
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OFFSET
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1,2
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COMMENTS
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All the multiply-perfect numbers (A007691) are terms of this sequence, since the continued fraction of their abundancy index contains a single element.
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LINKS
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EXAMPLE
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24 is a term since the sequence of elements of the abundancy index of 24, sigma(24)/24 = 5/2 = 2 + 1/2, is {2, 2}, which is palindromic.
42 is a term since the sequence of elements of the abundancy index of 42, sigma(42)/42 = 16/7 = 2 + 1/(3 + 1/2), is {2, 3, 2}, which is palindromic.
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MATHEMATICA
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q[n_] := PalindromeQ[ContinuedFraction[DivisorSigma[1, n]/n]]; Select[Range[1000], q]
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PROG
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(PARI) isok(k) = my(v=contfrac(sigma(k)/k)); v == Vecrev(v); \\ Michel Marcus, Nov 25 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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