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A125758
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Numbers congruent to 4 or 7 (mod 9).
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2
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4, 7, 13, 16, 22, 25, 31, 34, 40, 43, 49, 52, 58, 61, 67, 70, 76, 79, 85, 88, 94, 97, 103, 106, 112, 115, 121, 124, 130, 133, 139, 142, 148, 151, 157, 160, 166, 169, 175, 178, 184, 187, 193, 196, 202, 205, 211, 214, 220, 223, 229, 232, 238, 241, 247, 250, 256, 259, 265, 268
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OFFSET
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1,1
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COMMENTS
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For a given integer m, write its binary representation in reverse order, as in A125626, A125754, etc.; let a 0 mean "halving" and a 1 mean "k -> 3k+1". Then m specifies an operation on real numbers given by k -> f_m(k). Suppose the equation f_m(k) = k has a positive integer solution for some m. Then we conjecture that the values of k are precisely the terms of this sequence.
25 is a term because we have 25 -> 76 -> 38 -> 19 -> 58 -> 29 -> 88 -> 44 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 25.
In other words, we conjecture that this sequence coincides with A125757 sorted and with duplicates removed.
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 9.
G.f.: x*(4+3*x+2*x^2)/((1+x)*(x-1)^2). (End)
E.g.f.: 2 + ((9*x - 5/2)*exp(x) - (3/2)*exp(-x))/2. - David Lovler, Aug 21 2022
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MATHEMATICA
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Select[Range[300], MemberQ[{4, 7}, Mod[#, 9]]&] (* Harvey P. Dale, Mar 12 2011 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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