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 A125758 Numbers congruent to 4 or 7 (mod 9). 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. LINKS David Lovler, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (1,1,-1). FORMULA From R. J. Mathar, Apr 03 2009: (Start) a(n) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 9. a(n) + a(n+1) = A017185(n). 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 MATHEMATICA Select[Range[300], MemberQ[{4, 7}, Mod[#, 9]]&] (* Harvey P. Dale, Mar 12 2011 *) CROSSREFS Cf. A125626, A125754, A125755, A125756, A125757, A125710, A125711. Sequence in context: A310804 A310805 A310806 * A259566 A151788 A310807 Adjacent sequences: A125755 A125756 A125757 * A125759 A125760 A125761 KEYWORD nonn,easy AUTHOR N. J. A. Sloane and David Applegate, Feb 02 2007 STATUS approved

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Last modified August 9 20:04 EDT 2024. Contains 375044 sequences. (Running on oeis4.)