

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
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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



FORMULA

a(n) = a(n1) + a(n2)  a(n3) = a(n2) + 9.
G.f.: x*(4+3*x+2*x^2)/((1+x)*(x1)^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



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



