cp's OEIS Frontend

A125626 Numbers n whose reverse binary representation has the following property: let a 0 mean "halving" and a 1 mean "k -> 3k+1". The number describes an operation k -> f_n(k). If the equation f_n(k) = k has a positive solution, n is a term in the sequence.

%I A125626 #7 Jan 24 2022 08:01:32
%S A125626 4,8,16,32,33,34,36,40,48,64,65,66,68,72,80,96,128,129,130,131,132,
%T A125626 133,134,136,137,138,140,144,145,146,148,152,160,161,162,164,168,176,
%U A125626 192,193,194,196,200,208,224,256,257,258,259,260
%N A125626 Numbers n whose reverse binary representation has the following property: let a 0 mean "halving" and a 1 mean "k -> 3k+1". The number describes an operation k -> f_n(k). If the equation f_n(k) = k has a positive solution, n is a term in the sequence.
%C A125626 The terms in this sequence have the following characterization. Suppose the binary expansion of n contains i 1's and j 0's. Then it is easy to see that n is in the sequence if and only if 3^i < 2^j, or i/j < log 2 / log 3 = 0.630929753... - _David Applegate_ and _N. J. A. Sloane_, Feb 01 2007
%C A125626 Note that f_n(x) is always a linear function of x.
%C A125626 The reverse binary expansions of the first few terms are:
%C A125626 001
%C A125626 0001
%C A125626 00001
%C A125626 000001
%C A125626 100001
%C A125626 010001
%C A125626 001001
%C A125626 000101
%C A125626 000011
%C A125626 0000001
%C A125626 1000001
%C A125626 0100001
%C A125626 0010001
%C A125626 0001001
%C A125626 0000101
%C A125626 0000011
%C A125626 00000001
%C A125626 10000001
%C A125626 01000001
%C A125626 11000001
%C A125626 00100001
%C A125626 ...
%C A125626 Could be used in conjunction with the Collatz (or 3x+1) conjecture. If the positive solution k is an integer (most are not) then a cycle exists. If this cycle does not contain a 1 and the sequence of steps agrees with what Collatz's rule tells you to do when you start with k, then the Collatz conjecture would be false.
%e A125626 Consider the term 200: its binary representation is 11001000. Reversing this gives 00010011. We solve (3*(3*(((3*(((k/2)/2)/2)+1)/2)/2)+1)+1) = k and find k = 40. Since k is positive, 200 is a member of the sequence.
%o A125626 (C) #include <stdio.h> #include <stdlib.h> #include <math.h> void multiply(float *coef, float *cons) { (*coef) *= 3; (*cons) = 3*(*cons)+1; } void divide(float *coef, float *cons) { (*coef) /= 2; (*cons) /= 2; } int main() { int a, b, c, n; float coef, cons, final; char data[30], sequence[30]; for (a = 1; a < 500; a++) { coef = 1; cons = 0; c = a; sequence[0] = ''; for (b = 1; b < 12; b++) //12 is arbitrary; it allows for "a" up to 2^12 { if (c != 0) { if (c % 2) { sprintf(sequence, "%s1", sequence); multiply(&coef, &cons); } else { sprintf(sequence, "%s0", sequence); divide(&coef, &cons); } c = trunc(c/2); } else break; } if (coef >= 1.0) { coef -= 1.0; cons *= -1.0; } else coef = 1.0-coef; final = cons/coef; if (final > 0) { sprintf(data, "%10.3f %s %d ", final, sequence, a); printf(data); } } return 0; }
%Y A125626 For the values of n for which the fixed point k is a positive (or any) integer, see A125754-A125757.
%Y A125626 Cf. A112695, A125710, A125711.
%K A125626 easy,nonn
%O A125626 4,1
%A A125626 Nicholas Sanders (gummybean(AT)gmail.com), Jan 27 2007