A305753 A base-3/2 sorted Fibonacci sequence that starts with a(0) = 0 and a(1) = 1. The terms are interpreted as numbers written in base 3/2. To get a(n+2), add a(n) and a(n+1), write the result in base 3/2 and sort the "digits" into increasing order, omitting all zeros.
0, 1, 1, 2, 2, 12, 12, 112, 112, 1112, 1112, 11112, 11112, 111112, 111112, 1111112, 1111112, 11111112, 11111112, 111111112, 111111112, 1111111112, 1111111112, 11111111112, 11111111112, 111111111112, 111111111112, 1111111111112, 1111111111112, 11111111111112, 11111111111112
Offset: 0
Examples
Write decimal numbers as x_10, base-3/2 numbers as x_b (see A024629). We have a(1) = 1, a(2) = 2 (in both bases). Adding, we get 1+2 = 3_10 = 20_b, and sorting the digits gives a(3) = 2_b = 2_10. Adding 2 and 2 we get 4_10 = 21_b, and sorting the digits gives a(4) = 12_b = (7/2)_10. Adding 2 and 7/2 we get (11/2)_10 = 201_b, and sorting the digits gives a(5) = 12_b = (7/2)_10. Adding (7/2)_10 and (7/2)_10 we get 7_10 = 211_b, and sorting the digits gives a(6) = 112_b = (23/4)_10. Adding (7/2)_10 and (23/4)_10 we get (37/4)_10 = 2011_b, and sorting the digits gives a(7) = 112_b = (23/4)_10. And so on.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018.
- Index entries for linear recurrences with constant coefficients, signature (1,10,-10).
Crossrefs
Programs
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PARI
concat(0, Vec(x*(1 - 3*x)*(1 + 3*x) / ((1 - x)*(1 - 10*x^2)) + O(x^40))) \\ Colin Barker, Jun 19 2018
Formula
From Colin Barker, Jun 14 2018: (Start)
Generating function: x*(1 - 3*x)*(1 + 3*x) / ((1 - x)*(1 - 10*x^2)).
a(n) = (10^(n/2) + 80) / 90 for n>0.
a(n) = (10^((n-1)/2) + 8) / 9 for n>0.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) for n>4.
(End)
Extensions
Edited by N. J. A. Sloane, Jun 22 2018
Comments