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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.
2
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
0,4
COMMENTS
In base 10, the corresponding sequence is A069638 and is periodic.
LINKS
B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 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)
EXAMPLE
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.
PROG
(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
CROSSREFS
This is A047855 with terms repeated. - N. J. A. Sloane, Jun 19 2018
Sequence in context: A341432 A092144 A224497 * A181813 A059187 A350329
KEYWORD
nonn,base
AUTHOR
Tanya Khovanova and PRIMES STEP Senior group, Jun 09 2018
EXTENSIONS
Edited by N. J. A. Sloane, Jun 22 2018
STATUS
approved