%I #54 Jun 24 2018 14:42:26

%S 2211,2211,22211,22211,222211,222211,2222211,2222211,22222211,

%T 22222211,222222211,222222211,2222222211,2222222211,22222222211,

%U 22222222211,222222222211,222222222211,2222222222211,2222222222211,22222222222211,22222222222211

%N A base 3/2 reverse sorted Fibonacci sequence that starts with terms 2211 and 2211. 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 decreasing order, omitting all zeros.

%C a(2n-1) and a(2n) consist of n+1 2's followed by 2 1's.

%C If a reverse sorted Fibonacci sequence starts with any two numbers, then it eventually becomes either cyclic or turns into this sequence.

%C In base 10, the corresponding sequence is A069638 and is periodic.

%H Colin Barker, <a href="/A305880/b305880.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,10,-10).

%F From _Colin Barker_, Jun 19 2018: (Start)

%F G.f.: x*(2211 - 2110*x^2) / ((1 - x)*(1 - 10*x^2)).

%F a(n) = (2^((n+5)/2+3/2) * 5^((n+5)/2+1/2) - 101) / 9 for n even.

%F a(n) = (2^((n+9)/2) * 5^((n+7)/2) - 101) / 9 for n odd.

%F a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) for n>3.

%F (End)

%e 2211 + 2211 equals 210122 when all numbers are interpreted in base 3/2; after sorting and omitting 0's we obtain a(2) = 22211.

%e (A305753 has more detailed examples which may help explain the calculations here. - _N. J. A. Sloane_, Jun 22 2018)

%Y Cf. A000045, A024629, A069638, A237575, A305753.

%K nonn,base

%O 1,1

%A _Tanya Khovanova_ and PRIMES STEP Senior group, Jun 13 2018