

A305880


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.


2



2211, 2211, 22211, 22211, 222211, 222211, 2222211, 2222211, 22222211, 22222211, 222222211, 222222211, 2222222211, 2222222211, 22222222211, 22222222211, 222222222211, 222222222211, 2222222222211, 2222222222211, 22222222222211, 22222222222211
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OFFSET

1,1


COMMENTS

a(2n1) and a(2n) consist of n+1 2's followed by 2 1's.
If a reverse sorted Fibonacci sequence starts with any two numbers, then it eventually becomes either cyclic or turns into this sequence.
In base 10, the corresponding sequence is A069638 and is periodic.


LINKS

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,10,10).


FORMULA

From Colin Barker, Jun 19 2018: (Start)
G.f.: x*(2211  2110*x^2) / ((1  x)*(1  10*x^2)).
a(n) = (2^((n+5)/2+3/2) * 5^((n+5)/2+1/2)  101) / 9 for n even.
a(n) = (2^((n+9)/2) * 5^((n+7)/2)  101) / 9 for n odd.
a(n) = a(n1) + 10*a(n2)  10*a(n3) for n>3.
(End)


EXAMPLE

2211 + 2211 equals 210122 when all numbers are interpreted in base 3/2; after sorting and omitting 0's we obtain a(2) = 22211.
(A305753 has more detailed examples which may help explain the calculations here.  N. J. A. Sloane, Jun 22 2018)


CROSSREFS

Cf. A000045, A024629, A069638, A237575, A305753.
Sequence in context: A226562 A300166 A031772 * A031545 A191679 A238456
Adjacent sequences: A305877 A305878 A305879 * A305881 A305882 A305883


KEYWORD

nonn,base


AUTHOR

Tanya Khovanova and PRIMES STEP Senior group, Jun 13 2018


STATUS

approved



