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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
OFFSET
1,1
COMMENTS
a(2n-1) 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.
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(n-1) + 10*a(n-2) - 10*a(n-3) 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
KEYWORD
nonn,base
AUTHOR
Tanya Khovanova and PRIMES STEP Senior group, Jun 13 2018
STATUS
approved