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.
2211, 2211, 22211, 22211, 222211, 222211, 2222211, 2222211, 22222211, 22222211, 222222211, 222222211, 2222222211, 2222222211, 22222222211, 22222222211, 222222222211, 222222222211, 2222222222211, 2222222222211, 22222222222211, 22222222222211
Offset: 1
Examples
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)
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(n-1) + 10*a(n-2) - 10*a(n-3) for n>3.
(End)
Comments