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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 (list; graph; refs; listen; history; text; internal format)
 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. 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) 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 A349098 Adjacent sequences:  A305877 A305878 A305879 * A305881 A305882 A305883 KEYWORD nonn,base AUTHOR Tanya Khovanova and PRIMES STEP Senior group, Jun 13 2018 STATUS approved

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Last modified September 26 20:57 EDT 2022. Contains 357050 sequences. (Running on oeis4.)