A071679 Least k such that the maximum number of elements among the continued fractions for k/1, k/2, k/3, k/4, ..., k/k equals n.
1, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..4784
- Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (x)).
- Index entries for linear recurrences with constant coefficients, signature (1,1).
Programs
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Magma
[1] cat [Fibonacci(n+2): n in [2..50]]; // Vincenzo Librandi, Jul 12 2015
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Mathematica
Join[{1}, LinearRecurrence[{1, 1}, {3, 5}, 50]] (* Vincenzo Librandi, Jul 12 2015 *) -
PARI
a(n)=if(n>1,fibonacci(n+2),1) \\ Charles R Greathouse IV, Jan 17 2012
Formula
Smallest k such that n = Max_{ i=1..k: Card[ contfrac(k/i) ] }.
a(1) = 1; for n>1 a(n) = F(n+2) where F(n)=A000045(n) are the Fibonacci numbers.
G.f.: (1+x)^2/(1-x-x^2); a(n) = 3*F(n+1) - F(n-1) - 0^n. - Paul Barry, Jul 26 2004
a(n) = Fibonacci(n+2) for n > 1. - Charles R Greathouse IV, Jan 17 2012