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A345347 Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.

%I A345347 #29 Jan 02 2022 09:28:46
%S A345347 1,4,7,11,12,18,19,20,29,30,31,32,33,47,48,49,50,51,52,53,54,76,77,78,
%T A345347 79,80,81,82,83,84,85,86,87,88,123,124,125,126,127,128,129,130,131,
%U A345347 132,133,134,135,136,137,138,139,140,141,142,143,199,200,201,202,203,204
%N A345347 Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.
%C A345347 The terms consist of 1 together with numbers that appear in row m of the Wythoff array (A035513) if m is in the sequence.
%C A345347 a(0) = 1, otherwise a(n) is the number whose Zeckendorf representation is "10" followed by the Zeckendorf representation of n.
%C A345347 If we define an extended Zeckendorf representation to be the Zeckendorf representation with "01" appended, then the numbers in the sequence are exactly those whose extended representation starts 101... . This extended representation is a valid Fibonacci base representation if we specify the rightmost digit to have weight F(0) = 0.
%C A345347 Equivalently, for positive integer m, find the largest k with F(k) <= m, where F(k) is the k-th Fibonacci number. m is in the sequence if and only if m >= F(k) + F(k-2).
%C A345347 Numbers given to rabbits on Rabbit 1's branch of the generation tree described in the A035513 examples.
%C A345347 Equivalently, take the positive integers in turn, placing runs of them alternatively into 2 sets, with run lengths from A053602/A051792 (self-interleaved Fibonacci sequence) as follows:
%C A345347    set A: 1   0   1   1   2   3   5 ...
%C A345347    set B:   1   1   2   3   5   8 ...
%C A345347 The sequence lists the numbers in set A.
%H A345347 Encyclopedia of Mathematics, <a href="https://encyclopediaofmath.org/wiki/Zeckendorf_representation">Zeckendorf representation</a>.
%H A345347 OEIS Wiki, <a href="/wiki/Zeckendorf_representation">Zeckendorf representation</a>.
%H A345347 N. J. A. Sloane, <a href="/classic.html#WYTH">Classic Sequences</a>.
%F A345347 a(n) = A000045(A108852(n)+1) + n.
%F A345347 Union_{k >= 2} {m : F(k)+F(k-2) <= m < F(k+1)}, where F(k) = A000045(k).
%e A345347 The initial Fibonacci numbers are F(0)..F(5) = 0, 1, 1, 2, 3, 5.
%e A345347 For n = 0, the largest k with F(k) <= 0 is k = 0, so F(k+2) = F(2) = 1, so a(0) = 1 + 0 = 1.
%e A345347 For n = 1, the largest k with F(k) <= 1 is k = 2, so F(k+2) = F(4) = 3, so a(1) = 3 + 1 = 4.
%e A345347 For n = 4, the largest k with F(k) <= 4 is k = 4, so F(k+2) = F(6) = 8, so a(4) = 8 + 4 = 12.
%e A345347 In the paragraph that follows we use the Wythoff array-based definition from the start of the comments.
%e A345347 Every positive integer appears once (only) in the Wythoff array. 0 is not positive, so does not appear in the array, so is not in the sequence. 1 is in the sequence by definition. 2 appears in Wythoff row 0, and 0 is not in the sequence, so 2 is not in the sequence. 4 appears in Wythoff row 1, and 1 is in the sequence, so 4 is in the sequence.
%t A345347 kmax=12;Flatten[Table[Range[Fibonacci[k]+Fibonacci[k-2],Fibonacci[k+1]-1],{k,2,kmax}]] (* _Paolo Xausa_, Jan 02 2022 *)
%t A345347 A108852[n_]:=1+Floor[Log[GoldenRatio,1+n*Sqrt[5]]];
%t A345347 nterms=100;Table[n+Fibonacci[1+A108852[n]],{n,0,nterms-1}](* _Paolo Xausa_, Jan 02 2022 *)
%o A345347 (PARI) a(n) = my(k=0); while(fibonacci(k)<=n, k=k+1); n+fibonacci(k+1)
%Y A345347 Cf. A000045, A035513, A051792, A053602, A108852.
%Y A345347 Appears to be column 1 of A194030.
%K A345347 nonn,base,easy
%O A345347 0,2
%A A345347 _Peter Munn_, Jun 14 2021