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The part of n in base phi left of the decimal using a least-greedy algorithm representation.
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%I #16 May 27 2023 14:45:53

%S 0,1,1,10,11,101,111,1010,1011,1101,1110,1111,10101,10111,11010,11011,

%T 11101,11111,101010,101011,101101,101110,101111,110101,110111,111010,

%U 111011,111101,111110,111111,1010101,1010111,1011010,1011011,1011101

%N The part of n in base phi left of the decimal using a least-greedy algorithm representation.

%C Uses least-greedy algorithm (start with largest possible power of phi, writing a 1 only when required, then work downward).

%C a(n) is also the left portion of the base-phi representation of n in Knott's representation which uses the least number of 0's, the most 1's, and in which the right-hand portion (see A362919) is finite. - _N. J. A. Sloane_, May 27 2023

%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phigits.html">Phigits and the Base Phi representation</a>.

%H Jeffrey Shallit, <a href="https://arxiv.org/abs/2305.02672">Proving Properties of phi-Representations with the Walnut Theorem-Prover</a>, arXiv:2305.02672 [math.NT], 2023. [Note that this document has been revised multiple times.]

%e 6 = 111.01101010... in base phi using the least-greedy algorithm. The part to the left of the decimal is a(6) = 111.

%o constant (float): phi=(sqrt(5)+1)/2; variable (float): lphi=phi^floor[log(n)/log(phi)]; variable (float): rem=n; variable (integer): count=0; loop: while lphi>1 (count=count*10; lphi=lphi/phi; if(rem > lphi*phi) { rem=rem-lphi; count++;}}

%Y Cf. A055778, A104605, A118241, A105424, A362919, A362920.

%K nonn,base

%O 0,4

%A _Graeme McRae_, Apr 17 2006

%E a(1) corrected by _N. J. A. Sloane_, May 27 2023