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Number of digits in the minimal base-phi representation of n.
3

%I #21 Jun 20 2024 11:07:53

%S 1,4,5,5,8,8,9,9,9,9,9,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,

%T 13,13,13,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,17,17,17,

%U 17,17,17,17,17,17,17,17,17,17,17,17,17,17,17,17,17,17

%N Number of digits in the minimal base-phi representation of n.

%C See A130600(n) for the digits in the minimal base phi representation of n.

%C a(n) <= 2 * ceiling( log(n) / log(phi) ) for n > 1.

%D Michel Dekking and Ad van Loon. "On the representation of the natural numbers by powers of the golden mean." Fib. Quart. 61:2 (May 2023), 105-118.

%H T. D. Noe, <a href="/A190796/b190796.txt">Table of n, a(n) for n = 1..1000</a>

%H Michel Dekking and Ad van Loon, <a href="https://doi.org/10.48550/arXiv.2111.07544">On the representation of the natural numbers by powers of the golden mean</a>, arXiv:2111.07544 [math.NT], 15 Nov 2021.

%F a(n) = A055778(n) + A133775(n).

%F From _Michel Dekking_, Jun 19 2024: (Start)

%F Let (L(n)) = (2, 1, 3, 4, 7, 11, 18, 29, 47, ...) = A000032 be the Lucas numbers.

%F If L(2n) <= i <= L(2n+1), then a(i) = 4n+1; if L(2n+1)+1 <= i < L(2n+2), then a(i) = 4n+4.

%F This formula follows from Proposition 4.2. in "On the representation of the natural numbers by powers of the golden mean".

%F For example if n=1: L(2)=3, L(3)=4, L(4)=7, so a(3) = a(4) = 5, and a(5) = a(6) = 8.

%F Let (v(n)) = 1,4,5,8,9,12,... be the sequence of values taken by (a(n)). Then it follows directly from the Lucas formula for (a(n)) that v(n) = A042948(n) (where A042948 has been given offset 1, as it should; see also the comment by _Jianing Song_ in A042948).

%F (End)

%t nn = 100; len = 2*Ceiling[Log[GoldenRatio, nn]]; Table[d = RealDigits[n, GoldenRatio, len]; last1 = Position[d[[1]], 1][[-1, 1]]; last1, {n, 1, nn}]

%Y Cf. A130600, A055778, A133775, A042948.

%K nonn,base

%O 1,2

%A _T. D. Noe_, May 20 2011