A248808 - OEIS

%I #55 Dec 23 2024 14:53:44

%S 0,0,1,-2,2,-2,3,-1,-4,4,0,-4,5,1,-3,-6,6,2,-2,-6,7,3,-1,-5,-8,8,4,0,

%T -4,-8,9,5,1,-3,-7,-10,10,6,2,-2,-6,-10,11,7,3,-1,-5,-9,-12,12,8,4,0,

%U -4,-8,-12,13,9,5,1,-3,-7,-11,-14,14,10,6,2,-2,-6,-10,-14

%N Irregular triangle read by rows: row n gives golden ratio base representation of Fibonacci number F_n.

%C Each row can be viewed as either a sequence of powers of phi (the golden ratio) or as a sequence of indices of combinatorial Fibonacci numbers (Fibonacci numbers offset so that f(0)=1,f(1)=1) (see Benjamin & Quinn).

%C From _Wolfdieter Lang_, Oct 16 2014: (Start)

%C With F(n) = A000045(n) (extended to negative arguments by F(-n) = (-1)^(n+1)*F(n)) one has (proof by induction, using the F recurrence and phi^n = phi^(n-1) + phi^(n-2))

%C   F(2*k) = sum(phi^(2*(n-1) - 4*m), m = 0..k-1), k >= 1 and F(2*k+1) = sum(phi^(2*n-1 - 4*m), m = 0..k-1) + phi^(-2*n),  k >= 0.

%C This shows, using phi^n = F(n-1) + F(n)*phi for integer n, the identity sum(F(2*(k-1) - 4*m, m=0..k-1) = 0 (and similar ones).

%C The row sums [0, 0, -1, 0, -2, 0, -3, 0, ...] (offset 1) have o.g.f. -x^3/(1-x^2)^2. The alternating row sums [0, 0, 3, 4, 0, 0, 7, 8, 0, 0, 11, 12, ...] have o.g.f. x^3*(3 - 2*x + x^2)/(1 - x + x^2 -x^3)^2.

%C (End)

%C This representation of the Fibonacci numbers can be found in the paper of Frougny and Sakarovitch. - _Michel Dekking_, Feb 10 2020

%C See also A105424. - _Michel Dekking_, Feb 10 2020

%D Arthur T. Benjamin and Jennifer J. Quinn, Proofs that Really Count, Dolciani Mathematical Expositions (MAA), (2003).

%H Michel Marcus, <a href="/A248808/b248808.txt">Rows n = 1..100 of triangle, flattened</a>

%H George Bergman, <a href="http://www.jstor.org/stable/3029218">A number system with an irrational base</a>, Math. Mag. 31 (1957), pp. 98-110.

%H Christiane Frougny and Jacques Sakarovitch, <a href="https://www.researchgate.net/publication/2596608_Automatic_conversion_from_Fibonacci_representation_to_representation_in_base_ph_and_a_generalization">Automatic conversion from Fibonacci representation to representation in base phi and a generalization</a>, (1998).

%H Dale Gerdemann, <a href="https://www.youtube.com/watch?v=1LtjGV-nLG0">Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle</a> video (2014).

%H Dale Gerdemann, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2014-October/013807.html">Fibonacci numbers in Golden Ratio Base</a>, Seqfan post, Oct 12 2014.

%F T(2*k, m) = 2*(k-1)-4*m, m = 0, ..., k-1 for k >= 1, and T(2*k+1, m) = 2*k-1-4*m, m = 0, ..., k-1, and T(2*k+1, k) = -2*k for k >= 0. See a comment above for the proof. - _Wolfdieter Lang_, Oct 16 2014

%e There are two rows of length one, two rows of length 2, etc. For example, the 6th row, 4,0,-4, is the sequence of powers of phi ((1+sqrt(5))/2) used to represent the number 8.

%e [0] 1

%e [0] 1

%e [1, -2] 2

%e [2, -2] 3

%e [3, -1, -4] 5

%e [4,  0, -4] 8

%e [5,  1, -3, -6] 13

%e [6,  2, -2, -6] 21

%e [7,  3, -1, -5,  -8] 34

%e [8,  4,  0, -4,  -8] 55

%e [9,  5,  1, -3,  -7, -10] 89

%e [10, 6,  2, -2,  -6, -10] 144

%e [11, 7,  3, -1,  -5,  -9,  -12] 233

%e [12, 8,  4,  0,  -4,  -8,  -12] 377

%t T[n_, m_] := If[n == 2*m+1, -2*m, n-4*m-2];

%t Table[T[n, m], {n, 1, 15}, {m, 0, Floor[(n - 1)/2]}] // Flatten (* _Jean-François Alcover_, Jul 13 2016, after _Wolfdieter Lang_ *)

%o (PARI) T(n, m) = if (n == 2*m+1, -2*m, n-4*m-2);

%o tabf(nn) = for (n=1, nn, for (k=0, (n-1)\2, print1(T(n,k), ", "));); \\ _Michel Marcus_, May 28 2019

%Y Cf. A000045.

%K sign,tabf,nice

%O 1,4

%A _Dale Gerdemann_, Oct 14 2014

%E Rows 12 to 14 added by _Wolfdieter Lang_, Oct 23 2014

%E More terms from _Michel Marcus_, May 28 2019