A159990 Coefficients in sexagesimal expansion of the positive root of x^3 + 2*x^2 + 10*x = 20, first studied by Leonardo of Pisa (Fibonacci) in 1225.

1, 22, 7, 42, 33, 4, 38, 30, 50, 15, 43, 13, 56, 48, 24, 41, 0, 48, 22, 40, 39, 37, 23, 53, 55, 57, 45, 40, 5, 46, 50, 57, 28, 45, 46, 34, 2, 6, 7, 15, 25, 25, 13, 10, 59, 30, 13, 14, 7, 6, 15, 46, 23, 53, 59, 32, 24, 20, 11, 48, 35, 4, 4, 18, 33, 50, 7, 40, 16, 16, 1, 32, 24, 10, 43, 59, 23, 44, 51, 58, 11, 22, 26, 17

Offset

0,2

Comments

Leonardo of Pisa (Fibonacci) found a(0), ..., a(5) but gave a(6) as 40.

A159992(n)/A159993(n) = Sum_{k=0..n} a(k)/60^k = (Sum_{k=0..n} a(k)*60^(n-k))/60^n; let f(x) = x^3 + 2*x^2 + 10*x - 20, then for n > 0:

a(n) = Max(k: f(A159992(n-1)/A159993(n - 1) + k/60^n)) < 0),

a(n) + 1 = Min(k: f(A159992(n - 1)/A159993(n - 1) + k/60^n)) > 0);

A159994(n)/A159995(n) = f(A159992(n)/A159993(n)).

References

Cox, David A., Galois theory. Pure and Applied Mathematics (New York). Wiley-Interscience [John Wiley & Sons], Hoboken, NJ, 2004. xx+559 pp. ISBN: 0-471-43419-1 MR2119052 (2006a:12001). See page 9.

A. F. Horadam, Eight hundred years young, The Australian Mathematics Teacher 31 (1975) 123-134.

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 21.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..5000

Ezra Brown and Jason C. Brunson, Fibonacci's forgotten number

Stanislaw Glushkov, On approximation methods of Leonardo Fibonacci, Historia Mathematica 3 (1976), pp. 291-296.

J. J. O'Connor and E. F. Robertson, Fibonacci

Clark Kimberling, Fibonacci [containing the Horadam article]

Trond Steihaug, Fibonacci and digit-by-digit computation; An example of reverse engineering in computational mathematics, arXiv:2211.00504 [math.HO], 2022.

Wikipedia, Sexagesimal

Index entries for algebraic numbers, degree 3

Example

The root is 1 + 22/60 + 7/60^2 + 42/60^3 + 33/60^4 + 4/60^5 + 38/60^6 + 30/60^7 + 50/60^8 + ...
Leonardo's approximation 1;22.7.42.33.4.40 is to be read as 1 + 22/60 + 7/60^2 + 42/60^3 + 33/60^4 + 4/60^5 + 40/60^6 = A159992(5)/A159993(5) + 40/60^6 = 1596577777 / 1166400000 ~= 1.3688081078532235 and f(1596577777/1166400000) ~= +6.7193226361369/10^10; compare this to A159992(6)/A159993(6) = A159992(5)/A159993(5) + 38/60^6 = 31931555539 / 23328000000 ~= 1.3688081078103566 and f(31931555539/23328000000) ~= -2.3239469709985/10^10.
Assuming that Leonardo did similar calculations, the question may arise: why he didn't find a(6) = 38 instead of 40? Supposedly he just avoided the effort to calculate f(A159992(5)/A159993(5) + k/60^6) for k = 37, 38, or 39: 37/60^6 = 37/46656000000, 38/60^6 = 19/23328000000, or 39/60^6 = 13/15552000000; finally, he did calculate only f(A159992(5)/A159993(5) + k/60^6) for k = 36 and k = 40, the less complex cases concerning sexagesimal fractional arithmetic with 36/60^6 = 1/1296000000 and 40/60^6 = 1/1166400000: f(A159992(5)/A159993(5) + 36/60^6) ~= -1.9999999988632783, f(A159992(5)/A159993(5) + 40/60^6) ~= +0.0000000006719323.
The latter result looks precise enough and could explain and justify Leonardo's 'rounding'.

Mathematica

RealDigits[ Solve[x^3 + 2 x^2 + 10 x - 20 == 0, x][[3, 1, 2]], 60, 111][[1]] (* Robert G. Wilson v, May 11 2012 *)

Prog

(PARI) polrootsreal(x^3+2*x^2+10*x-20)[1] \\ Charles R Greathouse IV, Apr 14 2014

Crossrefs

Cf. A159991, A202300.

Sequence in context: A040467 A194708 A069285 * A243629 A040466 A133682

Adjacent sequences: A159987 A159988 A159989 * A159991 A159992 A159993

Keyword

nonn,base,cons

Author

Reinhard Zumkeller, May 01 2009

Status

approved