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

Author

Reinhard Zumkeller, May 01 2009

Keywords

• nonn
• base
• cons

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)).

Examples

			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'.
		

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

Crossrefs

Cf. A159991, A202300.

Programs

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 *)
RealDigits[Root[x^3+2x^2+10x-20,1],60,90][[1]] (* Harvey P. Dale, Jun 16 2025 *)

PARI

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