

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.


8



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
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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^(nk))/60^n; let f(x) = x^3 + 2*x^2 + 10*x  20, then for n > 0:
a(n) = Max(k: f(A159992(n1)/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). WileyInterscience [John Wiley & Sons], Hoboken, NJ, 2004. xx+559 pp. ISBN: 0471434191 MR2119052 (2006a:12001). See page 9.
A. F. Horadam, Eight hundred years young, The Australian Mathematics Teacher 31 (1975) 123134.
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. 291296.
J. J. O'Connor and E. F. Robertson, Fibonacci
Clark Kimberling, Fibonacci [containing the Horadam article]
Wikipedia, Sexagesimal


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*x20)[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



