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A159990
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Digits of sexagesimal fraction to approximate the positive root of x^3+2*x^2+10*x=20, first solved by Leonardo of Pisa.
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5
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| A159992(n)/A159993(n) = Sum(a(k)/60^k:0<=k<=n) = Sum(a(k)*60^(n-k):0<=k<=n)/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)).
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REFERENCES
| Stanislaw Glushkov, "On approximation methods of Leonardo Fibonacci", Historia Mathematica 3 (1976), pp. 291-296.
A. F. Horadam, Eight hundred years young, The Australian Mathematics Teacher 31 (1975) 123-134.
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LINKS
| Ezra Brown and Jason C. Brunson, Fibonacci's forgotten number
J. J. O'Connor and E. F. Robertson, Fibonacci
Clark Kimberling, Fibonacci [containing the Horadam article]
Wikipedia, Sexagesimal
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EXAMPLE
| Leonardo's approximation 1;22.7.42.33.4.40, 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 found 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 only
calculate 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'.
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CROSSREFS
| Cf. A159991, A202300.
Sequence in context: A040467 A194708 A069285 * A040466 A133682 A134911
Adjacent sequences: A159987 A159988 A159989 * A159991 A159992 A159993
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KEYWORD
| nonn,base
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AUTHOR
| Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 01 2009
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