0,2

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)).

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

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

Clark Kimberling, Fibonacci [containing the Horadam article]

Wikipedia, Sexagesimal

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. Supposably 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'.

A159991.

Sequence in context: A040468 A040467 A069285 this_sequence A040466 A133682 A134911

Adjacent sequences: A159987 A159988 A159989 this_sequence A159991 A159992 A159993

nonn

Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 01 2009