A081459 -id:A081459 - cp's OEIS frontend

A081460 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with a = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the denominators.

1, 4, 72, 23184, 2403763488, 25840354427429161536, 2986152136938872067784669198846010266752, 39878504028822311675150039382403961856254569551519724209276629577579916539865344

Offset: 1

Amarnath Murthy, Mar 22 2003

nonn

Related sequence pairs (numerator, denominator) can be obtained by choosing N = 2, 3, 6, etc.

The sequence satisfies the Pell equation A081459(n+1)^2 - 5*a(n+1)^2 = 1. - Vincenzo Librandi, Dec 20 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..11

Cf. A000032, A000045, A000129, A001333, A007283, A079613, A081459.

m:=8; f:=[ n eq 1 select 2 else (Self(n-1)+5/Self(n-1))/2: n in [1..m] ]; [ Denominator(f[n]): n in [1..m] ]; // Bruno Berselli, Dec 20 2011

Table[Fibonacci[2^(n - 1)*3], {n, 1, 8}]/2 (* Amiram Eldar, Apr 07 2023 *)

{r=2; N=5; for(n=1,8,a=numerator(r); b=denominator(r); print1(b,","); r=(1/2)*(r + N/r))}

a(n) = 2*a(n-1)*A081459(n-1). - Mario Catalani (mario.catalani(AT)unito.it), May 21 2003
a(n) = A000045(A007283(n-1))/2. - Ehren Metcalfe, Oct 07 2017
From Amiram Eldar, Apr 07 2023: (Start)
a(n) = A079613(n-1)/2.
a(n) = Product_{k=1..n-1} L(3*2^(n-1-k)), where L(k) is the k-th Lucas number (A000032). (End)
a(n) = A001076(2^(n-1)). - Robert FERREOL, Apr 18 2023

Edited and extended by Klaus Brockhaus and Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

a(8) corrected by Vincenzo Librandi, Dec 20 2011

A219511 Pierce expansion of 144 - 64*sqrt(5).

Original entry on oeis.org

1, 9, 40, 161, 648, 51841, 207368, 5374978561, 21499914248, 57780789062419261441, 231123156249677045768, 6677239169351578707225356193679818792961, 26708956677406314828901424774719275171848

Offset: 0

Peter Bala, Nov 23 2012

Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. This is the case p = 9. For other cases see A219508 (p = 3), A219509 (p = 5) and A219510 (p = 7).

G. C. Greubel, Table of n, a(n) for n = 0..20
J. Paradis, P. Viader, L. Bibiloni Approximation to quadratic irrationals and their Pierce expansions, The Fibonacci Quarterly, Vol.36 No. 2 (1998) 146-153.
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A081459, A219508 (p = 3), A219509 (p = 5), A219510 (p = 7).

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[144 - 64*Sqrt[5] , 7!], 10] (* G. C. Greubel, Nov 15 2016 *)

r=(9 + 4*sqrt(5))/16; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ G. C. Greubel, Nov 15 2016

a(2*n) = 2*{(2 + sqrt(5))^(2^n) + (2 - sqrt(5))^(2^n) + 2} for n >= 1.
a(2*n-1) = 1/2*{(2 + sqrt(5))^(2^n) + (2 - sqrt(5))^(2^n)} for n >= 1.
Recurrence equations: a(0) = 1, a(1) = 9 and for n >= 1, a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*(a(2*n-1))^2 - 1.
144 - 64*sqrt(5) = 1 - 1/9 + 1/(9*40) - 1/(9*40*161) + 1/(9*40*161*648) - ....
a(2*n) = 8*A081459(n)^2 for n >= 2.
a(2*n+1) = A081459(n+2) for n >= 0.

A244012 Numerators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

2, 11, 233, 108497, 23543191457, 1108563727961872518977, 2457827077905448997994482872789298261401217, 12081827889770476116093110581355561229584727594431650162181251776430351279198649072897

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...

R. Parimala, A Hasse principle for quadratic forms over function fields, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 3, 447--461. MR3196794.

Cf. A244013 (denominators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=7;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244013 Denominators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

1, 4, 88, 41008, 8898489952, 418997705236253480128, 928971316248341903257187589777603944778112, 4566501711345281867283814391125123371716411674583075407993026856131137508750543524608

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...

R. Parimala, A Hasse principle for quadratic forms over function fields, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 3, 447--461. MR3196794.

Cf. A244012 (numerators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=7;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244014 Numerators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

2, 5, 49, 4801, 46099201, 4250272665676801, 36129635465198759610694779187201, 2610701117696295981568349760414651575095962187244375364404428801

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...

Cf. A244015, A084765.

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=6;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244015 Denominators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

1, 2, 20, 1960, 18819920, 1735166549767840, 14749861913749949808286047759680, 1065814268211609269094400465471990022332221793124358274759711360

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...

Cf. A244014 (numerators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

m:=9; f:=[n eq 1 select 2 else (Self(n-1)+6/Self(n-1))/2: n in [1..m]]; [Denominator(f[n]): n in [1..m]]; // Vincenzo Librandi, Jan 12 2016

N:=6;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A083696 a(n) = Sum_{r=0..2^(n-1)} (5^r/(2r)!)*Product_{k=0..2r-1} (2^n - k).

Original entry on oeis.org

1, 6, 56, 6016, 72318976, 10460064284409856, 218825889667954898996994670329856, 95769539977943941232017762100658986141884645207653888255921750016

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), May 22 2003

Similar to A081459: a(n) is the numerator of the same mapping f(r) = (1/2)*(r + 5/r) but with initial value r=1.

G. C. Greubel, Table of n, a(n) for n = 0..10

Cf. A000032, A083697, A081459.

[2^(2^n -1)*Lucas(2^n): n in [0..8]]; // G. C. Greubel, Jan 14 2022

Table[Sum[Product[2^n - k, {k, 0, 2*r - 1}]5^r/(2*r)!, {r, 0, 2^(n - 1)}], {n, 0, 8}]
Table[2^(2^n - 1)*LucasL[2^n], {n, 0, 8}] (* Vaclav Kotesovec, Jan 08 2021 *)

[2^(2^n -1)*lucas_number2(2^n, 1, -1) for n in (0..8)] # G. C. Greubel, Jan 14 2022

a(n)/A083697(n) converges to sqrt(5).
a(n) = a(n-1)^2 + 5*A083697(n-1)^2.
a(n) = 2^(2^n - 1) * Lucas(2^n). - Vaclav Kotesovec, Jan 08 2021

A228933 Optimal ascending continued fraction expansion of phi-1=1/phi=(sqrt(5)-1)/2 .

Original entry on oeis.org

2, 4, -18, 322, 103682, 10749957122, 115561578124838522882, 13354478338703157414450712387359637585922, 178342091698891843163466683840822101223162205277179656650156983624835803932590082

Offset: 1

Giovanni Artico, Sep 10 2013

See A228929 for the definition of "optimal ascending continued fraction".

Conjecture: The golden ratio (phi) expansion exhibits from the fourth term the recurrence relation a(n) = a(n-1)^2 - 2 described in A228931.

			phi = 1+1/2*(1+1/4*(1-1/18*(1+1/322*(1+1/103682*(1+1/10749957122*(1+...))))))

Cf. A001622, A081459, A094214, A228929, A228931, A228932.

ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 8 terms of the expansion of 1/phi
ArticoExp((sqrt(5)-1)/2,8)

Flatten[{2, 4, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[3] == -18}, a, {n, 3, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)

a(n) = a(n-1)^2 - 2 for n>3.
For n>3, a(n) = (sqrt(5)+2)^(2^(n-2)) + (sqrt(5)-2)^(2^(n-2)). - Vaclav Kotesovec, Sep 20 2013
a(n) = 2*A081459(n-1) for n>3. - Amiram Eldar, Apr 07 2023

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

cp's OEIS Frontend

A081460 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with a = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the denominators.

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A219511 Pierce expansion of 144 - 64*sqrt(5).

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A244012 Numerators of rational approximations to sqrt(7) obtained from Newton's method.

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Maple

A244013 Denominators of rational approximations to sqrt(7) obtained from Newton's method.

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Maple

A244014 Numerators of rational approximations to sqrt(6) obtained from Newton's method.

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Maple

A244015 Denominators of rational approximations to sqrt(6) obtained from Newton's method.

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Magma

Maple

A083696 a(n) = Sum_{r=0..2^(n-1)} (5^r/(2r)!)*Product_{k=0..2r-1} (2^n - k).

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A228933 Optimal ascending continued fraction expansion of phi-1=1/phi=(sqrt(5)-1)/2 .

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