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

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

Offset

0,1

Comments

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)

Links

Table of n, a(n) for n=0..7.

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 + ...

Formula

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)

Example

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

Maple

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

Prog

(Python)
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

Crossrefs

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.

Sequence in context: A068693 A036930 A198085 * A209107 A015047 A339326

Adjacent sequences: A319746 A319747 A319748 * A319750 A319751 A319752

Keyword

nonn,frac,easy

Author

Paul Weisenhorn, Sep 27 2018

Extensions

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

Status

approved