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

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

Offset: 0

Author

Paul Weisenhorn, Sep 27 2018

Keywords

• nonn
• frac
• easy

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)

Examples

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

Links

• 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 + ...

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.

Programs

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

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

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)

Extensions

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