A041018 -id:A041018 - cp's OEIS frontend

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

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

A319750 a(n) is the denominator of the Heron sequence with h(0) = 3.

Original entry on oeis.org

1, 3, 33, 3927, 55602393, 11147016454528647, 448011292165037607943004375755833, 723685043824607606355691108666081531638582859833105061571146291527

Offset: 0

Paul Weisenhorn, Sep 27 2018

The numerators of the Heron sequence are in A319749.
There is the following relationship between the denominator of the Heron sequence and the denominator of the continued fraction A041018(n)/ A041019(n) convergent to sqrt(13).
n even: a(n) = A041019((5*2^n-5)/3).
n  odd: a(n) = A041019((5*2^n-1)/3).
General: all numbers c(n) = A078370(n) = (2*n+1)^2 + 4 have the same relationship between the denominator of the Heron sequence and the denominator of the continued fraction convergent to 2*n+1.
sqrt(c(n)) has the continued fraction [2*n+1; n, 1, 1, n, 4*n+2].
hn(n)^2 - c(n)*hd(n)^2 = 4 for n > 1.

			A078370(2) = 29.
hd(0) = A041047(0) = 1, hd(1) = A041047(3) = 5,
hd(2) = A041047(5) = 135, hd(3) = A041047(13) = 38145.

Cf. A041018, A041019, A078370, A010122, A041047, A319749.

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:

def aupton(nn):
    hn, hd, alst = 3, 1, [1]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hd)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 15 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).
a(0) = 1, a(1) = 3 and a(n) = 2*T(2^(n-2), 11/2)*a(n-1) for n >= 2, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 16 2022

a(5) corrected and terms a(6) and a(7) added by Peter Bala, Mar 15 2022

A341862 a(n) is the even term in the linear recurrence signature for numerators and denominators of continued fraction convergents to sqrt(n), or 0 if n is a square.

Original entry on oeis.org

0, 0, 2, 4, 0, 4, 10, 16, 6, 0, 6, 20, 14, 36, 30, 8, 0, 8, 34, 340, 18, 110, 394, 48, 10, 0, 10, 52, 254, 140, 22, 3040, 34, 46, 70, 12, 0, 12, 74, 50, 38, 64, 26, 6964, 398, 322, 48670, 96, 14, 0, 14, 100, 1298, 364, 970, 178, 30, 302, 198, 1060, 62, 59436

Offset: 0

Georg Fischer, Feb 22 2021

nonn

The Everest et al. link states that "the continued fraction expansion of a quadratic irrational is eventually periodic, which implies that the numerators px and denominators qx of its convergents satisfy linear recurrence relations".
Let k be the period length minus one of the continued fraction of sqrt(n). Then the linear recurrence signatures with constant coefficients have the form (0, 0, ..., 0, a(n), 0, 0, ..., 0, (-1)^(n+1)), with k zeroes before and behind a(n).
a(n) is twice the numerator of the convergent to sqrt(n) with index k (starting with 0).
These properties result from the mirrored structure of the period of such continued fractions.
The sequence has remarkably many terms in common with A180495 and with 2*A033313.

			The numerators for sqrt(13) begin with 3, 4, 7, 11, 18, 119, ... (A041018) and have the signature (0,0,0,0,36,0,0,0,0,1). The continued fraction has period [1,1,1,1,6], so k=4 and a(13) = 2*A041018(4) = 2*18 = 36. The signature ends with (-1)^4.
The numerators for sqrt(19) begin with 4, 9, 13, 48, 61, 170, 1421, ... (A041028) and have the signature (0,0,0,0,0,340,0,0,0,0,0,-1). The continued fraction has period [2,1,3,1,2,8], so k=5 and a(19) = 2*A041028(5) = 2*170 = 340. The signature ends with (-1)^5.

Jean-François Alcover, Table of n, a(n) for n = 0..10000
Graham Everest, Alf van der Poorten, Igor Shparlinski, and Thomas Ward, Recurrence Sequences, AMS Mathematical Surveys and Monographs, Volume 104 (2003) p. 8, 5th paragraph.

Cf. A006702, A033313, A180495.

a(n) = 2*A006702(n) if n is not square, otherwise 0.

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

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A319750 a(n) is the denominator of the Heron sequence with h(0) = 3.

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Author

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Comments

Examples

Crossrefs

Programs

Maple

Python

Formula

Extensions

A341862 a(n) is the even term in the linear recurrence signature for numerators and denominators of continued fraction convergents to sqrt(n), or 0 if n is a square.

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Keywords

Comments

Examples

Links

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Formula