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A319750
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a(n) is the denominator of the Heron sequence with h(0) = 3.
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1
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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Table of n, a(n) for n=0..7.
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FORMULA
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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
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EXAMPLE
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A078370(2) = 29.
hd(0) = A041047(0) = 1, hd(1) = A041047(3) = 5,
hd(2) = A041047(5) = 135, hd(3) = A041047(13) = 38145.
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MAPLE
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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:
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PROG
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(Python)
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
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CROSSREFS
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Cf. A041018, A041019, A078370, A010122, A041047, A319749.
Sequence in context: A194889 A126675 A038694 * A204687 A134477 A080985
Adjacent sequences: A319747 A319748 A319749 * A319751 A319752 A319753
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KEYWORD
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nonn,frac,easy
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AUTHOR
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Paul Weisenhorn, Sep 27 2018
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EXTENSIONS
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a(5) corrected and terms a(6) and a(7) added by Peter Bala, Mar 15 2022
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STATUS
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approved
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