A319749 -id:A319749 - cp's OEIS frontend

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

cp's OEIS Frontend

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

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