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%I #16 Apr 11 2022 14:52:45
%S 1,3,33,3927,55602393,11147016454528647,
%T 448011292165037607943004375755833,
%U 723685043824607606355691108666081531638582859833105061571146291527
%N a(n) is the denominator of the Heron sequence with h(0) = 3.
%C The numerators of the Heron sequence are in A319749.
%C 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).
%C n even: a(n) = A041019((5*2^n-5)/3).
%C n odd: a(n) = A041019((5*2^n-1)/3).
%C 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.
%C sqrt(c(n)) has the continued fraction [2*n+1; n, 1, 1, n, 4*n+2].
%C hn(n)^2 - c(n)*hd(n)^2 = 4 for n > 1.
%F h(n) = hn(n)/hd(n), hn(0) = 3, hd(0) = 1.
%F hn(n+1) = (hn(n)^2 + 13*hd(n)^2)/2.
%F hd(n+1) = hn(n)*hd(n).
%F A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
%F A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
%F 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
%e A078370(2) = 29.
%e hd(0) = A041047(0) = 1, hd(1) = A041047(3) = 5,
%e hd(2) = A041047(5) = 135, hd(3) = A041047(13) = 38145.
%p hn[0]:=3: hd[0]:=1:
%p for n from 1 to 6 do
%p hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
%p hd[n]:=hn[n-1]*hd[n-1]:
%p printf("%5d%40d%40d\n", n, hn[n], hd[n]):
%p end do:
%o (Python)
%o def aupton(nn):
%o hn, hd, alst = 3, 1, [1]
%o for n in range(nn):
%o hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
%o alst.append(hd)
%o return alst
%o print(aupton(7)) # _Michael S. Branicky_, Mar 15 2022
%Y Cf. A041018, A041019, A078370, A010122, A041047, A319749.
%K nonn,frac,easy
%O 0,2
%A _Paul Weisenhorn_, Sep 27 2018
%E a(5) corrected and terms a(6) and a(7) added by _Peter Bala_, Mar 15 2022
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