cp's OEIS Frontend

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

%I A319749 #26 Apr 11 2022 14:54:19
%S A319749 3,11,119,14159,200477279,40191139395243839,
%T A319749 1615327685887921300502934267457919,
%U A319749 2609283532796026943395592527806764363779539144932833602430435810559
%N A319749 a(n) is the numerator of the Heron sequence with h(0)=3.
%C A319749 The denominator of the Heron sequence is in A319750.
%C A319749 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).
%C A319749 n even: a(n)=A041018((5*2^n-5)/3).
%C A319749 n  odd: a(n)=A041018((5*2^n-1)/3).
%C A319749 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.
%C A319749 sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
%C A319749 hn(n)^2-c(n)*hd(n)^2=4 for n>1.
%C A319749 From _Peter Bala_, Mar 29 2022: (Start)
%C A319749 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)
%H A319749 P. Liardet and P. Stambul, <a href="http://www.numdam.org/item?id=JTNB_2000__12_1_37_0">Séries d'Engel et fractions continuées</a>, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
%H A319749 Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%H A319749 Wikipedia, <a href="https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method">Methods of computing square roots</a>
%H A319749 <a href="/index/Aa#AHSL">Index entries for sequences of form a(n+1)=a(n)^2 + ...</a>
%F A319749 h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
%F A319749 hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
%F A319749 hd(n+1) = hn(n)*hd(n).
%F A319749 A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
%F A319749 A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
%F A319749 From _Peter Bala_, Mar 16 2022: (Start)
%F A319749 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.
%F A319749 a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
%F A319749 a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
%F A319749 a(n+1) = a(n)^2 - 2 for n >= 1.
%F A319749 a(n) = A057076(2^(n-1)) for n >= 1.
%F A319749 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.)
%F A319749 sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
%F A319749 sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)
%e A319749 A078370(2)=29.
%e A319749 hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
%e A319749 hn(3)=A041046(13)=528527.
%p A319749 hn[0]:=3:  hd[0]:=1:
%p A319749 for n from 1 to 6 do
%p A319749 hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
%p A319749 hd[n]:=hn[n-1]*hd[n-1]:
%p A319749    printf("%5d%40d%40d\n", n, hn[n], hd[n]):
%p A319749 end do:
%p A319749 #alternative program
%p A319749 a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
%p A319749 seq(a(n), n = 0..7); # _Peter Bala_, Mar 16 2022
%o A319749 (Python)
%o A319749 def aupton(nn):
%o A319749     hn, hd, alst = 3, 1, [3]
%o A319749     for n in range(nn):
%o A319749         hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
%o A319749         alst.append(hn)
%o A319749     return alst
%o A319749 print(aupton(7)) # _Michael S. Branicky_, Mar 16 2022
%Y A319749 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).
%Y A319749 Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.
%K A319749 nonn,frac,easy
%O A319749 0,1
%A A319749 _Paul Weisenhorn_, Sep 27 2018
%E A319749 a(6) and a(7) added by _Peter Bala_, Mar 16 2022