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%I #77 Aug 04 2021 02:24:57
%S 2,3,19,64,538,2833,29169,210308,2572158,23595915,334778571,
%T 3732092084,60305234822,791741083537,14359827157009,217037153818264,
%U 4366918714540522,74685204276602819,1651116684587556019,31524723785455714840,759659139498065625218,16017463672140861567617
%N Numerators of the n-th iteration of the alternating continued fraction formed from the positive integers, starting with (1 - ...).
%C As n->inf, a(n)/A263295(n) converges to 0.57663338973... (A346590); this number has a surprisingly elegant standard continued fraction representation of [0; 1, 1, 2, 1, 3, 4, 1, 5, 6, 1, 7, 8, ...].
%C From _Robert Israel_, Dec 22 2015: (Start)
%C a(n) is the numerator of b(n)/c(n) where
%C b(1) = 2, b(2) = 3, c(1) = 3, c(2) = 5,
%C b(n+1) = ((-1)^n*(n-1)+n*(n+2))*b(n) - (1+(-1)^n*(n+1))*b(n-1))/(n-(-1)^n),
%C c(n+1) = ((-1)^n*(n-1)+n*(n+2))*c(n) - (1+(-1)^n*(n+1))*c(n-1))/(n-(-1)^n).
%C Conjecture: b(n) and c(n) are coprime for all n, so that a(n) = b(n).
%C I have verified this for n <= 10000. (End)
%H Robert Israel, <a href="/A262957/b262957.txt">Table of n, a(n) for n = 1..448</a>
%H Peter Bala, <a href="/A262957/a262957.pdf">A note on A262957 and A263295</a>
%e (1-1/(2+1)) = 2/3, so a(1) = 2;
%e (1-1/(2+1/(3-1))) = 3/5, so a(2) = 3;
%e (1-1/(2+1/(3-1/(4+1)))) = 19/33, so a(3) = 19;
%e (1-1/(2+1/(3-1/(4+1/(5-1))))) = 64/111, so a(4) = 64.
%p P[1]:= 2: P[2]:= 3:
%p Q[1]:= 3; Q[2]:= 5;
%p for i from 2 to 100 do
%p P[i+1]:= ((-1)^i*(i-1) + i^2 + 2*i)/(i-(-1)^i)*P[i] + (1 + (i+1)*(-1)^i)/((-1)^i-i)*P[i-1];
%p Q[i+1]:= ((-1)^i*(i-1) + i^2 + 2*i)/(i-(-1)^i)*Q[i] + (1 + (i+1)*(-1)^i)/((-1)^i-i)*Q[i-1];
%p od:
%p seq(numer(P[i]/Q[i]),i=1..100); # _Robert Israel_, Dec 22 2015
%o (PARI) a(n) = if(n%2==0, s=-1, s=1); t=1; while(n>-1, t=n+1+s/t; n--; s=-s); denominator(t=1/t)
%o vector(30, n, a(n)) \\ _Mohamed Sabba_, Dec 22 2015
%Y Same principle as A244279 and A244280 - except here we begin with subtraction rather than addition.
%Y Cf. A263295 (denominators), A346590.
%K nonn,frac
%O 1,1
%A _Mohamed Sabba_, Nov 19 2015
%E More terms from _Mohamed Sabba_, Dec 22 2015
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