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 A248048 Numerator of u(n) where u(n) = (u(n-1) + u(n-2)) * (u(n-2) + u(n-3)) / u(n-4) with u(0) = -1, u(1) = u(2) = u(3) = 1. 0

%I #15 Feb 08 2021 23:02:35

%S -1,1,1,1,-4,-6,30,-240,1260,35700,1256640,-199020360,-202839655480,

%T 1124753048683264,-181696576073176468304,1026657060420588391021488976,

%U 919637161132261232937508950440493056,839430326120023909391651548323223480699275649536

%N Numerator of u(n) where u(n) = (u(n-1) + u(n-2)) * (u(n-2) + u(n-3)) / u(n-4) with u(0) = -1, u(1) = u(2) = u(3) = 1.

%C u(144) has denominator 2.

%C The sequence terms with u(0)=x and the same recursion is a Laurent polynomial in x. This sequence is what you get when you replace x with -1. Note that u(n) = 0 if -7 <= n <= -1. Usually u(n) for n <= -5 would not be defined using the backward recursion due to division by 0. - _Michael Somos_, Mar 04 2020

%C Again, the u(n) sequence with u(0)=x is given by u(n) = v(n)/(2^b(n-13) * x^b(n-3)) where b is A023434 and v(n) is a polynomial in x. For example, u(6) = (2*x^2 + 20*x + 48)/x^2. Also, A284049(n) has the same recursion as u(n) and initial values except A248049(0)=2. Note, log(abs(u(n))) ~ r^n where r = A060006 the plastic constant. - _Michael Somos_, Mar 04 2020

%F a(n) = a(-8-n) for all n in Z.

%F u(n) * u(n+4) = (u(n+1) + u(n+2)) * (u(n+2) + u(n+3)) for all n in Z.

%F a(7*n) < 0, a(7*n + 4) < 0, a(7*n + 5) < 0, a(7*n + 1) > 0, a(7*n + 2) > 0, a(7*n + 3) > 0, a(7*n + 6) > 0 for all n >= 0.

%o (PARI) {a(n) = if( n<-4, n=-8-n); if( n<0, 0, n<4, (-1)^(n==0), (a(n-1) + a(n-2)) * (a(n-2) + a(n-3)) / a(n-4))};

%Y Cf. A023434, A060006, A248049.

%K sign,frac

%O 0,5

%A _Michael Somos_, Sep 30 2014

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Last modified March 4 23:31 EST 2024. Contains 370537 sequences. (Running on oeis4.)