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.

-1, 1, 1, 1, -4, -6, 30, -240, 1260, 35700, 1256640, -199020360, -202839655480, 1124753048683264, -181696576073176468304, 1026657060420588391021488976, 919637161132261232937508950440493056, 839430326120023909391651548323223480699275649536

Offset

0,5

Comments

u(144) has denominator 2.

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

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

Links

Table of n, a(n) for n=0..17.

Formula

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

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

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.

Prog

(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))};

Crossrefs

Cf. A023434, A060006, A248049.

Sequence in context: A090530 A154667 A093121 * A209298 A075590 A088255

Adjacent sequences: A248045 A248046 A248047 * A248049 A248050 A248051

Keyword

sign,frac

Author

Michael Somos, Sep 30 2014

Status

approved