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July 22

The Fibonacci-like sequence x(1)=x(2)=1, x(n+1)=2*agm(x(n), x(n-1))

Hi, as you know, the Fibonacci sequence has f(n+1)=f(n)+f(n-1) = 2*(the arithmetic mean of f(n) and f(n-1)), and lim as n-->+oo of the ratios f(n)/f(n-1) is M = Golden mean, and the lim as n-->-oo is -1/M. I've played with sequences like y(n+1) = 2*(harmonic mean [and similarly rms mean] of y(n) and y(n-1)), but I'm really wondering about the arithmetic-geometric mean(agm) variation: what are the numbers that the ratios x(n)/x(n-1) approaches as n-->+-oo, where {x(n)} is defined in the title above? Are the ratios well known in another context, like perhaps pi/2? Numerical approximations would also be helpful. Thanks. 64.134.223.214 (talk) 00:55, 22 July 2017 (UTC)[reply]

Writing a quick one-liner (almost) in Haskell, the ratios of consecutive terms of your sequence seem to be converging to around 1.60207620749888.... Checking the OEIS for this number gives no hits, so no idea if it's expressible in terms of known constants. --Deacon Vorbis (talk) 04:07, 22 July 2017 (UTC)[reply]

Thanks. Does your program tell what the ratio x(n)/x(n-1) looks like for large negative n?64.134.223.214 (talk) 04:14, 22 July 2017 (UTC)[reply]

Never mind. I'm such an idiot!64.134.223.214 (talk) 03:17, 23 July 2017 (UTC)[reply]

It is not too difficult to find an equation satisfied by the limit ratio. I think the following gives 100 digits of your constant using Mathematica (assuming I got the argument of the elliptic function right):

In[30]:= FindRoot[(1+x)/x^2 == 2/Pi*EllipticK[(1-x)^2/(1+x)^2],{x,1.6},WorkingPrecision->100]
Out[30]= {x->1.602076207498882854166181178384113469249890744070779427538587439268519411756346198948331083813417404}

No idea whether this number is related to anything else. —Kusma (t·c) 10:01, 24 July 2017 (UTC)[reply]

[edit conflict] This looks like a job for isc rather than OEIS. But I didn't find a match there either.

FWIW, I've put more digits of the ratio at https://pastebin.com/qSkERhCF, it should be correct to the sf specified. -- Meni Rosenfeld (talk) 10:23, 24 July 2017 (UTC)[reply]

I guess the process should converge for any "reasonable" mean and give a limit ratio between

{\sqrt {2}}

and 2. —Kusma (t·c) 11:23, 24 July 2017 (UTC)[reply]

dang i had an idea this morning that it would be the agm of the limit for x(n+1)=2(x(n)*x(n-1))^.5(which is the series based on the geometric mean), which is 4^(1/3), and the limit for Fibonacci, which is M. But agm(4^1/3, M)~1.6026809, not 1.602076.Not sure why this didn't work.144.35.45.45 (talk) 18:51, 26 July 2017 (UTC)[reply]

I had hoped homogeneity would allow the agm and the limits to commute)144.35.45.63 (talk) 21:09, 26 July 2017 (UTC)[reply]