Asymptote of A003242
hv at crypt.org
hv at crypt.org
Tue Aug 5 04:08:32 CEST 2008
Maple gives
1.7502412917183090312497386246398158787782058181381590561316586131751935167152
No chance of a closed form, I think.
Brendan.
* hv at crypt.org <hv at crypt.org> [080805 13:11]:
> By my calculation, A003242(n) should tend to k^n, where k is given by:
> 1 = sum_i=1^inf{ 1/(k^i + 1) }
>
> By numerical methods I find k ~= 1.750241291718309, can anyone suggest
> how to solve that equation to get a closed form?
>
> Actual values seem to support this; here are some selected cases:
>
> n: a(n)^(1/n)
> ====================
> 1: 1.0
> 2: 1.0
> 4: 1.41421356237
> 8: 1.58082238074
> 16: 1.66658717253
> 32: 1.70785645051
> 64: 1.7289190436
> 128: 1.73954749867
> 256: 1.7448862029
> 512: 1.7475616961
> 1024: 1.74890098071
> 2048: 1.74957100787
> 4096: 1.7499061177
> 8192: 1.75007369668
> 16384: 1.7501574922
>
> Hugo
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