F(33) Fermat 33-th

[enzocreti]

is there any software available (for "normal" human beings) that could verify the 33-th Fermat number for primality at least in principle ?

[xilman]

Originally Posted by enzocreti

is there any software available (for "normal" human beings) that could verify the 33-th Fermat number for primality at least in principle ?

Yes, in principle, as long as you are prepared to wait for a few millennia for the answer.

[ixfd64]

Ernst's thread may be of interest: https://mersenneforum.org/showthread.php?t=18748

[retina]

Originally Posted by xilman

Yes, in principle, as long as you are prepared to wait for a few millennia for the answer.

That is hardly a normal person. Or maybe you know something you aren't revealing?

:link to a vid of Queen's "Who wants to live forever?":

[Mysticial]

Right now I estimate that it's possible to do a 2^33-bit convolution in about 250ms on a high-end computer. So that's 4 iter/sec.

2^33 iterations would be 2^31 seconds or about 63 years.

So not quite a few millennia, but not something we'd want to try now.

[axn]

Originally Posted by Mysticial

2^33 iterations would be 2^31 seconds or about 63 years.

Did you do this conversion in your head? Because I get 68 years (with a calculator, natch).

[LaurV]

We know (from the past, when we implemented timers and other things in electronic toys we produce here) that there are about 2^25 seconds in an year. So, without calculator, I would have said in a blink, 64 years (31-25=6, 2^6=64).
But the result is indeed 68 (checked axn with my super-accurate windows 7 calcuator, he is lucky this time, I mean axn, not the calculator, I didn't catch him with the wrong answer, but my time will come... hehe)

[Mysticial]

Originally Posted by axn

Did you do this conversion in your head? Because I get 68 years (with a calculator, natch).

No, but I got lazy and used 8 billion instead of 2^33.

[rudy235]

Originally Posted by Mysticial

No, but I got lazy and used 8 billion instead of 2^33.

An interesting mnemonic device for me is 1 year =π *10⁷ seconds.

Actual number is 3.1577*10⁷

[ATH]

Originally Posted by rudy235

An interesting mnemonic device for me is 1 year =π *10⁷ seconds.

Actual number is 3.1577*10⁷

Actually it is 3.1536*10⁷ or 3.16224*10⁷ for leap year or 3.15576*10⁷ for 4-year average.

[rudy235]

Originally Posted by rudy235

An interesting mnemonic device for me is 1 year =π *10⁷ seconds.

Actual number is 3.1577*10⁷

Originally Posted by ATH

Actually it is 3.1536*10⁷ or 3.16224*10⁷ for leap year or 3.15576*10⁷ for 4-year average.

1 Year = (365 + 97/400) days
1 day =86,400 seconds


(365+97/400)*86400 =3.1557 * 10⁷

=0.45% over π*10⁷