Oops, let's stick with: Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6), for what n in 1000...1999 is .5 <= f(n) < .55? There should only be four or five answers. Dan Hoey
At 03:36 PM 2/11/2003, Dan Hoey wrote:
Oops, let's stick with:
Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6), for what n in 1000...1999 is .5 <= f(n) < .55?
There should only be four or five answers.
I get 1091 0.527777777777778 - 19/36 1210 0.537037037037037 - 29/54 1730 0.500000000000000 - 1/2 1748 0.509259259259259 - 55/108 1947 0.509259259259259 - 55/108
For what it's worth, a graph of the first million terms of Ulam[n]- 13.5 n can be seen at http://www.wolframscience.com/preview/nks_pages/?NKS0908.gif The method for coming up with 13.5 was to divide the millionth term by a million. As can be seen, the Ulam sequence doesn't have a predictable pattern, so far. What is the 5 millionth term? I am willion to predict that Ulam[5000000]/5000000 will not equal 13.5 or 21.6. It'll be some other number. --Ed Pegg Jr.
----- Original Message ----- From: "Jud McCranie" <judmccr@bellsouth.net> To: "Dan Hoey" <Hoey@aic.nrl.navy.mil> Cc: <math-fun@mailman.xmission.com> Sent: Tuesday, February 11, 2003 4:21 PM Subject: Re: [math-fun] Ulam(1,2)
At 03:36 PM 2/11/2003, Dan Hoey wrote:
Oops, let's stick with:
Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6), for what n in 1000...1999 is .5 <= f(n) < .55?
There should only be four or five answers.
I get
1091 0.527777777777778 - 19/36 1210 0.537037037037037 - 29/54 1730 0.500000000000000 - 1/2 1748 0.509259259259259 - 55/108 1947 0.509259259259259 - 55/108
For n = 1730, u(n) = 21870, my f(n) fell a hair below .5, it should have been .5 exactly. This put 1730 into the wrong bucket, and that probably accounts for the discrepancy between my bucket counts and others'.
If anyone is interested, I have a list of the Ulam(1,2) numbers <= 5,000,000 and I expect to have one for <= 10,000,000 tomorrow. If you want me to send you a copy, let me know. The former is over 3MB, storing the numbers in an ASCII text file. It is 900KB zipped, or 1.5MB stored as 32-bit (binary) integers.
At 08:29 PM 2/11/03 -0500, you wrote:
If anyone is interested, I have a list of the Ulam(1,2) numbers <= 5,000,000 and I expect to have one for <= 10,000,000 tomorrow. If you want me to send you a copy, let me know. The former is over 3MB, storing the numbers in an ASCII text file. It is 900KB zipped, or 1.5MB stored as 32-bit (binary) integers.
I bet that if you encode the list with first-order differences, the file size shrinks dramatically. Especially if you then zip, because so many of the differences recur so frequently, the zip compressor will have a lot of repetition to compress. -A
I bet that if you encode the list with first-order differences, the file size shrinks dramatically. Especially if you then zip, because so many of the differences recur so frequently, the zip compressor will have a lot of repetition to compress.
Yes, if anyone needs an especially small file sent, I can do the differences.
----- Original Message ----- From: "Jud McCranie" <judmccr@bellsouth.net> To: <math-fun@mailman.xmission.com> Sent: Tuesday, February 11, 2003 11:06 PM Subject: Re: [math-fun] Ulam(1,2)
For n = 1730, u(n) = 21870, my f(n) fell a hair below .5, it should have been .5 exactly.
If you divided by 21.6, that could be the cause.
Yeah, that was it. When I change to $n * 108 / 5, it worked. I'm usually not that careless. Also, Perl uses decimal strings to represent numbers, so there is probably some lost precision there too.
participants (5)
-
Allan C. Wechsler -
Dan Hoey -
David Wilson -
Ed Pegg Jr -
Jud McCranie