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Chuck
Daedalian Member

 Posted: Sun Apr 14, 2002 3:29 am    Post subject: 1 What are the next three numbers in this sequence? 2 9 28 73 152 1751 2881 ..... ..... ..... Don't bother. No one could reasonably be expected to solve this. Define function F(x) by applying the following procedure to x. Square x From this subtract the highest cube that won't leave a negative result. Repeat the highest cube subtraction on this new number. The result is the function F of x. Example for x = 25: Square 25 giving 625. Subtract the highest cube, 512, giving 113. Subtract the new highest cube, 64, giving 49. So F(25) = 49 For some values of x, F(x) = x. Example for x = 73: Square 73 giving 5329. Subtract the highest cube, 4913, giving 416. Subtract the new highest cube, 343, giving 73. So F(73) = 73. F(x) = x for these numbers: 2 9 28 73 152 1751 2881 3614 4031 18908 21609 24193 24472 27703 29178 30800 32617 53001 69338 81680 142849 151802 157914 183897 234955 313173 344961 517501 538232 694712 727208 852777 908568 951482 1492993 1534502 1807208 3394593 3673531 3748544 3803896 4069117 4150334 Are these all there are? I've check past 2,000,000,000 so far without another hit. Why did they stop after 4,150,334? Will they ever start again? I don't have the answer. I'm posting it because it seems peculiar.
CrystyB
Misunderstood Guy

 Posted: Sun Apr 14, 2002 5:00 pm    Post subject: 2 what number type did you use?
Da5id
Daedalian Member

 Posted: Sun Apr 14, 2002 5:15 pm    Post subject: 3 I assume you're not using a program that would introduce rounding errors?
Chuck
Daedalian Member

 Posted: Sun Apr 14, 2002 5:29 pm    Post subject: 4 I used integers in UBasic which uses 2600 digit numbers. To subtract the highest cube I took the cube root of the number and cubed the integer part, and then subtracted.
CrystyB
Misunderstood Guy

 Posted: Sun Apr 14, 2002 6:14 pm    Post subject: 5 i should have seen that coming. So it's not the data. Then it's in the numbers. Sketch of a proof: for n sufficiently big, n^2 and G(n)^3 are somewhat near to each other. Do the cube-substraction AGAIN and we end up with a number significantly lower than the original (could you please test if i'm wrong and there's any nb>2M for which F(n)>n ? )
Chuck
Daedalian Member

 Posted: Sun Apr 14, 2002 6:44 pm    Post subject: 6 For x > 13,919,148 F(x) seems to produce only lower numbers. I guess you're right. I'll let it run until I have to go home in a hour or so.
Chuck
Daedalian Member

 Posted: Sun Apr 14, 2002 10:58 pm    Post subject: 7 I'm past 52,000,000 with no more F(x) > x. That should be enough evidence to prove the theorem.
sodasipper
Guest

 Posted: Mon Apr 15, 2002 1:10 am    Post subject: 8 Maybe we should declare 4,150,334 as the "Chuck Constant"...
Da5id
Daedalian Member

 Posted: Mon Apr 15, 2002 3:05 am    Post subject: 9 I wonder if there is a "Chuck constant" for f(x) if you use cubes and quads. I couldn't go beyond 100000 with excel.
Chuck
Daedalian Member

 Posted: Mon Apr 15, 2002 3:14 pm    Post subject: 10 Yes. I want my constant named after me. I also want 13,919,148 to be known as The Chuck Limit.
cubestudent
3D Member

 Posted: Mon Apr 15, 2002 9:54 pm    Post subject: 11 Well, 'constant' normally implies that it's used in an equation - the coefficient of some interesting formula. I'd vote instead for 'the Chuck Number'
ctrlaltdel
Member of the Daedalians

 Posted: Mon Apr 15, 2002 10:03 pm    Post subject: 12 constants usually also have a symbol - like 'e', 'c', 'G', 'kappa' etc. so whats your choice chuck?
Chuck
Daedalian Member

 Posted: Mon Apr 15, 2002 10:39 pm    Post subject: 13 I'd like to use © so people won't steal it.
El Blobbo revived
Daedalian Member

 Posted: Thu Apr 18, 2002 12:23 am    Post subject: 14 The cubes all return zero, correct? Unless you mean the highest cube that returns a positive result, not a non-negative one. [F(1)=1 or 0]?
Chuck
Daedalian Member

 Posted: Thu Apr 18, 2002 12:28 am    Post subject: 15 Yes, cubes all return zero.
Da5id
Daedalian Member

 Posted: Thu Apr 18, 2002 6:19 am    Post subject: 16 I tried uBasic out. I didn't get the syntax. But I was wondering Are there any x where (x^3 - the largest y^4 minus the largest z^4)=x ? I couldn't find any using excel but the digits aren't really there to calculate it.
Chuck
Daedalian Member

 Posted: Thu Apr 18, 2002 2:36 pm    Post subject: 17 I tried cubing and subtracting fourth powers twice. No equality to about 3,000,000.
dave10000
Tinhorn

 Posted: Thu Apr 18, 2002 4:26 pm    Post subject: 18 How about squaring and then subtracting, twice, the highest third *or higher* power? That might always provide enough variety to allow an infinite number of equalities.
Chuck
Daedalian Member

 Posted: Thu Apr 18, 2002 9:46 pm    Post subject: 19 The problem was that for high enough numbers F(x) was always less than x. Wouldn't giving it more powers to choose from make it more difficult?
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