A number series that seems to terminate

[Chuck]

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]

what number type did you use?

[Da5id]

I assume you're not using a program that would introduce rounding errors?

[Chuck]

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]

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]

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]

I'm past 52,000,000 with no more F(x) > x. That should be enough evidence to prove the theorem.

[sodasipper]

Maybe we should declare 4,150,334 as the "Chuck Constant"...

[Da5id]

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]

Yes. I want my constant named after me. I also want 13,919,148 to be known as The Chuck Limit.

[cubestudent]

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]

constants usually also have a symbol - like 'e', 'c', 'G', 'kappa' etc. so whats your choice chuck?

[Chuck]

I'd like to use © so people won't steal it.

[El Blobbo revived]

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]

Yes, cubes all return zero.

[Da5id]

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]

I tried cubing and subtracting fourth powers twice. No equality to about 3,000,000.

[dave10000]

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]

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?