# Defining sequences in mathematica

1. May 7, 2006

### heartless

Hello,
I'd like to define a sequence in mathematica and let it go, but I'm not sure how to tell mathematica to look for the previous number in sequence and then derive the new one.

Something like this,

$$a_1, a_2, a_3, a_4, a_5 =$$

$$3, 7, 23, 87, 343 =$$

$$0+3, 3+4, 7+16, 23+64, 87+256$$

where we start with 3 and add every previous number to the consecutive power of 4.

Thanks,
If you know, please tell as fast as you're able, I'd really need this today.

Thanks ^ ∞

wait and something else,

how to do this
I have a function f.ex f(x) = log[x]
but now I'd like to graph the functions
f(f(f(f(f(x)))))) up to a certain number of times, for example 100 such functions
and then give off all the values starting with x=10.

Any ideas?

PLease help me, I really need this, or otherwise I shall be tortured to the end of my little life.

Last edited: May 7, 2006
2. May 7, 2006

### heartless

Alright, I know. But anyway, I'd still like to see your way.

3. May 7, 2006

### GregA

Don't use mathematica (I know it has a programming language tho), and am just learning the very basics of programming myself but isn't there some way of doing this?:

make variables a,b,c,d
make an array
a = 3
c = 1
d = 4
put 3 into the first position in the array

make a loop for c < 100
Print a
Assign to b the value of a + d then increment c by one
Let d = d*4
put the value of b into the (c)th position in the array
a=b
start over until c =100
print a
Plot the values in the array against its position in the array

Disclaimer: Most of this may be garbage...I apologise if it is but I'm a novice at the moment

*edit* changed (c+1)th to (c)th

Last edited: May 7, 2006
4. May 7, 2006

### matt grime

That would define an array with 100 entries. The n'th entry would have the correct property, but that array is no more the sequence than the set of numbers {1,..,100} is the set of natural numbers.

A complete guess would be that a sensible programming language converts an infinite list to either a function f(n)=3 + (4^n - 1)/3 -1 in this case, or stores the recursive formual x(1)=3, x(n)=x(n-1) + 4^(n-1), and will call some procedure for evaluating it at some n as required. Though how you do that in any particular language if at all is not something I know.

5. May 7, 2006

### GregA

Aye that is true ...apart from letting c be a number such that the computer would perish long before it was ever reached...or the user just gives up, I couldn't suggest a way to let it just keep going...but even then I'm not defining the complete sequence, just a rule to make it print *some* of them.

Last edited: May 7, 2006