Fibonacci challenge

Question

create a series of numbers in which each number is the multiple of the two preceding numbers.

for example :
1, 2 , 2 ,4 ,8 ,32 .....

conditions :
#Numbers should start from one
#you can refer Fibonacci series as reference
0 ,1 ,1 ,2, 3 ,5, 8 ,13.......but these are sums of two preceding numbers
#reach the highest number

Go for it.....it might become headech

Answer 1

Your example is a little bit vague, so you mean start with 1 & 2?

Answer 2

You need at least two first term ^^, and if both are equals to 1, the serie will be infinite 1,1,....,1
More accurate to describe your serie as f(0) == 1, f(1) == n != 1, so serie is 1, n, 1*n, n*n, n*n*n, n*n*n*n*n... or rather: n^0, n^1, n^2, n^3, n^5, n^8... So you can say that your serie is rather done by:

f(0) == n^0 == 1
f(1) == n^1 == n
f(x) == f(x-2)*f(x-1) == n^(x-2) * n^(x-1) == n^((x-2)+(x-1)) == n^(2x-3) for x > 1

If I doesn't make mistake ;P (my mathematic skills were far far away ^^)

Answer 3

Well the sequence grow exponentially fast in this case, I can get up to 11th term only before the integer overflow.

f(0) = 2^0
f(1) = 2^1
f(2) = 2^1
f(3) = 2^2
f(4) = 2^3
f(5) = 2^5
f(6) = 2^8
f(7) = 2^13
f(8) = 2^21
f(9) = 2^34
f(10) = 2^55
f(11) = 2^89
* 12th term failed to fit into 64-bit)

💡 Demo
https://code.sololearn.com/cPvxMQAt80Ac/?ref=app

Answer 4

You can use recursion to make life easy.

Answer 5

def sequence(m):
if m==1:
return 2
elif m==2:
return 4
else:
return sequence(m-1)*sequence(m-2)
#2 and 4 are example starting numbera

Answer 6

@visph Recursion is the easiest way here.
It can work for any 2 inputted starting numbers

Answer 7

Cmon Visph you can do it! :)

Answer 8

Hmm the next number going to generate is based on the previous 2 numbers, so we will start with

f(0) = 2 // 1st term
f(1) = 2 // 2nd term
.. and so on

Am I right?

Answer 9

that was just an example.... your first term should be 1 .....

Answer 10

ok ...your
f(0) = 1 //1st term
if that helps

Answer 11

"that was just an example"
conditions specifically says
numbers should "strat" From 1 (one).
the second number can be anything ......
just give me the highest number
and number of terms rather creating an infinite series ;)

Answer 12

you can try .....
or any different methods .....to do it
but it should give you series ....as expected.

Answer 13

you miss the 4 ;(

1x2=2
2x2=4
4x2=8
8x4=32.......and so on the next value should be multiple of preceding two numbers

Answer 14

Arf!

Not still working ^^
It must be fixed again... I will try ;)

Answer 15

#just enter the number of terms you want
t=int(input(""))
n1,n2=1,2
while(t>0):
print(n1)
n=n1*n2
n1=n2
n2=n
t-=1;
see the answer without
any recursion
© wahid shaikh

Answer 16

With Python, you doesn't have "highest" number, as Python natively handle big integers (stored as string, so theorically unlimited in size -- but not in memory to handle them ^^)

Answer 17

@$Vengat:

No need for recursion, as the serie can be computed as a simple function ;) (but using only integer values, so not a continuous function ^^)

Answer 18

def sequence(n,m):
if '.' in str(m) or '.' in str(n):
print('Value Error')
if m == 0:
print(1)
elif m == 1:
print(n)
else:
print(n**(2*m-3))

for i in range (100):
sequence(2,i)

Answer 19

@Wahid:

Are you talking to me?
My code only require the f(1) item, as f(0) is 1... run the code, and you will see that third value (f(2)) is 4 (if you give n==2 as f(1) item) ^^

Answer 20

# Wow... my mistake, that's not working for f(2), you're right: need to add an elif statement for f(2) case:

def sequence(n,m):
if '.' in str(m) or '.' in str(n):
print('Value Error')
if m == 0:
print(1)
elif m == 1:
print(n)
elif m == 2:
print(n*n)
else:
print(n**(2*m-3))

for i in range (100):
sequence(2,i)