## Verify whether s list of number is part of the fibonacci series

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I have made a function which takes a list as an input and returns a list too.

e.g. input is `[4,6,8,10,12]` and output should be `[0,0,1,0,0]`

because 8 belongs in fibonacci series

my code is

```for i in input1:
phi=0.5+0.5*math.sqrt(5.0)
a=phi*i
out =[ i == 0 or abs(round(a) - a) < 1.0 / i];
return out;
```

This will work:

```input1 = [4,6,8,10,12]
out=[]
for i in input1:
phi=0.5+0.5*math.sqrt(5.0)
a=phi*i
out.append(i == 0 or abs(round(a) - a) < 1.0 / i);
```

To convert bool to int

```import numpy
y=numpy.array(out)
new_output = y*1
```

Checking if a number is fibonacci – Ritambhara Technologies, How do you check whether a number is in Fibonacci series or not? Python program to check Fibonacci number # python program to check if given # number is a Fibonacci number import math # function to check perferct square def checkPerfectSquare (n): sqrt = int (math. sqrt (n)) if pow (sqrt, 2) == n: return True else: return False # function to check Fibonacci number def isFibonacciNumber (n): res1 = 5 * n * n + 4 res2 = 5 * n * n -4 if checkPerfectSquare (res1) or checkPerfectSquare (res2): return True else: return False # main code num = int (input ("Enter

I would think that the best way is probably to write a function called `is_fibonacci`, which takes a numerical input and returns `True` if the input is a fibonacci number, otherwise `False`. Then you can just do a list comprehension on your initial list `input1`: `return [1 if is_fibonacci(num) else 0 for num in input1]`. (Of course, `is_fibonacci` could automatically return `1` or `0` instead of a Boolean, in which case the list comprehension is even simpler.)

Writing the `is_fibonacci` function is an interesting exercise that I will leave to you :) (But happy to help if you are struggling with it.)

How to Calculate the Fibonacci Sequence (with Pictures), since (5*3*3 + 4) is 49 which is 7*7. For every Fibonacci number, check if it is prime or not. If prime, then print it. An efficient solution is to use Sieve to generate all Prime numbers up to n. After we have generated prime numbers, we can quickly check if a prime is Fibonacci or not by using the property that a number is Fibonacci if it is of the form 5i 2 + 4 or in the form 5i 2 – 4

This should solve I guess

```import math

# A function that returns true if x is perfect square
def isPerfectSquare(x):
s = int(math.sqrt(x))
return s * s == x

# Returns true if n is a Fibinacci Number, else false

def isFibonacci(n):
return isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4)

i = [4, 6, 8, 10, 12]
print(i)
j = []
# A utility function to test above functions
for item in i:
if (isFibonacci(item) == True):
j.append(1)

else:
j.append(0)
print(j)
```

Output:

```[4, 6, 8, 10, 12]
[0, 0, 1, 0, 0]
```

What is the 10th number in the Fibonacci sequence?, , you should think of 0 coming before 1 (the first term), so 1 + 0 = 1. Active 3 years, 4 months ago. Viewed 18k times. 37. We are given a sequence of numbers, as a vector foo. The task is to find is foo is monotonically increasing - every item is less than or equal to the next one - or monotonically decreasing - every item greater than or equal than the next one.

This does what you want

```def isFibonaccy(inputList):
out = []
for i in inputList:
phi = 0.5 + 0.5 * math.sqrt(5.0)
a = phi * i
out.append(int(i == 0 or abs(round(a) - a) < 1.0 / i))

return out

print(isFibonaccy([4, 6, 8, 10, 12])) # -> [0, 0, 1, 0, 0]
```

What is the Fibonacci Sequence (aka Fibonacci Series)?, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946… This code is tested in Keil uVision 4. Developed for ARM LPC2148 by Abhay Kagalkar ARM Code : AREA Prime_or_Not,code,readonly ENTRY MOV R0,#15 ;Number which you want to test CMP R0,#01 ;Comparing with 01 BEQ PRIME ;If equal declare directly as prime CMP R0,#02 ;Compare with 02 BEQ PRIME ;If equal declare directly […]

Verify whether s list of number is part of the fibonacci series, This will work: input1 = [4,6,8,10,12] out=[] for i in input1: phi=0.5+0.5*math.sqrt(​5.0) a=phi*i out.append(i == 0 or abs(round(a) - a) < 1.0 / i);. Enter a positive number: 12331 The reverse of the number is: 13321 The number is not a palindrome. In the above program, use is asked to enter a positive number which is stored in the variable num. The number is then saved into another variable n to check it when the original number has been reversed.

How to Check if a given Number is Fibonacci number?, A number is given to you, how will you check if that number is a Fibonacci L and S within a given fixed length results in the Fibonacci numbers; the number of​  1. Select the list you want to check the certain value from, and click Kutools > Select > Select Specific Cells. See screenshot: 2. In the Select Specific Cells dialog, select Equals from the first drop down list in Specific type section, and then enter the value you want to check and locate into the next textbox.

The Mathematical Magic of the Fibonacci Numbers, Algorithm · Array · Strings · Linked List · Tree A simple way is to generate Fibonacci numbers until the generated number is about Fibonacci numbers that can also be used to check if a given number is Fibonacci or not. int s = sqrt (x);  The k-Fibonacci sequence is defined by the numbers which satisfy the second order recurrence relation F k,n = kF k,n−1 + F k,n−2 with the initial conditions F k,0 = 0 and F k,1 = 1. Falcon  defined the k-Lucas sequence which is companion sequence of k-Fibonacci sequence defined with the k-Lucas numbers which are defined with

• Here is a working shorter version of your code: `phi=0.5+0.5*math.sqrt(5.0)` and then `out = [1 if abs(round(phi*i) - phi*i) < (1/i) else 0 for i in input1]` using list comprehension. I replaced `a` by `phi*i`. If something unclear, feel free to ask