bin
echo "Enter How many numbers:"
read 100
num1=0
num2=1
echo "Fibonacci series:"
echo $num1
echo $num2
count=2
while [ $count -le $num ]
do
num3=`expr $num1 + $num2`
echo $num3
num1=$num2
num2=$num3
count=`expr $count + 1`
done
--------------------------------------
ANSWER:
The above script didn't work at all for me so I wrote this one. That said, the author of the script above appears to be familiar with the subject matter and the challenges involved, but was possibly busy or distracted at the time the answer was written.
NOTE: One obvious omission is the lack of an absolute PATH for the 'shebang' line of the script, it is this which leads me suspect his or her
mind may not have been on the task at hand when the script was written. I believe the author above is experienced at script writing however.
#!/bin/bash
#
## Written by John Horn, June, 2013.
#
# set -x
if [ -z ${1} ]; then
echo "Pass the number of fibonacci numbers you wish generated as the argument to this script."
exit
fi
#
## Decided to add comma insertion every 3rd digit to make the
## output easier to read. Friday the 13th of September, 2013.
## Friday the 13th? That isn't very auspicious. :(
#
## This subroutine inserts a comma every 3rd digit to the arg passed.
#
commify () {
typeset text=${1}
typeset bdot=${text%%.*}
typeset adot=${text#${bdot}}
typeset i commified
(( i = ${#bdot} - 1 ))
while (( i>=3 )) && [[ ${bdot:i-3:1} == [0-9] ]]; do
commified=",${bdot:i-2:3}${commified}"
(( i -= 3))
done
echo "${bdot:0:i+1}${commified}${adot}"
}
#
## Initialize some variables for whatever reason (simplicity maybe).
#
LIMIT=${1}
COUNT=0
N1=0
N2=1
FIBONACCI=0
#
## Might as well just print the first two as they can't possibly
## have any other value than 0 and 1.
#
echo "And our first ${LIMIT} FIBONACCI numbers are...:"
echo "0"
echo "1"
while [ ${COUNT} -le ${LIMIT} ]; do
#
## Fork the 'bc' command to perform the actual addition operation.
#
FIBONACCI=`echo "${N1} + ${N2}"|bc -l`
#
## Call the commify() subroutine to format the integer with commas.
#
COMMA_FORMATTED_RESULT=`commify "${FIBONACCI}"`
#
## Print our latest FIBONACCI number to STDOUT.
#
echo "${COMMA_FORMATTED_RESULT}"
#
## Overwrite our variable contents with our new values.
#
N1=${N2}
N2=${FIBONACCI}
#
## Increment our counter so we'll know when we've reached our $LIMIT.
#
(( COUNT++ ))
done
echo -e "\n\nAll done.\n"
----------------------------
On my Ubuntu Linux 12.04 LTS machine the above script works perfectly. Your
mileage may vary. Good luck!
JMH
i dn't know. haha
Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
2,3,5,7,9,11,13,17,19,23,29,31,37,39,41,43,47,49
#include #include void main() { clrscr() int a=0,b=1,c,i,n; coutn cout
write a c program to accept a number and generate a square root cube and exponential values
I don't program in just C (I use C++) so you may have to change some of the code:int main(){int n;int last = 1;int slast = 0;int numbers[n];numbers[0] = 1;for(int i = 1; i < n; i++){numbers[i] = last + slast;slast = last;last = numbers[i];}return 0;}int n needs to be set to however many Fibonacci numbers you want to make.int last is the last number in the sequence (so far)int slast is the second last number in the sequence (so far)int numbers[n] is the int array which holds your numbersint i is the count for the loop, it is set to 1 because before the loop, I set numbers[0] to 1
(rand()&50+1)*2
Unlike some other types of numbers like prime numbers, calculating large Fibonacci numbers can be done quite easily with even a standard household computer. The process involves only repeated addition (rather than the intense division processes involved with large prime numbers). Beyond that, large Fibonacci numbers do not serve as much purpose as other large numbers (like primes). Because of this, these large numbers are generally left for quick calculation by machine if ever necessary. An example of a computer program that could calculate the nth Fibonacci number (n greater than 1 and counting the first 1 in the sequence as the second term) is given below in pseudo-code: Function Fibonacci(n) a = 0 b = 1 k = 2 While n > k ( a + b = c a = b b = c k = k + 1 ) Print b A very large Fibonacci number is the 250th in the sequence which has a value of: 12776523572924732586037033894655031898659556447352249. The 1000th term in the sequence is: 4346655768693745643568852767504062580256466051737178040248172908953655 5417949051890403879840079255169295922593080322634775209689623239873322 471161642996440906533187938298969649928516003704476137795166849228875. Much, much larger values (even beyond the 10,000,000th term) can be calculated quite quickly with a simple, well-written program. See related links for a site which can quickly calculate large Fibonacci numbers (using the form Fibonacci n).
not sure of the exact syntax but its quite hard to generate them, but you can go through a list of numbers, that could be 0-10000000 say, and check if it is prime by dividing it by all the numbers that are smaller than it
In c: int fibr(int n) { // Find nth Fibonacci number using recursion. if (n<=2) return 1; // the first two Fibonacci numbers are 1 and 1 return (fibr(n-2)+fibr(n-1)); } int fibi(int n) { // Find nth Fibonacci number using iteration. int temp,last=1,f=1; int i; for (i=3;i<n;++i) { // the first two Fibonacci numbers are 1 and 1 temp=f; f+=last; last=temp; } return f; }
In a Fibonacci sequence, sum of two successive terms gives the third term.... here is the Fibonacci sequence: 0,1,1,2,3,5,8,13,21,34,55,89,144........ General formula to generate a Fibonacci sequence is """Fn= Fn-1 + Fn-2""" To check whether a number is Fibonacci or not follow the following steps: 1) Get the number as input from user. 2) Fix the first two numbers of sequence as 0 and 1. 3) put a sentinel loop with upper limit being the input number. 4)in the body of loop generate the next number in sequence in each iteration and continue swapping the values as follows: a=0 b=1 next=a+b while (next< input) a=b b=next next=a+b wend 5) lastly when program exits the loop compare the last number of sequence with the input number if they are equal then number is Fibonacci otherwise not. otherwise the last term of sequence will be less than the input number.
Exactly what do you mean by 'C program in Java'