When n is odd, H = 3N + 13, when n is even, H = NX1 / 2x1 / 2 Where h is an odd number. How many times does 7 equal after 2010?

7 the first time: 7 * 3 + 13 = 34 the second time: 34 * 1 / 2 = 17 the third time: 17 * 3 + 13 = 64 the fourth time: 64 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 the fifth time: 1 * 3 + 13 = 16 the sixth time: 16 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 From the fourth time, they are all in the cycle of 1 and 16, 1 for the even number time and 16 for the odd number time

If n is a positive integer, then the odd number is?

If n is a positive integer, the odd number is 2N-1

According to nocomachns theorem, the cube of any positive integer n can be expressed as the sum of n continuous odd numbers
For example: 1 ^ 3 = 1
2^3=3+5
3^3=7+9+11
4^3=13+15+17+19
Here, if the minimum odd number in each formula is called X, then after n is given, please write the relationship expression between X and N:_______________________
Teach me, I really can't understand the class!

x(n+1)-xn=2n
x(n+1)=xn+2n

Finding the sum of the first n even numbers of a positive integer column

Sn=(a1+an)*n/2=(2+2n)*n/2=n*(n+1)

Finding the sum of the first n numbers in a positive integer column
Finding the sum of the first n even numbers in a whole sequence

1： The sum of the first n positive integers: 1 + + 2 + 3 + 4 + 5 +. + n =?
N + n-1 + n-2 +. + 1 =? They are equal
If the sum of the first term and the last term n + 1 has n groups and has been calculated twice, the final result is:
n(n+1)/2
2： Similar to the above method, from 2 + 4 +. + 2n =?
2n+2（n-1...+2=?
The sum of the first term and the last term is 2n + 2, which has the sum of N groups n (2n + 2) / 2 = n (n + 1)

The equation for finding the sum of all even numbers from 1 to n (n is a positive integer)

If n is even
(1+n/2)*n/2
If n is odd, then:
[1+(n-1)/2]*(n-1)/2

1. If I give you a true integer n, can you tell whether n + N and N + N + 1 are odd or even? Let's give an example to verify your conclusion
2. If n is a positive integer, is 2n odd or even? What about 2n + 1? What about 2N-1? Fast, I only need the first three respondents to get a reward!

1. N + n is even, N + N + 1 is odd
(1) When n is even, N + 1 is odd. For example: n = 2,4,6., N + n = even + even = even, N + N + 1 = even + odd = odd
(2) When n is odd, N + 1 is even. For example: n = 1,3,5., N + n = odd + odd = even, N + N + 1 = odd + even = odd
2. If n is a positive integer, then 2n is even, 2n + 1 is odd, and 2N-1 is odd
(1) When n is even, N + 1 is odd, and N-1 is odd, for example: n = 2,4,6., 2n = n + n = even + even = even, 2n + 1 = n + N + 1 = even + odd = odd, 2N-1 = n + n-1 = even + odd = odd
(2) When n is odd, N + 1 is even, and N-1 is even, for example: n = 1,3,5., 2n = n + n = odd + odd = even, 2n + 1 = n + N + 1 = odd + even = odd, 2N-1 = n + n-1 = odd + even = odd

The "H operation" of positive integer n is: 1. When n is odd, H = 3N + 13; 2. When n is even, H = n * 0.5 * 0.5 * (where h is an odd number). For example, the result of one "H operation" for number 5 is 28, that of two "H operations" is 7, and that of three "H operations" is 34. Then the result of 257 "H operations" is 28

16
Let the result of M times "H operation" of 257 be a m, then a 1 = 784, a 2 = 49, a 3 = 160, a 4 = 5, a 5 = 28, a 6 = 7, a 7 = 34, a 8 = 17, a 9 = 64, a 10 = 1, a 11 = 16, a 12 = 1 ,a 257 = 16.

C language programming data statistics arbitrary input n integers, respectively, the odd number of statistics, odd number of statistics, even number of statistics, even number of statistics
Input: n integers
Output: odd sum, odd number, even number and even number
Prompt: ① input the number of data first
② Input n numbers by circulation;
③ Define four variables to store the statistical results, and put the specific statistical statements in the loop
Tip: there must be statements for input data and statements for counting the number of input data in the loop

#include
#include
#define MAX 1000
int main()
{ int a[MAX] , i , j , n , p , q , s1 , s2;
p = 0 ; q = 0 ; s1 = 0 ; s2 = 0;
Printf ("please enter the number of values to be entered):";
scanf("%d",&n);
Printf ("please enter the numerical value of these numbers): (n));
for(i = 0 ; i < n ; i++ )
scanf("%d",&a[i]);
for(j = 0 ; j < n ; j++)
if(a[j]%2 == 0 ){
p = p + 1 ; s1 = s1 + a[j];}
else {
q = q + 1 ; s2 = s2 + a[j];}
Printf ("even number is% D, P)";
Printf ("sum of even numbers is% D, S1)";
Printf ("odd number is% D, n", q));
Printf ("odd sum is% D, S2)";
system("PAUSE");
return 0;
}
Hope to help you

Arbitrary input n integers, respectively, the odd number of statistics, odd number, even number and even number. C language

#include <stdio.h>;int main(int argc, char **argv) { int i, n, m, jh = 0, oh = 0, jg ...

When n is odd, H = 3N + 13, when n is even, H = NX1 / 2x1 / 2 Where h is an odd number. How many times does 7 equal after 2010?