趣味数学与编程｜任何一个整数的立方都可以写成一串连续奇数的和

对于任一正整数a，不论a 是奇数还是偶数，整数（a×a−a+1）必然为奇数。

构造一个等差数列，数列的首项为（a×a−a+1），等差数列的差值为2（奇数数列），则前a 项的和为：

a× ((a×a−a+1))+2×a(a−1)/2

=a×a×a−a×a+a+a×a−a

=a×a×a

证明过程可知所要求的奇数数列的首项为（a×a−a+1），长度为a。

#include<stdio.h>void vrNico(int a)// 任何一个整数的立方都可以写成一串连续奇数的和{ int b,c,d; b=a*a*a; /*求整数的三次方*/ printf(' >> %d*%d*%d=%d=',a,a,a,b); for(d=0,c=0;c<a;c++) /*输出数列，首项为a*a-a+1,等差值为2*/ { d+=a*a-a+1+c*2; /*求数列的前a项的和*/ printf(c?'+%d':'%d',a*a-a+1+c*2); } if(d==b)printf(' Satisfy!\n'); /*若条件满足则输出'Y'*/ else printf(' Dissatisfy!\n'); /*否则输出'N'*/}void main(){ int n=1; // 尼科彻斯定理，即任何一个整数的立方都可以写成一串连续奇数的和。 puts('******************************************************'); puts('* This program is to verify Theorem of Nicoqish. *'); puts('* That is the cube of any integer can be represented *'); puts('* as the sum of some continue odd numbers. *'); puts('* For example, 8^3=512=57+59+61+63+65+67+69+71. *'); puts('******************************************************'); while(n!=0) { printf(' >> Please input a integer to verify(0 to quit): '); scanf('%d',&n); /*输入任一整数*/ if(n==0) break; printf(' >> ------ Results of verification: ------------\n'); vrNico(n); /*调用函数进行验证*/ printf(' >> ---------------------------------------------\n'); } puts('\n Press any key to quit...'); getchar();}/******************************************************** This program is to verify Theorem of Nicoqish. ** That is the cube of any integer can be represented ** as the sum of some continue odd numbers. ** For example, 8^3=512=57+59+61+63+65+67+69+71. ******************************************************* >> Please input a integer to verify(0 to quit): 5 >> ------ Results of verification: ------------ >> 5*5*5=125=21+23+25+27+29 Satisfy! >> --------------------------------------------- >> Please input a integer to verify(0 to quit):*/

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