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  1. #include <stdio.h>
  2.  
  3. // 課題3で作成した再帰関数をそのまま利用
  4. int rec_kadai3(int n){
  5. if (n == 0) return 3;
  6. else if (n == 1) return 0;
  7. else if (n == 2) return 2;
  8. else return rec_kadai3(n - 2) + rec_kadai3(n - 3);
  9. }
  10.  
  11. int main(void) {
  12. int max_n = 50; // 50項まで調べる
  13. int an;
  14.  
  15. printf("課題4: a[n]がnで割り切れるときのnの値:\n");
  16.  
  17. // n=1からmax_nまで調べる
  18. for (int n = 1; n <= max_n; n++) {
  19. an = rec_kadai3(n);
  20.  
  21. // if文を用いて、(a[n] % n == 0) を満たすか判定
  22. if (an % n == 0) {
  23. printf("n = %d (a[%d] = %d)\n", n, n, an);
  24. }
  25. }
  26.  
  27. return 0;
  28. }
Success #stdin #stdout 0.01s 5276KB
stdin
Standard input is empty
stdout
課題4: a[n]がnで割り切れるときのnの値:
n = 1 (a[1] = 0)
n = 2 (a[2] = 2)
n = 3 (a[3] = 3)
n = 5 (a[5] = 5)
n = 7 (a[7] = 7)
n = 11 (a[11] = 22)
n = 13 (a[13] = 39)
n = 17 (a[17] = 119)
n = 19 (a[19] = 209)
n = 23 (a[23] = 644)
n = 29 (a[29] = 3480)
n = 31 (a[31] = 6107)
n = 37 (a[37] = 33004)
n = 41 (a[41] = 101639)
n = 43 (a[43] = 178364)
n = 47 (a[47] = 549289)