由递归T(0)=0,T(1)=T(2)=1,T(3)=2和T定义的类斐波那契数(n个)=T(n个-1) +T(+温度)(n个-2) +T(+温度)(n个-3) +T(+温度)(n个-4).更多
tetranacci的数量高达1015:
1,2,4,8,15,29,56,108,208,401,773,1490,2872,5536,10671,20569,39648,76424,147312,283953,547337,1055026,2033628,3919944,7555935,14564533,28074040,54114452,104308960,201061985,387559437,747044834,1439975216,2775641472,5350220959,10312882481,19878720128,38317465040,73859288608,142368356257,274423830033,528968939938,1019620414836,1965381541064,3788394725871,7302365621709,14075762303480,27131904192124,52298426843184,100808458960497,194314552299285,374553342295090,721974780398056. 剩余物的分配当这个家族中的数字除以n个=2, 3,..., 11.(我考虑了100000个值,从1到2‰1028500).
n\r公司 | 0 | 1 |
2 | 60000 | 40000 | 2 |
三 | 34614 | 30771 | 34615 | 3 |
4 | 50000 | 30000 | 10000 | 10000 | 4 |
5 | 23076 | 19230 | 19227 | 19234 | 19233 | 5 |
6 | 20768 | 12308 | 20769 | 13846 | 18463 | 13846 | 6 |
7 | 14327 | 14626 | 16664 | 12574 | 12575 | 12572 | 16662 | 7 |
8 | 40000 | 20000 | 10000 | 0 | 10000 | 10000 | 0 | 10000 | 8 |
9 | 15384 | 14104 | 11539 | 7692 | 10257 | 11538 | 11538 | 6410 | 11538 | 9 |
10 | 13844 | 7693 | 11537 | 7692 | 11540 | 9232 | 11537 | 7690 | 11542 | 7693 | 10 |
11 | 9165 | 14167 | 8332 | 8335 | 9166 | 6669 | 9999 | 8333 | 10833 | 4167 | 10834 |
上表的图示
想象一下用一个数字来划分这个家庭的成员n个并计算余数。如果它们均匀分布,则每个余数从0到n个-1将在大约(1)中获得/n个)-第个案例。此结果由白色方块表示。红色(分别为蓝色)方块表示剩余物,其出现频率高于(分别低于)1/n个.