(In reply to
re: partial answer by brianjn)
The n+1 terminal digit, where the Fibonacci number ends in 999 {n=3}, has a cyclic period of 10 with the tenth occurance being a 9, itself. I.e., 1497+15000(x-1), where x is the xth occurance of the Fibonacci number ending in 999.. The n+1 terminal digit, where the Fibonacci number ends in 9999 {n=4}, has a cyclic period of 10, with the tenth occurance being a 9. I.e., 1497+150000*(x-1). For n=3, the n+1 terminal digit is equal to [7*(x-1)+6] modulo 10.
Edited on November 12, 2012, 6:36 am
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Posted by Dej Mar
on 2012-11-12 06:25:23 |
re(2): partial answer
